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\chapter{Conditional Probability}\label{chp 4}

\section{Discrete Conditional Probability}\label{sec 4.1}

\subsection*{Conditional Probability}
\par
In this section we ask and answer the following question.  Suppose we assign a
distribution function to a sample space and then learn that an event $E$ has
occurred.  How should we change the probabilities of the remaining events?  We
shall call the new probability for an event $F$ the \emx {conditional
probability\index{probability!conditional}\index{conditional probability} of $F$ given $E$} and
denote it by
$P(F|E)$.

\begin{example}\label{exam 4.1}
An experiment consists of rolling a die once.  Let $X$ be the outcome.  Let $F$ be the event
$\{X = 6\}$, and  let $E$ be the event $\{X > 4\}$.  We assign the
distribution function
$m(\omega) = 1/6$ for $\omega = 1, 2, \ldots , 6$.  Thus, $P(F) =
1/6$.  Now suppose that the die is rolled and we are told that the event $E$ has occurred.
This leaves only two possible outcomes: 5 and~6.  In the absence of any other
information, we would still regard these outcomes to be equally likely, so the
probability of $F$ becomes 1/2, making $P(F|E) = 1/2$.
\end{example}

\begin{example}\label{exam 4.1.5}
In the Life Table (see Appendix~C), one finds that in a population of 100{,}000 females, 89.835\%
can expect to live to age 60, while 57.062\% can expect to live to age 80.  Given that a woman is
60, what is the probability that she lives to age 80?
\par
This is an example of a conditional probability.  In this case, the original sample space
can be thought of as a set of 100{,}000 females.  The events $E$ and $F$ are the subsets of the
sample space consisting of all women who live at least 60 years, and at least 80 years,
respectively.  We consider $E$ to be the new sample space, and note that $F$ is a subset of $E$.
Thus, the size of $E$ is 89{,}835, and the size of $F$ is 57{,}062.  So, the probability in
question equals $57{,}062/89{,}835 = .6352$.  Thus, a woman who is 60 has a 63.52\% chance of
living to age 80.
\end{example}

\begin{example}\label{exam 4.2}
Consider our voting example from Section~\ref{sec 1.2}: three candidates A, B,
and C are running for office.  We decided that A and B have an equal chance of
winning and C is only 1/2 as likely to win as A.  Let $A$ be the event ``A
wins," $B$ that ``B wins," and $C$ that ``C wins."  Hence, we assigned
probabilities $P(A) = 2/5$, $P(B) = 2/5$, and $P(C) = 1/5$.
\par
Suppose that before the election is held, $A$ drops out of the race.  As in Example~\ref{exam
4.1}, it would be natural to assign new probabilities to the events $B$ and $C$ which are
proportional to the original probabilities.  Thus, we would have $P(B|~A) = 2/3$, and
$P(C|~A) = 1/3$.  It is important to note that any time we assign probabilities to real-life
events, the resulting distribution is only useful if we take into account all relevant
information.  In this example, we may have knowledge that most voters who favor $A$ will vote
for $C$ if $A$ is no longer in the race.  This will clearly make the probability that
$C$ wins greater than the value of 1/3 that was assigned above.
\end{example}

In these examples we assigned a distribution function and then were given new
information that determined a new sample space, consisting of the outcomes that
are still possible, and caused us to assign a new distribution function to this
space.

We want to make formal the procedure carried out in these examples.  Let
$\Omega = \{\omega_1,\omega_2,\dots,\omega_r\}$ be the original sample space
with distribution function $m(\omega_j)$ assigned.  Suppose we learn that the
event $E$ has occurred.  We want to assign a new distribution function
$m(\omega_j|E)$ to $\Omega$ to reflect this fact.  Clearly, if a sample point
$\omega_j$ is not in $E$, we want $m(\omega_j|E) = 0$.  Moreover, in the
absence of information to the contrary, it is reasonable to assume that the
probabilities for $\omega_k$ in $E$ should have the same relative magnitudes
that they had before we learned that $E$ had occurred.  For this we require
that
$$
m(\omega_k|E) = cm(\omega_k)
$$
for all $\omega_k$ in $E$, with $c$ some positive constant.  But we must also
have
$$
\sum_E m(\omega_k|E) = c\sum_E m(\omega_k) = 1\ .
$$
Thus,
$$
c = \frac 1{\sum_E m(\omega_k)} = \frac 1{P(E)}\ .
$$
(Note that this requires us to assume that $P(E) > 0$.)  Thus, we will
define
$$
m(\omega_k|E) = \frac {m(\omega_k)}{P(E)}
$$
for $\omega_k$ in $E$.  We will call this new distribution the \emx {conditional
distribution}\index{conditional distribution} given $E$.  For a general event $F$, this gives
$$
P(F|E) = \sum_{F \cap E} m(\omega_k|E) = \sum_{F \cap E}\frac {m(\omega_k)}{P(E)} =
\frac {P(F \cap E)}{P(E)}\ .
$$

We call $P(F|E)$ the \emx {conditional probability of $F$ occurring given that $E$ occurs,}
and compute it using the formula
$$
P(F|E) = \frac {P(F \cap E)}{P(E)}\ .
$$

\begin{example}(Example~\ref{exam 4.1} continued){\label{exam 4.3}}
Let us return to the example of rolling a die.  Recall that $F$ is the event $X = 6$, and
$E$ is the event $X > 4$.  Note that $E \cap F$ is the event $F$.  So, the above formula gives
\begin{eqnarray*}
P(F|E) & = & \frac {P(F \cap E)}{P(E)} \\
& = & \frac {1/6}{1/3} \\
& = & \frac 12\ ,\\
\end{eqnarray*}
in agreement with the calculations performed earlier.
\end{example}

\begin{example}\label{exam 4.4}
We have two urns, I and II.  Urn I contains 2 black balls and 3 white balls.
Urn II contains 1 black ball and 1 white ball.  An urn is drawn at random and a ball
is chosen at random from it.  We can represent the sample space of this experiment as the paths
through a tree as shown in Figure~\ref{fig 4.1}. The probabilities assigned to the paths are
also shown.
\par
Let $B$ be the event ``a black ball is drawn," and $I$ the event ``urn I is
chosen."  Then the branch weight 2/5, which is shown on one branch in the figure, can now be
interpreted as the conditional probability $P(B|I)$.

\putfig{4.5truein}{PSfig4-1}{Tree diagram.}{fig 4.1}

\par
Suppose we wish to calculate $P(I|B)$.  Using the formula, we obtain
\begin{eqnarray*}
P(I|B) & = & \frac {P(I \cap B)}{P(B)} \\
       & = & \frac {P(I \cap B)}{P(B \cap I) + P(B \cap II)} \\
       & = & \frac {1/5}{1/5 + 1/4} = \frac 49\ .
\end{eqnarray*}
\end{example}

\subsection*{Bayes Probabilities}

Our original tree measure gave us the probabilities for drawing a ball of a
given color, given the urn chosen.  We have just calculated the \emx {inverse
probability} that a particular urn was chosen, given the color of the ball.
Such an inverse probability is called a \emx {Bayes probability}\index{Bayes
probability}\index{probability!Bayes} and may be obtained by a formula that we shall develop
later.  Bayes probabilities can also be obtained by simply constructing the tree measure for the
two-stage experiment carried out in reverse order.  We show this tree in Figure~\ref{fig 4.2}.

\putfig{4.5truein}{PSfig4-2}{Reverse tree diagram.}{fig 4.2}

The paths through the reverse tree are in one-to-one correspondence with those in the forward
tree, since they correspond to individual outcomes of the experiment, and so they are assigned
the same probabilities.  From the forward tree, we find that the probability of a black ball is
$$
\frac 12 \cdot \frac 25 + \frac 12 \cdot \frac 12 = \frac 9{20}\ .
$$

The probabilities for the branches at the second level are found by simple
division.  For example, if $x$ is the probability to be assigned to the top
branch at the second level, we must have
$$
\frac 9{20} \cdot x = \frac 15
$$
or $x = 4/9$.  Thus, $P(I|B) = 4/9$, in agreement with our previous
calculations.  The reverse tree then displays all of the inverse, or Bayes,
probabilities.

\begin{example}\label{exam 4.5}
We consider now a problem called the \emx {Monty Hall}\index{Monty Hall problem} problem.   This
has long been a favorite problem but was revived by a letter from Craig Whitaker\index{WHITAKER,
C.} to Marilyn vos Savant\index{vos SAVANT, M.} for consideration in her column in \emx {Parade
Magazine}.\footnote{Marilyn vos Savant, Ask Marilyn, \emx {Parade Magazine}, 9
September;  2 December;  17  February 1990, reprinted in Marilyn vos Savant, \emx {Ask
Marilyn}, St. Martins, New York, 1992.}  Craig wrote:
\begin{quote}  Suppose you're on Monty Hall's \emx {Let's Make a Deal!} You are given the choice of
three doors, behind one door is a car\index{car}, the others, goats\index{goat}.  You pick a door,
say 1, Monty opens another door, say 3, which has a goat.  Monty says to you ``Do you want to
pick door 2?" Is it to your advantage to switch your choice of doors?
\end{quote}
\par
Marilyn gave a solution concluding that you should switch, and if you do, your probability of
winning is 2/3.  Several irate readers, some of whom identified themselves as having a PhD in
mathematics, said that this is absurd since after Monty has ruled out one door there are only two
possible doors and they should still each have the same probability 1/2 so there is no advantage to
switching.  Marilyn stuck to her solution and encouraged her readers to simulate the game and draw
their own conclusions from this.  We also encourage the reader to do this (see
Exercise~\ref{exer 4.1.9.6}).
\par
Other readers complained that Marilyn had not described the problem completely.  In particular, the
way in which certain decisions were made during a play of the game were not specified.  This aspect
of the problem will be discussed in Section~\ref{sec 4.3}.  We will assume that the car was put
behind a door by rolling a three-sided die which made all three choices equally likely.  Monty
knows where the car is, and always opens a door with a goat behind it.  Finally, we assume that if
Monty has a choice of doors (i.e., the contestant has picked the door with the car behind it), he
chooses each door with probability 1/2.  Marilyn
clearly expected her readers to assume that the game was played in this manner.
\par
As is the case with most apparent paradoxes, this one can be resolved through careful analysis.
We begin by describing a simpler, related question.  We say that a contestant is using
the ``stay" strategy if he picks a door, and, if offered a chance to switch to another door,
declines to do so (i.e., he stays with his original choice).  Similarly, we say that the
contestant is using the ``switch" strategy if he picks a door, and, if offered a chance to switch to
another door, takes the offer.  Now suppose that a contestant decides in advance to play the
``stay" strategy.  His only action in this case is to pick a door (and decline an invitation to
switch, if one is offered).  What is the probability that he wins a car?  The same
question can be asked about the ``switch" strategy.
\par
Using the ``stay" strategy, a contestant will win the car with probability 1/3, since 1/3 of the
time the door he picks will have the car behind it.  On the other hand, if a contestant plays
the ``switch" strategy, then he will win whenever the door he originally picked does not have
the car behind it, which happens 2/3 of the time.
\par
This very simple analysis, though correct, does not quite solve the problem that Craig posed.  Craig
asked for the conditional probability that you win if you switch, given that you have chosen door 1
and that Monty has chosen door 3.  To solve this problem, we set up the problem before
getting this information and then compute the conditional probability given this information.
This is a process that takes place in several stages; the car is put behind a door, the contestant
picks a door, and finally  Monty opens a door.  Thus it is natural to analyze this using a tree
measure.  Here we make an additional assumption that if Monty has a choice of doors (i.e., the
contestant has picked the door with the car behind it) then he picks each door with probability
1/2.  The assumptions we have made determine the branch probabilities and these in turn determine
the tree measure. The resulting tree and tree measure are shown in Figure~\ref{fig 4.4}.  It is tempting to
reduce the tree's size by making certain assumptions such as: ``Without loss of generality, we
will assume that the contestant always picks door 1."  We have chosen not to make any such
assumptions, in the interest of clarity.

\putfig{4.5truein}{PSfig4-0}{The Monty Hall problem.}{fig 4.4}
\par
Now the given information, namely that the contestant chose door 1 and Monty chose door 3, means
only two paths through the tree are possible (see Figure~\ref{fig 4.4.5}).
\putfig{4.5truein}{PSfig4-4-5}{Conditional probabilities for the Monty Hall problem.}{fig 4.4.5}
For one of these paths, the car is behind door 1 and for the other it is behind door 2.
The path with the car behind door 2 is twice as likely as the one with the car behind door 1.
Thus the conditional probability is 2/3 that the car is behind door 2 and 1/3 that it is
behind door 1, so if you switch you have a 2/3 chance of winning the car, as Marilyn claimed.
\par
At this point, the reader may think that the two problems above are the same, since they have the
same answers.  Recall that we assumed in the original problem if the contestant
chooses the door with the car, so that Monty has a choice of two doors, he chooses each of them with
probability 1/2.  Now suppose instead that in the case that he has a choice, he chooses the door
with the larger number with probability 3/4.  In the ``switch" vs. ``stay" problem, the
probability of winning with the ``switch" strategy is still 2/3.  However, in
the original problem, if the contestant switches, he wins with probability 4/7.  The reader can
check this by noting that the same two paths as before are the only two possible paths in the tree.
The path leading to a win, if the contestant switches, has probability 1/3, while the path which
leads to a loss, if the contestant switches, has probability 1/4.
\end{example}

\subsection*{Independent Events}

It often happens that the knowledge that a certain event $E$ has occurred has no effect on the
probability that some other event $F$ has occurred, that is, that $P(F|E) = P(F)$.  One would
expect that in this case, the equation
$P(E|F) = P(E)$ would also be true.  In fact (see Exercise~\ref{exer 4.1.1}),
each equation implies the other.  If these equations are true, we might say
the $F$ is \emx {independent} of $E$.  For example, you would not expect the
knowledge of the outcome of the first toss of a coin to change the probability
that you would assign to the possible outcomes of the second toss, that is, you
would not expect that the second toss depends on the first.  This idea is
formalized in the following definition of independent events.

\begin{definition}\label{def 4.2}
Let $E$ and $F$ be two events.  We say that they are \emx {independent}\index{events!independent}\index{independence of
events} if either 1) both events have positive probability and
$$
P(E|F) = P(E)\ {\rm and}\ P(F|E) = P(F)\ ,
$$
or 2) at least one of the events has probability 0.
\end{definition}

As noted above, if both $P(E)$ and $P(F)$ are positive, then each of the above equations
imply the other, so that to see whether two events are independent, only one of these
equations must be checked (see Exercise~\ref{exer 4.1.1}).
\par
The following theorem provides another way to check for independence.

\begin{theorem}\label{thm 4.1}
Two events $E$ and $F$ are independent if and only if
$$P(E\cap F) = P(E)P(F)\ .$$
\proof
If either event has probability 0, then the two events are independent and the above equation
is true, so the theorem is true in this case.  Thus, we may assume that both events have positive
probability in what follows. Assume that $E$ and $F$ are independent.  Then $P(E|F) = P(E)$, and so
\begin{eqnarray*}
P(E\cap F) & = & P(E|F)P(F) \\
       & = & P(E)P(F)\ .
\end{eqnarray*}

Assume next that $P(E\cap F) = P(E)P(F)$.  Then
$$
P(E|F) = \frac {P(E \cap F)}{P(F)} = P(E)\ .
$$
Also,
$$
P(F|E) = \frac {P(F \cap E)}{P(E)} = P(F)\ .
$$
Therefore, $E$ and $F$ are independent.
\end{theorem}

\begin{example}\label{exam 4.6}
Suppose that we have a coin which comes up heads with probability $p$, and tails with probability
$q$.  Now suppose that this coin is tossed twice.  Using a frequency interpretation of probability, it is reasonable to assign to the
outcome $(H,H)$ the probability $p^2$, to the outcome $(H, T)$ the probability $pq$, and so on.
Let $E$ be the event that heads turns up on the first toss and $F$ the event that tails turns up on
the second toss.  We will now check that with the above probability assignments, these two events are
independent, as expected.  We have $P(E) = p^2 + pq = p$, $P(F) = pq + q^2 = q$.  Finally $P(E\cap
F) = pq$, so $P(E\cap F) = \linebreak[4] P(E)P(F)$.
\end{example}

\begin{example}\label{exam 4.7}
It is often, but not always, intuitively clear when two events are
independent.  In Example~\ref{exam 4.6}, let $A$ be the event ``the
first toss is a head" and $B$ the event ``the two outcomes are the same." Then
$$
P(B|A) = \frac {P(B \cap A)}{P(A)} = \frac {P\{\mbox {HH}\}}{P\{\mbox {HH,HT}\}} =
\frac {1/4}{1/2} = \frac 12 = P(B).
$$
Therefore, $A$ and $B$ are independent, but the result was not so obvious.
\end{example}

\begin{example}\label{exam 4.8}
Finally, let us give an example of two events that are not independent.
In Example~\ref{exam 4.6}, let $I$ be the event ``heads on the first toss" and $J$
the event ``two heads turn up."  Then $P(I) = 1/2$ and $P(J) = 1/4$.  The event
$I \cap J$ is the event ``heads on both tosses" and has probability $1/4$.
Thus, $I$ and $J$ are not independent since $P(I)P(J) = 1/8 \ne P(I \cap J)$.
\end{example}

We can extend the concept of independence to any finite set of events
$A_1$,~$A_2$, \dots,~$A_n$.

\begin{definition}\label{def 4.3}
A set of events $\{A_1,\ A_2,\ \ldots,\ A_n\}$ is said to be \emx {mutually
independent}\index{independence of events!mutual}\index{mutually independent events} if for any
subset
$\{A_i,\ A_j,\ldots,\ A_m\}$ of these events we have
$$
P(A_i \cap A_j \cap\cdots\cap A_m) = P(A_i)P(A_j)\cdots P(A_m),
$$
or equivalently, if for any sequence $\bar A_1$,~$\bar A_2$,
\dots,~$\bar A_n$ with $\bar A_j = A_j$ or $\tilde A_j$,
$$
P(\bar A_1 \cap \bar A_2 \cap\cdots\cap \bar A_n) =
P(\bar A_1)P(\bar A_2)\cdots P(\bar A_n).
$$
(For a proof of the equivalence in the case $n = 3$,
see Exercise~\ref{exer 4.1.31}.)
\end{definition}
Using this terminology, it is a fact that any sequence  $(\mbox S,\mbox S,\mbox F,\mbox
F, \mbox S, \dots,\mbox S)$ of possible outcomes of a Bernoulli trials process forms a sequence of mutually
independent events.
\par
It is natural to ask: If all pairs of a set of events are independent, is the whole set
mutually independent?  The answer is \emx {not necessarily,} and an example is given in
Exercise~\ref{exer 4.1.7}.
\par
It is important to note that the statement
$$
P(A_1 \cap A_2 \cap \cdots \cap A_n) = P(A_1)P(A_2) \cdots P(A_n)
$$
does not imply that the events $A_1$,~$A_2$, \dots,~$A_n$ are mutually independent (see
Exercise~\ref{exer 4.1.8}).

\subsection*{Joint Distribution Functions and Independence of Random Variables}

It is frequently the case that when an experiment is performed, several different
quantities concerning the outcomes are investigated.

\begin{example}\label{exam 4.91}
Suppose we toss a coin three times.  The basic random variable ${\bar X}$ corresponding
to this experiment has eight possible outcomes, which are the ordered triples consisting
of H's and T's.  We can also define the random variable $X_i$, for $i = 1, 2, 3$, to be
the outcome of the $i$th toss.  If the coin is fair, then we should assign the probability
1/8 to each of the eight possible outcomes.  Thus, the distribution functions of $X_1$,
$X_2$, and $X_3$ are identical; in each case they are defined by $m(H) = m(T) = 1/2$.
\end{example}

If we have several random variables $X_1, X_2, \ldots, X_n$ which correspond to a given
experiment, then we can consider the joint random variable ${\bar X} = (X_1, X_2,
\ldots, X_n)$ defined by taking an outcome $\omega$ of the experiment, and writing, as an
$n$-tuple, the corresponding $n$ outcomes for the random variables $X_1, X_2, \ldots, X_n$.
Thus, if the  random variable $X_i$ has, as its set of possible outcomes the set $R_i$, then
the set of possible outcomes of the joint random variable ${\bar X}$ is the Cartesian
product of the $R_i$'s, i.e., the set of all $n$-tuples of possible outcomes of the $X_i$'s.

\begin{example}(Example~\ref{exam 4.91} continued){\label{exam 4.92}}
In the coin-tossing example above, let $X_i$ denote the outcome of the $i$th toss.  Then the
joint random variable ${\bar X} = (X_1, X_2, X_3)$ has eight possible outcomes.
\par
Suppose that we now define $Y_i$, for $i = 1, 2, 3$, as the number of heads which occur in
the first $i$ tosses.  Then $Y_i$ has $\{0, 1, \ldots, i\}$ as possible outcomes, so at
first glance, the set of possible outcomes of the joint random variable ${\bar Y} = (Y_1,
Y_2, Y_3)$ should be the set
$$\{(a_1, a_2, a_3)\ :\ 0 \le a_1 \le 1, 0 \le a_2 \le 2, 0 \le a_3 \le 3\}\ .$$
However, the outcome $(1, 0, 1)$ cannot occur, since we must have $a_1 \le a_2 \le a_3$.  The
solution to this problem is to define the probability of the outcome $(1, 0, 1)$ to be 0.  In addition, we must
have $a_{i+1} - a_i \le 1$ for $i = 1, 2$.
\par
We now illustrate the assignment of probabilities to the various outcomes for the joint
random variables ${\bar X}$ and ${\bar Y}$.  In the first case, each of the
eight outcomes should be assigned the probability 1/8, since we are assuming that we have a
fair coin.  In the second case, since $Y_i$ has $i+1$ possible outcomes, the set of possible
outcomes has size 24.    Only eight of these 24 outcomes can actually occur, namely the ones
satisfying $a_1 \le a_2 \le a_3$.  Each of these outcomes corresponds to exactly one of the
outcomes of the random variable ${\bar X}$, so it is natural to assign probability 1/8
to each of these.  We assign probability 0 to the other 16 outcomes.  In each case, the
probability function is called a joint distribution function.
\end{example}

We collect the above ideas in a definition.

\begin{definition}\label{def 4.4}
Let $X_1, X_2, \ldots, X_n$ be random variables associated with an experiment.  Suppose
that the sample space (i.e., the set of possible outcomes) of $X_i$ is the set $R_i$.  Then
the joint random variable\index{random variable!joint}\index{joint random variable} ${\bar
X} = (X_1, X_2,
\ldots, X_n)$ is defined to be the random variable whose outcomes consist of ordered $n$-tuples
of outcomes, with the
$i$th coordinate lying in the set $R_i$.  The sample space $\Omega$ of ${\bar X}$ is the
Cartesian product of the $R_i$'s:
$$\Omega = R_1 \times R_2 \times \cdots \times R_n\ .$$
The joint distribution function\index{distribution function!joint}\index{joint distribution
function} of
${\bar X}$ is the function which gives the probability of each of the outcomes of
${\bar X}$.
\end{definition}

\begin{example}(Example~\ref{exam 4.91} continued){\label{exam 4.93}}
We now consider the assignment of probabilities in the above example.  In the case of the
random variable ${\bar X}$, the probability of any outcome $(a_1, a_2, a_3)$ is just the
product of the probabilities $P(X_i = a_i)$, for $i = 1, 2, 3$.  However, in the case of
${\bar Y}$, the probability assigned to the outcome $(1, 1, 0)$ is not the product of
the probabilities $P(Y_1 = 1)$, $P(Y_2 = 1)$, and $P(Y_3 = 0)$.  The difference between these
two situations is that the value of $X_i$ does not affect the value of $X_j$, if $i \ne j$,
while the values of $Y_i$ and $Y_j$ affect one another.  For example, if $Y_1 = 1$, then
$Y_2$ cannot equal 0.  This prompts the next definition.
\end{example}

\begin{definition}\label{def 4.4.5}
The random variables $X_1$,~$X_2$, \ldots,~$X_n$ are \emx {mutually
independent}\index{independence of random variables!mutual}\index{mutually independent
random\\ variables}\index{random variables!mutual independence of} if
\begin{eqnarray*}
&&P(X_1 = r_1, X_2 = r_2, \ldots, X_n = r_n)   \\
&& \;\;\;\;\;\;\;\;\;\;\;\;\;\;\; = P(X_1 = r_1) P(X_2 = r_2) \cdots  P(X_n = r_n)
\end{eqnarray*}
for any choice of $r_1, r_2, \ldots, r_n$.  Thus, if $X_1,~X_2, \ldots,~X_n$ are mutually
independent, then the joint distribution function of the random variable
$${\bar X} = (X_1, X_2, \ldots, X_n)$$
is just the product of the individual distribution
functions.  When two random\linebreak[4] variables are mutually independent, we shall say more
briefly that they are \emx { indepen-\linebreak[4] dent.}\index{independence of random\\
variables}\index{random variables!independence of}
\end{definition}

\begin{example}\label{exam 4.94} In a group of 60 people, the numbers who do or do not smoke
and do or do not have cancer are reported as shown in Table~\ref{table 4.1}.
\begin{table}
\centering
\begin{tabular}{l|cc|c}
             &\hspace{.15in}Not smoke\hspace{.15in}&\hspace{.15in}Smoke\hspace{.15in}&\hspace{.15in}Total\hspace{.15in}  \\ \hline
 Not cancer  & 40             & 10             & 50 \\
 Cancer      & \hspace{.1in}7 &\hspace{.075in}3& 10 \\ \hline
 Totals      & 47          & 13     & 60 \\
\end{tabular}
\caption{Smoking and cancer.}
\label{table 4.1}
\end{table}
Let $\Omega$ be the sample space consisting of these 60 people.  A person is chosen at random
from the group.  Let $C(\omega) = 1$ if this person has cancer and~0 if not, and $S(\omega) =
1$ if this person smokes and~0 if not. Then the joint distribution of $\{C,S\}$ is given in
Table~\ref{table 4.2}.
\begin{table}
\centering
\begin{tabular}{lr|lcr}
  &    &            &  S      \\
  &    & \hspace{.15in} 0&  & 1\hspace{.15in}\\ \hline
  & 0  &  40/60     &  &10/60  \\
C &    &            &  &       \\
  & 1  & \hspace{.1in}7/60  &  &3/60   \\
\end{tabular}
\caption{Joint distribution.}
\label{table 4.2}
\end{table}
\noindent For example $P(C = 0, S = 0) = 40/60$, $P(C = 0, S = 1) = 10/60$, and so forth.  The
distributions of the individual random variables are called \emx {marginal
distributions.}\index{marginal distribution function}\index{distribution function!marginal}  The
marginal distributions of
$C$ and $S$ are:
$$ p_C = \pmatrix{ 0     & 1     \cr 50/60 & 10/60 \cr},
$$

$$ p_S = \pmatrix{ 0     & 1     \cr 47/60 & 13/60 \cr}.
$$
The random variables $S$ and $C$ are not independent, since
\begin{eqnarray*}
 P(C = 1,S = 1)   &=& \frac 3{60} = .05\ , \\
P(C = 1)P(S = 1) &=& \frac {10}{60} \cdot \frac
{13}{60} = .036\ .
\end{eqnarray*}
Note that we would also see this from the fact that
\begin{eqnarray*} P(C = 1|S = 1) &=& \frac 3{13} = .23\ , \\
P(C = 1) &=& \frac 16 = .167\ .
\end{eqnarray*}

\end{example}

\subsection*{Independent Trials Processes}

The study of random variables proceeds by considering special classes of random variables.
One such class that we shall study is the class of \emx {independent trials.}

\begin{definition}\label{def 5.5} A sequence of random variables $X_1$,~$X_2$, \dots,~$X_n$
that are mutually independent and that have the same distribution is called a sequence of
independent trials or an \emx {independent trials process.}\index{independent trials process}

Independent trials processes arise naturally in the following way.  We have a single
experiment with sample space $R = \{r_1,r_2,\dots,r_s\}$ and a distribution function
$$m_X = \pmatrix{ r_1 & r_2 & \cdots & r_s \cr p_1 & p_2 & \cdots & p_s\cr}\ .
$$

We repeat this experiment $n$ times.  To describe this total experiment, we choose as sample
space the space
$$
\Omega = R \times R \times\cdots\times R,
$$ consisting of all possible sequences $\omega = (\omega_1,\omega_2,\dots,\omega_n)$ where
the value of each $\omega_j$ is chosen from $R$.  We assign a distribution function to be
the \emx {product distribution}
$$ m(\omega) = m(\omega_1)\cdot\ \ldots\ \cdot m(\omega_n)\ ,
$$
with $m(\omega_j) = p_k$ when $\omega_j = r_k$.  Then we let $X_j$ denote the $j$th coordinate
of the outcome $(r_1, r_2, \ldots, r_n)$.  The random variables
$X_1$,~\dots,~$X_n$ form an independent trials process.
\end{definition}

\begin{example}\label{exam 5.6.1} An experiment consists of rolling a die three times.  Let
$X_i$ represent the outcome of the $i$th roll, for $i = 1, 2, 3$.  The common distribution
function is
$$ m_i = \pmatrix{ 1 & 2 & 3 & 4 & 5 & 6 \cr 1/6 & 1/6 & 1/6 & 1/6 & 1/6 & 1/6 \cr}.
$$

The sample space is $R^3 = R \times R \times R$ with $R = \{1,2,3,4,5,6\}$.  If
$\omega = (1,3,6)$, then $X_1(\omega) = 1$, $X_2(\omega) = 3$, and
$X_3(\omega) = 6$ indicating that the first roll was a 1, the second was a 3, and the third
was a 6.  The probability assigned to any sample point is
$$ m(\omega) = \frac16 \cdot \frac16 \cdot \frac16 = \frac1{216}\ .
$$
\end{example}

\begin{example}\label{exam 5.7} Consider next a Bernoulli trials process with probability $p$
for success on each experiment.  Let $X_j(\omega) = 1$ if the $j$th outcome is success and
$X_j(\omega) = 0$ if it is a failure.  Then $X_1$,~$X_2$, \dots,~$X_n$ is an independent
trials process.  Each $X_j$ has the same distribution function
$$ m_j = \pmatrix{ 0 & 1 \cr q & p  \cr},
$$ where $q = 1 - p$.

If $S_n = X_1 + X_2 +\cdots + X_n$, then
$$ P(S_n = j) = {n \choose j} p^{j} q^{n - j}\ ,
$$ and $S_n$ has, as distribution, the binomial distribution $b(n,p,j)$.
\end{example}

\subsection*{Bayes' Formula}

In our examples, we have considered conditional probabilities of the following
form: Given the outcome of the second stage of a two-stage experiment, find the
probability for an outcome at the first stage.  We have remarked that these
probabilities are called \emx {Bayes probabilities.}

We return now to the calculation of more general Bayes probabilities.  Suppose
we have a set of events $H_1,$~$H_2$, \dots,~$H_m$ that are pairwise disjoint
and such that the sample space $\Omega$ satisfies the equation
$$
\Omega = H_1 \cup H_2 \cup\cdots\cup H_m\ .
$$
We call these events \emx {hypotheses.}\index{hypotheses}  We also have an event $E$ that gives us
some information about which hypothesis is correct.  We call this event \emx {evidence.}

Before we receive the evidence, then, we have a set of \emx {prior
probabilities}\index{prior probabilities} $P(H_1)$,                \hfill\break
~$P(H_2)$, \dots,~$P(H_m)$ for the hypotheses.
If we know the correct hypothesis, we know the probability for the evidence.  That
is, we know $P(E|H_i)$ for all $i$.  We want to find the probabilities for the
hypotheses given the evidence.  That is, we want to find the conditional
probabilities $P(H_i|E)$.  These probabilities are called the \emx {posterior
probabilities.}\index{posterior probabilities}

To find these probabilities, we write them in the form

\begin{equation}
P(H_i|E) = \frac{P(H_i \cap E)}{P(E)}\ . \label{eq 4.1}
\end{equation}
We can calculate the numerator from our given information by
\begin{equation}
P(H_i \cap E) = P(H_i)P(E|H_i)\ . \label{eq 4.2}
\end{equation}
Since one and only one of the events $H_1$,~$H_2$, \dots,~$H_m$ can occur, we
can write the probability of $E$ as

$$
P(E) = P(H_1 \cap E)  + P(H_2 \cap E)  + \cdots + P(H_m \cap E)\ .
$$
Using Equation~\ref{eq 4.2}, the above expression can be seen to equal

\begin{equation}
   P(H_1)P(E|H_1) + P(H_2)P(E|H_2) + \cdots + P(H_m)P(E|H_m) \ .
\label{eq 4.3}
\end{equation}
Using (\ref{eq 4.1}), (\ref{eq 4.2}), and (\ref{eq 4.3})
yields \emx {Bayes' formula}:\index{Bayes' formula}
$$
P(H_i|E) = \frac{P(H_i)P(E|H_i)}{\sum_{k = 1}^m P(H_k)P(E|H_k)}\ .
$$

Although this is a very famous formula, we will rarely use it.  If the number
of hypotheses is small, a simple tree measure calculation is easily carried
out, as we have done in our examples.  If the number of hypotheses is large, then we should
use a computer.
\par
Bayes probabilities are particularly appropriate for medical diagnosis.  A
doctor is anxious to know which of several diseases a patient might have.  She
collects evidence in the form of the outcomes of certain tests.  From
statistical studies the doctor can find the prior probabilities of the various
diseases before the tests, and the probabilities for specific test outcomes,
given a particular disease.  What the doctor wants to know is the posterior
probability for the particular disease, given the outcomes of the tests.

\begin{example}\label{exam 4.10}
A doctor is trying to decide if a patient has one of three diseases
$d_1$,~$d_2$, or~$d_3$.  Two tests are to be carried out, each of which results
in a positive $(+)$ or a negative $(-)$ outcome.  There are four possible test
patterns $+{}+$,~$+{}-$, $-{}+$, and~$-{}-$.  National records have indicated
that, for 10{,}000 people having one of these three diseases, the distribution of
diseases and test results are as in Table~\ref{table 4.3}.

%\Mplace{\Table \paste{maintab}4 \endTable}
\begin{table}
\centering
\begin{tabular}{c|c|cccc}
             & Number having & \multicolumn{4}{c}{\underbar{The results}}  \\
  Disease    & this disease\hspace{.25in} &\hspace{.2in}+\hspace{.1in}+\hspace{.2in}
&\hspace{.1in}+\hspace{.1in}--\hspace{.2in}&\hspace{.2in}--\hspace{.1in}+\hspace{.2in}
&\hspace{.1in}--\hspace{.1in}--\hspace{.2in}\\ \hline
  $ d_{1} $  &\hspace{.075in}3215&2110            &\hspace{.15in}301&\hspace{.1in}704  & \hspace{.1in}100\\
  $ d_{2} $  &\hspace{.075in}2125&\hspace{.1in}396&\hspace{.15in}132&1187  & \hspace{.1in}410\\
  $ d_{3} $  &\hspace{.075in}4660&\hspace{.1in}510&\hspace{.075in}3568&\hspace{.15in}73  & \hspace{.1in}509\\ \hline
Total        &10000  &&&&\\ \hline
\end{tabular}
\caption{Diseases data.}
\label{table 4.3}
\end{table}

From this data, we can estimate the prior probabilities for each of the
diseases and, given a particular disease, the probability of a particular test
outcome.  For example, the prior probability of disease $d_1$ may be estimated
to be $3215/10{,}000 = .3215$.  The probability of the test result $+{}-$, given
disease $d_1$, may be estimated to be $301/3215 = .094$.
\par
We can now use Bayes' formula to compute various posterior probabilities.  The
computer program {\bf Bayes}\index{Bayes (program)} computes these posterior probabilities.  The
results for this example are shown in Table~\ref{table 4.35}.

\begin{table}
\centering
\begin{tabular}{lccc}
                                    &  $d_1$ &  $d_2$ &  $d_3$  \\
+\hspace{.1in}+                     & .700   & .131   & .169    \\
+\hspace{.125in}--                    & .075   & .033   & .892    \\
\hspace{.025in}--\hspace{.125in}+     &
.358   & .604   & .038    \\
\hspace{.025in}--\hspace{.15in}--  &
.098   & .403   & .499    \\
\end{tabular}
\caption{Posterior probabilities.}
\label{table 4.35}
\end{table}
We note from the outcomes that, when the test result is $++$, the disease $d_1$
has a significantly higher probability than the other two.  When the outcome is $+-$,
this is true for disease $d_3$.  When the outcome is $-+$, this is true for disease $d_2$.
Note that these statements might have been guessed by looking at the data. If the outcome is $--$,
the most probable cause is $d_3$, but the probability that a patient has $d_2$ is only slightly smaller.
If one looks at the data in this case, one can see that it might be hard to guess which of the two
diseases $d_2$ and $d_3$ is more likely.
\end{example}

Our final example shows that one has to be careful when the prior probabilities
are small.

\begin{example}\label{exam 4.11}
A doctor gives a patient a test for a particular cancer\index{cancer}.  Before the results of
the test, the only evidence the doctor has to go on is that 1 woman in~1000 has
this cancer.  Experience has shown that, in 99 percent of the cases in which
cancer is present, the test is positive; and in 95 percent of the cases in
which it is not present, it is negative.  If the test turns out to be positive,
what probability should the doctor assign to the event that cancer is present?
An alternative form of this question is to ask for the relative frequencies of false
positives and cancers.

We are given that %${\rm prior(cancer)} = .001$ and
$\mbox{prior(cancer)} = .001$ and
$\mbox{prior(not\ cancer)} = .999$.  We know also that $P(+| \mbox{cancer}) =
.99$, $P(-|\mbox{cancer}) = .01$, $P(+|\mbox{not\ cancer}) = .05$, and
$P(-|\mbox{not\ cancer}) = .95$.  Using this data gives the result shown in
Figure~\ref{fig 4.5}.

\putfig{5truein}{PSfig4-5}{Forward and reverse tree diagrams.}{fig 4.5}

We see now that the probability of cancer given a positive test has only
increased from .001 to .019.  While this is nearly a twenty-fold increase, the
probability that the patient has the cancer is still small.  Stated in another way,
among the positive results, 98.1 percent are false positives, and 1.9 percent are
cancers.  When a group of second-year medical students was asked this question, over half of the
students incorrectly guessed the probability to be greater than .5.
\end{example}

\subsection*{Historical Remarks}

Conditional probability was used long before it was formally defined.  Pascal and Fermat
considered the \emx {problem of points}:\index{problem of points} given that team A has won $m$
games and team B has won $n$ games, what is the probability that A will win the series?  (See
Exercises~\ref{exer 5.1.11}--\ref{exer 5.1.13}.)  This is clearly a conditional probability
problem.
\par
In his book,  Huygens\index{HUYGENS, C.} gave a number of problems, one of which was:
\begin{quote}
\indent Three gamblers, A, B and C, take 12 balls of which 4 are white and 8
black.  They play with the rules that the drawer is blindfolded, A is to draw
first, then B and then C, the winner to be the one who first draws a white
ball.  What is the ratio of their chances?\footnote{Quoted in F.~N. David,
\emx {Games, Gods and Gambling} (London: Griffin, 1962), p.~119.}
\end{quote}
\par
From his answer it is clear that Huygens meant that each ball is replaced after
drawing.  However, John Hudde\index{HUDDE, J.}, the mayor of Amsterdam, assumed that he meant to
sample without replacement and corresponded with Huygens about the difference
in their answers.  Hacking remarks that ``Neither party can understand what the
other is doing."\index{HACKING, I.}\footnote{I.~Hacking, \emx {The Emergence of Probability}
(Cambridge: Cambridge University Press, 1975), p.~99.}
\par
By the time of de~Moivre's\index{de MOIVRE, A.} book, \emx {The Doctrine of Chances,} these
distinctions were well understood.  De~Moivre defined independence and
dependence as follows:
\begin{quote}
\indent Two Events are independent, when they have no connexion one with the
other, and that the happening of one neither forwards nor obstructs the
happening of the other.
\par
Two Events are dependent, when they are so connected together as that the
Probability of either's happening is altered by the happening of the
other.\footnote{A.~de~Moivre, \emx {The Doctrine of Chances,} 3rd~ed. (New York:
Chelsea, 1967), p.~6.}
\end{quote}

De~Moivre used sampling with and without replacement to illustrate that the
probability that two independent events both happen is the product of their
probabilities, and for dependent events that:
\begin{quote}
The Probability of the happening of two Events dependent, is the product of the
Probability of the happening of one of them, by the Probability which the other
will have of happening, when the first is considered as having happened; and
the same Rule will extend to the happening of as many Events as may be
assigned.\footnote{ibid, p.~7.}
\end{quote}
\par
The formula that we call Bayes' formula, and the idea of computing the
probability of a hypothesis given evidence, originated in a famous essay of
Thomas Bayes\index{BAYES, T.}.  Bayes was an ordained minister in Tunbridge Wells near London.
His mathematical interests led him to be elected to the Royal Society in~1742,
but none of his results were published within his lifetime.  The work upon
which his fame rests, ``An Essay Toward Solving a Problem in the Doctrine of
Chances," was published in~1763, three years after his
death.\footnote{T.~Bayes, ``An Essay Toward Solving a Problem in the Doctrine
of Chances," \emx {Phil.\ Trans.\ Royal Soc.\ London,} vol.~53 (1763),
pp.~370--418.}  Bayes reviewed some of the basic concepts of probability and
then considered a new kind of inverse probability problem requiring the use of
conditional probability.
\par
Bernoulli\index{BERNOULLI, J.}, in his study of processes that we now call Bernoulli trials, had
proven his famous law of large numbers which we will study in Chapter~\ref{chp
8}.  This theorem assured the experimenter that if he knew the probability $p$
for success, he could predict that the proportion of successes would approach
this value as he increased the number of experiments.  Bernoulli himself
realized that in most interesting cases you do not know the value of $p$ and
saw his theorem as an important step in showing that you could determine $p$ by
experimentation.
\par
To study this problem further, Bayes started by assuming that the probability
$p$ for success is itself determined by a random experiment.  He assumed in
fact that this experiment was such that this value for $p$ is equally likely to
be any value between 0 and 1.\choice{\footnote{Bayes is using continuous
probabilities as discussed in Chapter 2 of the complete Grinstead-Snell book.}}{}  Without knowing this value we
carry out
$n$ experiments and observe $m$ successes.  Bayes proposed the problem of finding
the conditional probability that the unknown probability $p$ lies between $a$
and $b$.  He obtained the answer:
$$
P(a \leq p < b | m {\mbox{\,\,successes\,\, in}}\,\,n \,\,{\mbox{trials}})
= \frac {\int_a^b x^m(1 - x)^{n - m}\,dx}{\int_0^1 x^m(1 - x)^{n - m}\,dx}\ .
$$
\par
\choice{}{We shall see in the next section how this result is obtained.}  Bayes clearly
wanted to show that the conditional distribution function, given the outcomes of
more and more experiments, becomes concentrated around the true value of $p$.
Thus, Bayes was trying to solve an \emx {inverse problem.}  The computation of
the integrals was too difficult for exact solution except for small values
of~$j$ and~$n$, and so Bayes tried approximate methods.  His methods were not
very satisfactory and it has been suggested that this discouraged him from
publishing his results.
\par
However, his paper was the first in a series of important studies carried out
by Laplace, Gauss, and other great mathematicians to solve inverse problems.
They studied this problem in terms of errors in measurements in astronomy.  If
an astronomer were to know the true value of a distance and the nature of the
random errors caused by his measuring device he could predict the probabilistic
nature of his measurements.  In fact, however, he is presented with the inverse
problem of knowing the nature of the random errors, and the values of the
measurements, and wanting to make inferences about the unknown true value.
\par
As Maistrov remarks, the formula that we have called Bayes' formula does not
appear in his essay.  Laplace gave it this name when he studied these inverse
problems.\index{MAISTROV, L.}\footnote{L.~E. Maistrov, \emx {Probability Theory: A Historical
Sketch,} trans.~and ed.~Samual Kotz (New York: Academic Press, 1974), p.~100.}
The computation of inverse probabilities is fundamental to statistics and has
led to an important branch of statistics called Bayesian analysis, assuring
Bayes eternal fame for his brief essay.

\exercises
\begin{LJSItem}

\i\label{exer 4.1.1} Assume that $E$ and $F$ are two events with positive
probabilities.  Show that if $P(E|F) = P(E)$, then $P(F|E) = P(F)$.

\i\label{exer 4.1.2} A coin is tossed three times.  What is the probability that exactly two
heads occur, given that
\begin{enumerate}
\item the first outcome was a head?

\item the first outcome was a tail?

\item the first two outcomes were heads?

\item the first two outcomes were tails?

\item the first outcome was a head and the third outcome was a head?
\end{enumerate}

\i\label{exer 4.1.3} A die is rolled twice.  What is the probability that the sum of the faces
is greater than 7, given that
\begin{enumerate}
\item the first outcome was a 4?

\item the first outcome was greater than 3?

\item the first outcome was a 1?

\item the first outcome was less than 5?
\end{enumerate}

\i\label{exer 4.1.4} A card is drawn at random from a deck of cards.  What is the probability
that
\begin{enumerate}
\item it is a heart, given that it is red?

\item it is higher than a 10, given that it is a heart? (Interpret J,~Q, K,~A
as 11,~12, 13,~14.)

\item it is a jack, given that it is red?
\end{enumerate}

\i\label{exer 4.1.5} A coin is tossed three times.  Consider the following events\newline
$A$: Heads on the first toss.\newline
$B$: Tails on the second.\newline
$C$: Heads on the third toss.\newline
$D$: All three outcomes the same (HHH or TTT).\newline
$E$: Exactly one head turns up.
\begin{enumerate}
\item Which of the following pairs of these events are independent?\newline
(1) $A$, $B$\newline
(2) $A$, $D$\newline
(3) $A$, $E$\newline
(4) $D$, $E$

\item Which of the following triples of these events are
independent?\newline
(1) $A$, $B$, $C$\newline
(2) $A$, $B$, $D$\newline
(3) $C$, $D$, $E$
\end{enumerate}

\i\label{exer 4.1.6} From a deck of five cards numbered 2,~4, 6, 8, and~10, respectively, a
card is drawn at random and replaced.  This is done three times.  What is the
probability that the card numbered 2 was drawn exactly two times, given that
the sum of the numbers on the three draws is~12?

\i\label{exer 4.1.7} A coin is tossed twice.  Consider the
following events.\newline
$A$: Heads on the first toss.\newline
$B$: Heads on the second toss.\newline
$C$: The two tosses come out the same.
\begin{enumerate}
\item Show that $A$,~$B$,~$C$ are pairwise independent but not
independent.

\item Show that $C$ is independent of $A$ and $B$ but not of $A \cap B$.
\end{enumerate}

\i\label{exer 4.1.8} Let $\Omega = \{a,b,c,d,e,f\}$.  Assume that $m(a) = m(b) = 1/8$ and
$m(c) = m(d) = m(e) = m(f) = 3/16$.  Let $A$,~$B$, and~$C$ be the events $A = \{d,e,a\}$,
$B = \{c,e,a\}$, $C = \{c,d,a\}$.  Show that $P(A \cap B \cap C) = P(A)P(B)P(C)$ but no two of these
events are independent.

\i\label{exer 4.1.9} What is the probability that a family of two children has
\begin{enumerate}
\item two boys given that it has at least one boy?

\item two boys given that the first child is a boy?
\end{enumerate}

\i\label{exer 4.1.9.5}  In Example~\ref{exam 4.1.5}, we used the Life Table (see Appendix~C)
to compute a conditional probability.  The number 93{,}753 in the table, corresponding to
40-year-old males, means that of all the males born in the United States in 1950, 93.753\% were
alive in 1990.  Is it reasonable to use this as an estimate for the probability of a male, born
this year, surviving to age 40?

\i\label{exer 4.1.9.6} Simulate the Monty Hall problem.  Carefully state any assumptions
that you have made when writing the program.  Which version of the problem do you think that
you are simulating?

\i\label{exer 4.1.10} In Example~\ref{exam 4.11}, how large must the prior probability of
cancer be to give a posterior probability of .5 for cancer given a positive
test?

\i\label{exer 4.1.11} Two cards are drawn from a bridge deck.  What is the probability that
the second card drawn is red?

\i\label{exer 4.1.12} If $P(\tilde B) = 1/4$ and $P(A|B) = 1/2$, what is $P(A \cap B)$?

\i\label{exer 4.1.13}
\begin{enumerate}
\item What is the probability that your bridge partner has exactly two aces,
given that she has at least one ace?

\item What is the probability that your bridge partner has exactly two aces,
given that she has the ace of spades?
\end{enumerate}

\i\label{exer 4.1.14} Prove that for any three events $A$,~$B$,~$C$, each having positive
probability, and with the property that $P(A \cap B) > 0$,
$$
P(A \cap B \cap C) = P(A)P(B|A)P(C|A \cap B)\ .
$$

\i\label{exer 4.1.15} Prove that if $A$ and $B$ are independent so are
\begin{enumerate}
\item $A$ and $\tilde B$.

\item $\tilde A$ and $\tilde B$.
\end{enumerate}

\i\label{exer 4.1.16} A doctor assumes that a patient has one of three diseases $d_1$,~$d_2$,
or~$d_3$.  Before any test, he assumes an equal probability for each disease.
He carries out a test that will be positive with probability .8 if the patient
has $d_1$, .6 if he has disease $d_2$, and .4 if he has disease $d_3$.  Given that
the outcome of the test was positive, what probabilities should the doctor now
assign to the three possible diseases?

\i\label{exer 4.1.17} In a poker hand, John has a very strong hand and bets 5 dollars.  The
probability that Mary has a better hand is .04.  If Mary had a better hand she
would raise with probability .9, but with a poorer hand she would only raise
with probability .1.  If Mary raises, what is the probability that she has a
better hand than John does?

\i\label{exer 4.1.18} The Polya urn model\index{Polya urn model} for contagion is as follows: We
start with an urn which contains one white ball and one black ball.  At each
second we choose a ball at random from the urn and replace this ball and add one
more of the color chosen.  Write a program to simulate this model, and see if
you can make any predictions about the proportion of white balls in the urn
after a large number of draws.  Is there a tendency to have a large fraction of
balls of the same color in the long run?

\i\label{exer 4.1.19} It is desired to find the probability that in a bridge deal each player
receives an ace.  A student argues as follows.  It does not matter where the
first ace goes.  The second ace must go to one of the other three players and
this occurs with probability 3/4.  Then the next must go to one of two, an
event of probability 1/2, and finally the last ace must go to the player who
does not have an ace.  This occurs with probability 1/4.  The probability that
all these events occur is the product $(3/4)(1/2)(1/4) = 3/32$.  Is this
argument correct?

\i\label{exer 4.1.20} One coin in a collection of 65 has two heads.  The rest are fair.  If a
coin, chosen at random from the lot and then tossed, turns up heads 6 times in
a row, what is the probability that it is the two-headed coin?

\i\label{exer 4.1.21} You are given two urns and fifty balls.  Half of the balls are white and
half are black.  You are asked to distribute the balls in the urns with no
restriction placed on the number of either type in an urn.  How should you
distribute the balls in the urns to maximize the probability of obtaining a
white ball if an urn is chosen at random and a ball drawn out at random?
Justify your answer.

\i\label{exer 4.1.22} A fair coin is thrown $n$ times.  Show that the conditional probability
of a head on any specified trial, given a total of $k$ heads over the $n$
trials, is $k/n$ $(k > 0)$.

\i\label{exer 4.1.23} (Johnsonbough\index{JOHNSONBOUGH, R.}\footnote{R. Johnsonbough, ``Problem
\#103," \emx {Two Year College Math Journal,} vol.~8 (1977), p.~292.})  A coin with probability
$p$ for heads is tossed $n$ times.  Let $E$ be the event ``a head is obtained on the
first toss' and $F_k$ the event `exactly $k$ heads are obtained."  For which
pairs $(n,k)$ are $E$ and $F_k$ independent?

\i\label{exer 4.1.24} Suppose that $A$ and $B$ are events such that $P(A|B) = P(B|A)$ and $P(A
\cup B) = 1$ and $P(A \cap B) > 0$.  Prove that $P(A) > 1/2$.

\i\label{exer 4.1.25} (Chung\index{CHUNG, K. L.}\footnote{K. L. Chung, \emx {Elementary
Probability Theory With Stochastic Processes, 3rd ed.} (New York:  Springer-Verlag, 1979), p.~152.})
In London, half of the days have some rain.  The weather forecaster is correct 2/3 of the time,
i.e., the probability that it rains, given that she has predicted rain, and the probability that it
does not rain, given that she has predicted that it won't rain, are both equal to 2/3.  When rain is
forecast, Mr.\ Pickwick\index{Pickwick, Mr.} takes his umbrella.  When rain is not forecast, he
takes it with probability 1/3.  Find
\begin{enumerate}
\item the probability that Pickwick has no umbrella, given that it rains.

\item the probability that he brings his umbrella, given that it doesn't rain.
\end{enumerate}

\i\label{exer 4.1.26} Probability theory was used in a famous court
case: \emx {People v. Collins.}\index{Collins, People v.}\index{People v. Collins}\footnote{M.~W.
Gray, ``Statistics and the Law," \emx {Mathematics Magazine,} vol.~56 (1983), pp.~67--81.}  In
this case a purse was snatched from an elderly person in a Los Angeles suburb.  A couple seen
running from the scene were described as a black man with a beard\index{beard} and a
mustache\index{mustache} and a blond girl with hair in a ponytail\index{ponytail}.  Witnesses said
they drove off in a partly yellow car.  Malcolm and Janet Collins were arrested.  He was black and
though clean shaven when arrested had evidence of recently having had a beard
and a mustache.  She was blond and usually wore her hair in a ponytail.  They
drove a partly yellow Lincoln.  The prosecution called a professor of mathematics
as a witness who suggested that a conservative set of probabilities for the
characteristics noted by the witnesses would be as shown in Table~\ref{table 4.4}.
\begin{table}[h]
\centering
\begin{tabular}{lc}
{\rm man with mustache}    & 1/4          \cr
{\rm girl with blond hair} & 1/3          \cr
{\rm girl with ponytail}   & 1/10         \cr
{\rm black man with beard} & 1/10          \cr
{\rm interracial couple in a car} & 1/1000 \cr
{\rm partly yellow car}    & 1/10    \cr
\end{tabular}
\caption{Collins case probabilities.}
\label{table 4.4}
\end{table}
\par
The prosecution then argued that the probability that all of these
characteristics are met by a randomly chosen couple is the product of the
probabilities or 1/12{,}000{,}000, which is very small.  He claimed this was proof
beyond a reasonable doubt that the defendants were guilty.  The jury agreed
and handed down a verdict of guilty of second-degree robbery.

If you were the lawyer for the Collins couple how would you have countered the
above argument?  (The appeal of this case is discussed in
Exercise~\ref{sec 5.1}.\ref{exer 9.2.23}.)

\i\label{exer 4.1.27} A student is applying to Harvard and Dartmouth.  He estimates that he
has a probability of .5 of being accepted at Dartmouth and .3 of being accepted at
Harvard.  He further estimates the probability that he will be accepted by both
is .2.  What is the probability that he is accepted by Dartmouth if he is
accepted by Harvard?  Is the event ``accepted at Harvard" independent of the
event ``accepted at Dartmouth"?

\i\label{exer 4.1.28} Luxco, a wholesale lightbulb manufacturer, has two factories.  Factory A
sells bulbs in lots that consists of 1000 regular and 2000 \emx {softglow}
bulbs each.  Random sampling has shown that on the average there tend to be
about 2 bad regular bulbs and 11 bad softglow bulbs per lot.  At factory B the
lot size is reversed---there are 2000 regular and 1000 softglow per lot---and
there tend to be 5 bad regular and 6 bad softglow bulbs per lot.

The manager of factory A asserts, ``We're obviously the better producer; our
bad bulb rates are .2 percent and .55 percent compared to B's .25 percent and
.6 percent.  We're better at both regular and softglow bulbs by half of a tenth
of a percent each."

``Au contraire," counters the manager of B, ``each of our 3000 bulb lots
contains only 11 bad bulbs, while A's 3000 bulb lots contain 13.  So our .37
percent bad bulb rate beats their .43 percent."
\par
Who is right?

\i\label{exer 4.1.29} Using the Life Table for 1981 given in Appendix~C, find the
probability that a male of age~60 in 1981 lives to age~80.  Find the same
probability for a female.

\i\label{exer 4.1.30}
\begin{enumerate}
\item There has been a blizzard and Helen is trying to drive from Woodstock\index{Woodstock}
to Tunbridge\index{Tunbridge}, which are connected like the top graph in Figure~\ref{fig
4.51}.  Here $p$ and $q$ are the probabilities that the two roads are passable.  What
is the probability that Helen can get from Woodstock to Tunbridge?

\item Now suppose that Woodstock and Tunbridge are connected like the middle graph
in Figure~\ref{fig 4.51}.
What now is the probability that she can get from $W$ to $T$?  Note that if we
think of the roads as being components of a system, then in
(a) and (b) we have computed the
\emx {reliability}\index{reliability of a system} of a system whose components are (a)~\emx {in
series} and (b)~\emx {in parallel.}

\item Now suppose $W$ and $T$ are connected like the bottom graph in Figure~\ref{fig 4.51}.
Find the probability of Helen's getting from $W$ to $T$.  \emx {Hint}: If the
road from $C$ to $D$ is impassable, it might as well not be there at all; if it
is passable, then figure out how to use part (b) twice.
\end{enumerate}

\putfig{3.5truein}{PSfig4}{From Woodstock to Tunbridge.}{fig 4.51} %%3.5truein

\i\label{exer 4.1.31} Let $A_1$,~$A_2$, and $A_3$ be events, and let $B_i$ represent
either $A_i$ or its complement $\tilde A_i$.  Then there are eight possible choices for the
triple $(B_1, B_2, B_3)$.  Prove that the events $A_1$,~$A_2$,~$A_3$ are independent if and
only if $$P(B_1 \cap B_2 \cap B_3) = P(B_1)P(B_2)P(B_3)\ ,$$
for all eight of the possible choices for the triple $(B_1, B_2, B_3)$.

\i\label{exer 5.1.4} Four women, A,~B, C, and~D, check their hats, and the hats are returned
in a random manner.  Let $\Omega$ be the set of all possible permutations of A,~B, C,~D.  Let
$X_j = 1$ if the $j$th woman gets her own hat back and 0 otherwise.  What is the
distribution of $X_j$?  Are the $X_i$'s mutually independent?

\i\label{exer 5.1.5} A box has numbers from 1 to 10.  A number is drawn at random.  Let
$X_1$ be the number drawn.  This number is replaced, and the ten numbers mixed.  A second
number $X_2$ is drawn.  Find the distributions of $X_1$ and $X_2$.  Are
$X_1$ and $X_2$ independent?  Answer the same questions if the first number is not replaced
before the second is drawn.

\i\label{exer 5.1.6} A die is thrown twice.  Let $X_1$ and $X_2$ denote the outcomes.
Define $X = \min(X_1, X_2)$.  Find the distribution of $X$.

\istar\label{exer 5.1.7} \begin{sloppypar} Given that $P(X = a) = r$, $P(\max(X,Y) = a) = s$,
and
$P(\min(X,Y) = a) = t$, show that you can determine $u = P(Y = a)$ in terms of $r$,~$s$,
and~$t$.\end{sloppypar}

\i\label{exer 5.1.8} A fair coin is tossed three times.  Let $X$ be the number of heads that
turn up on the first two tosses and $Y$ the number of heads that turn up on the third toss.
Give the distribution of
\begin{enumerate}
\item the random variables $X$ and $Y$.

\item the random variable $Z = X + Y$.

\item the random variable $W = X - Y$.
\end{enumerate}

\i\label{exer 5.1.10} Assume that the random variables $X$ and $Y$ have the joint distribution
given in Table~\ref{table 4.5}.
\begin{table}
\centering
\begin{tabular}{lr|rrrrr}
     &     &  $Y$  &       &        &        \\
     &     &   -1  &  0    &  1     &  2     \\ \hline
$X$  & -1  &   0   & 1/36  &  1/6   & 1/12   \\
     &  0  & 1/18  &  0    &  1/18  &  0     \\
     &  1  &   0   & 1/36  &  1/6   & 1/12   \\
     &  2  & 1/12  &  0    &  1/12  & 1/6    \\
\end{tabular}
\caption{Joint distribution.}
\label{table 4.5}
\end{table}

\begin{enumerate}
\item What is $P(X \geq 1\ \mbox {and\ } Y \leq 0)$?

\item What is the conditional probability that $Y \leq 0$ given that $X = 2$?

\item Are $X$ and $Y$ independent?

\item What is the distribution of $Z = XY$?
\end{enumerate}

\i\label{exer 5.1.11} In the \emx {problem of points}\index{problem of points}, discussed in
the historical remarks in Section~\ref{sec 3.2}, two players, A and B, play a series of points in a
game with player A winning each point with probability $p$ and player B winning each point with
probability
$q = 1 - p$.  The first player to win $N$ points wins the game.  Assume that $N = 3$.  Let $X$
be a random variable that has the value~1 if player A wins the series and 0 otherwise.  Let
$Y$ be a random variable with value the number of points played in a game.  Find the
distribution of $X$ and $Y$ when $p = 1/2$.  Are $X$ and $Y$ independent in this case?  Answer
the same questions for the case $p = 2/3$.

\i\label{exer 5.1.12} The letters between Pascal\index{PASCAL, B.} and Fermat\index{FERMAT, P.}, which
are often credited with having started probability theory, dealt mostly with the \emx {problem of
points} described in Exercise~\ref{exer 5.1.11}.  Pascal and Fermat considered the problem of
finding a fair division of stakes if the game must be called off when the first player has won
$r$ games and the second player has won $s$ games, with $r < N$ and $s < N$.  Let $P(r,s)$ be the
probability that player A wins the game if he has already won $r$ points and player B has won
$s$ points.  Then
\begin{enumerate}
\item $P(r,N) = 0$ if $r < N$,
\item $P(N,s) = 1$ if $s < N$,
\item $P(r,s) = pP(r + 1,s) + qP(r,s + 1)$ if $r < N$ and $s < N$;
\end{enumerate} and (1),~(2), and~(3) determine $P(r,s)$ for $r \leq N$ and $s \leq N$.  Pascal
used these facts to find $P(r,s)$ by working backward: He first obtained $P(N - 1,j)$ for $j =
N - 1$,~$N - 2$, \dots,~0; then, from these values, he obtained
$P(N - 2,j)$ for $j = N - 1$,~$N - 2$, \dots,~0 and, continuing backward, obtained all the
values $P(r,s)$.  Write a program to compute $P(r,s)$ for given $N$,~$a$, $b$, and~$p$.  \emx {
Warning}: Follow Pascal and you will be able to run $N = 100$; use recursion and you will
not be able to run $N = 20$.

\i\label{exer 5.1.13} Fermat solved the \emx {problem of points} (see Exercise~\ref{exer
5.1.11}) as follows: He realized that the problem was difficult because the possible ways the
play might go are not equally likely.  For example, when the first player needs two more games
and the second needs three to win, two possible ways the series might go for the first player
are WLW and LWLW.  These sequences are not equally likely.  To avoid this difficulty, Fermat
extended the play, adding fictitious plays so that the series went the maximum number of games
needed (four in this case).  He obtained equally likely outcomes and used, in effect, the
Pascal triangle to calculate $P(r,s)$.  Show that this leads to a \emx {formula} for $P(r,s)$
even for the case $p \ne 1/2$.

\i\label{exer 5.1.14} The Yankees are playing the Dodgers in a world series.  The Yankees win
each game with probability .6.  What is the probability that the Yankees win the series?  (The
series is won by the first team to win four games.)

\i\label{exer 5.1.15} C.~L. Anderson\index{ANDERSON, C. L.}\footnote{C.~L. Anderson, ``Note on the
Advantage of First Serve," \emx {Journal of Combinatorial Theory,} Series~A, vol.~23 (1977),
p.~363.} has used Fermat's argument for the \emx {problem of points} to prove the following result
due to J.~G. Kingston\index{KINGSTON, J. G.}.  You are playing the \emx {game of points} (see
Exercise~\ref{exer 5.1.11}) but, at each point, when you serve you win with probability $p$, and
when your opponent serves you win with probability $\bar{p}$.  You will serve first, but you
can choose one of the following two conventions for serving: for the first convention you
alternate service (tennis)\index{tennis}, and for the second the person serving continues to serve
until he loses a point and then the other player serves (racquetball)\index{racquetball}.  The
first player to win $N$ points wins the game.  The problem is to show that the probability of
winning the game is the same under either convention.
\begin{enumerate}

\item Show that, under either convention, you will serve at most $N$ points and your opponent
at most $N - 1$ points.

\item Extend the number of points to $2N - 1$ so that you serve $N$ points and your opponent
serves $N - 1$.  For example, you serve any additional points necessary to make $N$ serves and
then your opponent serves any additional points necessary to make him serve $N - 1$ points.
The winner is now the person, in the extended game, who wins the most points.  Show that
playing these additional points has not changed the winner.

\item Show that (a) and (b) prove that you have the same probability of winning the game under
either convention.
\end{enumerate}

\i\label{exer 5.1.15.5}
In the previous problem, assume that $p = 1 - \bar{p}$.
\begin{enumerate}
\item  Show that under either service convention, the first player will win more often than the second
player if and only if $p > .5$.
\item
In volleyball,\index{volleyball} a team can only win a point while it is serving.  Thus, any
individual ``play" either ends with a point being awarded to the serving team or with the service
changing to the other team.  The first team to win $N$ points wins the game.  (We ignore here the
additional restriction that the winning team must be ahead by at least two points at the end of the
game.)  Assume that each team has the same probability of winning the play when it is serving, i.e.,
that $p = 1 - \bar{p}$.  Show that in this case, the team that serves first will win more than half the
time, as long as $p > 0$.  (If $p = 0$, then the game never ends.)  \emx {Hint}: Define $p'$ to be
the probability that a team wins the next point, given that it is serving.  If we write $q = 1 -
p$, then one can show that
$$p' = \frac p{1-q^2}\ .$$
If one now considers this game in a slightly different way, one can see that the second service
convention in the preceding problem can be used, with $p$ replaced by $p'$.
\end{enumerate}

\i\label{exer 5.1.19} A poker hand consists of 5 cards dealt from a deck of 52 cards.  Let $X$
and $Y$ be, respectively, the number of aces and kings in a poker hand.  Find the joint
distribution of $X$ and $Y$.

\i\label{exer 5.1.24.5} Let $X_1$ and $X_2$ be independent random variables and let $Y_1 =
\phi_1(X_1)$ and $Y_2 = \phi_2(X_2)$.
\begin{enumerate}

\item Show that
$$ P(Y_1 = r, Y_2 = s) = \sum_{\phi_1(a) = r \atop \phi_2(b) = s} P(X_1 = a, X_2 = b)\ .
$$

\item Using (a), show that $P(Y_1 = r, Y_2 = s) = P(Y_1 = r)P(Y_2 = s)$ so that $Y_1$ and
$Y_2$ are independent.
\end{enumerate}

\i\label{exer 4.1.32} Let $\Omega$ be the sample space of an experiment.  Let $E$ be an
event with $P(E) > 0$ and define $m_E(\omega)$ by
$m_E(\omega) = m(\omega|E)$.  Prove that $m_E(\omega)$ is a distribution function on
$E$, that is, that $m_E(\omega) \geq 0$ and that $\sum_{\omega\in\Omega}
m_E(\omega) = 1$.  The function $m_E$ is called the \emx {conditional distribution given $E$.}

\i\label{exer 4.1.33} You are given two urns each containing two biased coins.  The coins in
urn~I come up heads with probability~$p_1$, and the coins in urn~II come up
heads with probability $p_2 \ne p_1$.  You are given a choice of (a)~choosing
an urn at random and tossing the two coins in this urn or (b)~choosing one coin
from each urn and tossing these two coins.  You win a prize if both coins turn
up heads.  Show that you are better off selecting choice (a).

\i\label{exer 4.1.34} Prove that, if $A_1$,~$A_2$, \dots,~$A_n$ are independent events
defined on a sample space $\Omega$ and if $0 < P(A_j) < 1$ for all $j$, then
$\Omega$ must have at least $2^n$ points.

\i\label{exer 4.1.35} Prove that if
$$
P(A|C) \geq P(B|C) \mbox{\,\,and\,\,} P(A|\tilde C) \geq P(B|\tilde C)\ ,
$$
then $P(A) \geq P(B)$.

\i\label{exer 4.1.36} A coin is in one of $n$ boxes.  The probability that it is in the $i$th
box is $p_i$.  If you search in the $i$th box and it is there, you find it with
probability $a_i$.  Show that the probability~$p$ that the coin is in the $j$th
box, given that you have looked in the $i$th box and not found it, is
$$
p =  \left \{ \matrix{
                 p_j/(1-a_ip_i),&\,\,\, \mbox{if} \,\,\, j \ne i,\cr
                 (1 - a_i)p_i/(1 - a_ip_i),&\,\,\,\mbox{if} \,\, j = i.\cr}\right.
$$

\i\label{exer 4.1.37} George Wolford\index{WOLFORD, G.} has suggested the following variation on
the Linda problem (see Exercise~\ref{sec 1.2}.\ref{exer 1.2.24}).
The registrar is carrying John and Mary's registration cards and drops them in a
puddle.  When he pickes them up he cannot read the names but on the first card
he picked up he can make out Mathematics 23 and Government 35, and on the second
card he can make out only Mathematics 23.  He asks you if you can help him
decide which card belongs to Mary.  You know that Mary likes government but does
not like mathematics.  You know nothing about John and assume that he is just a
typical Dartmouth student.  From this you estimate:
\vskip .1in
$$\begin{array}{ll}
P(\mbox {Mary\ takes\ Government\ 35})  &= .5\ , \\
P(\mbox {Mary\ takes\ Mathematics\ 23}) &= .1\ , \\
P(\mbox {John\ takes\ Government\ 35})  &= .3\ , \\
P(\mbox {John\ takes\ Mathematics\ 23}) &= .2\ .
\end{array}
$$
\vskip .1in
Assume that their choices for courses are independent events.  Show that the
card with Mathematics 23 and Government 35 showing is more likely to be Mary's
than John's.  The conjunction fallacy referred to in the Linda problem would be
to assume that the event ``Mary takes Mathematics 23 and Government 35" is
more likely than the event ``Mary takes Mathematics 23."  Why are we not
making this fallacy here?

\i\label{exer 4.1.38} (Suggested by Eisenberg and Ghosh\index{EISENBERG,
B.}\index{GHOSH, B. K.}\footnote{B. Eisenberg and B.~K. Ghosh, ``Independent Events in a Discrete
Uniform Probability Space," \emx {The American Statistician,} vol.~41, no.~1 (1987),
pp.~52--56.})  A deck of playing cards can be described as a Cartesian product
$$
\mbox{Deck} =  \mbox{Suit} \times \mbox{Rank}\ ,
$$
where $\mbox{Suit} = \{\clubsuit,\diamondsuit,\heartsuit,\spadesuit\}$ and %
$\mbox{Rank} = \{2,3,\dots,10,{\mbox J},{\mbox Q},{\mbox K},{\mbox A}\}$.  This
just means that every card may be thought of as an ordered pair like
$(\diamondsuit,2)$.  By a \emx {suit event}\index{suit event} we mean any event $A$ contained in
Deck which is described in terms of Suit alone.  For instance, if $A$ is ``the
suit is red," then
$$
A = \{\diamondsuit,\heartsuit\} \times \mbox{Rank}\ ,
$$
so that $A$ consists of all cards of the form $(\diamondsuit,r)$ or
$(\heartsuit,r)$ where $r$ is any rank.  Similarly, a \emx {rank event}\index{rank event} is any
event described in terms of rank alone.
\begin{enumerate}
\item Show that if $A$ is any suit event and $B$ any rank event, then $A$
and $B$ are \emx {independent.}  (We can express this briefly by saying that
suit and rank are independent.)

\item Throw away the ace of spades.  Show that now no nontrivial (i.e.,
neither empty nor the whole space) suit event $A$ is independent of any
nontrivial rank event $B$.  \emx {Hint}: Here independence comes down to
$$
c/51 = (a/51) \cdot (b/51)\ ,
$$
where $a$,~$b$,~$c$ are the respective sizes of $A$,~$B$ and~$A \cap B$.  It
follows that 51 must divide $ab$, hence that 3 must divide one of $a$ and $b$,
and 17 the other.  But the possible sizes for suit and rank events preclude
this.

\item Show that the deck in (b) nevertheless does have
pairs $A$,~$B$ of nontrivial independent events.  \emx {Hint}: Find 2 events
$A$ and $B$ of sizes 3 and 17, respectively, which intersect in a single point.

\item Add a joker to a full deck.  Show that now there is no pair $A$,~$B$
of nontrivial independent events.  \emx {Hint}: See the hint in
(b); 53 is prime.
\end{enumerate}
\par
\medskip
\noindent
The following problems are suggested by Stanley Gudder\index{GUDDER, S.} in his article
``Do Good Hands Attract?"\footnote{S. Gudder, ``Do Good Hands Attract?" \emx {
Mathematics Magazine,} vol.~54, no.~1 (1981), pp.~13--16.}  He says that
event $A$ \emx {attracts}\index{events!attraction of} event $B$ if $P(B|A) > P(B)$ and \emx {
repels}\index{events!repulsion of}
$B$ if $P(B|A) < P(B)$.

\i\label{exer 4.1.39} Let $R_i$ be the event that the $i$th player in a poker game has a
royal flush.  Show that a royal flush (A,K,Q,J,10 of one suit) attracts another royal flush,
that is $P(R_2|R_1) > P(R_2)$.  Show that a royal flush repels full houses.

\i\label{exer 4.1.40} Prove that $A$ attracts $B$ if and only if $B$ attracts $A$.  Hence we
can say that $A$ and $B$ are \emx {mutually attractive} if $A$ attracts $B$.

\i\label{exer 4.1.41} Prove that $A$ neither attracts nor repels $B$ if and only if $A$ and
$B$ are independent.

\i\label{exer 4.1.42} Prove that $A$ and $B$ are mutually attractive if and only if $P(B|A) >
P(B|\tilde A)$.

\i\label{exer 4.1.43} Prove that if $A$ attracts $B$, then $A$ repels $\tilde B$.

\i\label{exer 4.1.44} Prove that if $A$ attracts both $B$ and $C$, and $A$ repels $B \cap C$,
then $A$ attracts $B \cup C$.  Is there any example in which $A$ attracts both
$B$ and $C$ and repels $B \cup C$?

\i\label{exer 4.1.45} Prove that if $B_1$,~$B_2$, \dots,~$B_n$ are mutually disjoint and
collectively exhaustive, and if $A$ attracts some $B_i$, then $A$ must repel
some $B_j$.

\i\label{exer 4.1.46}
\begin{enumerate}
\item
Suppose that you are looking in your desk for a letter from some time ago.  Your desk has eight
drawers, and you assess the probability that it is in any particular drawer is 10\% (so there is
a 20\% chance that it is not in the desk at all).  Suppose now that you start searching
systematically through your desk, one drawer at a time.  In addition, suppose that you have not found
the letter in the first $i$ drawers, where $0 \le i \le 7$.  Let $p_i$ denote the probability that
the letter will be found in the next drawer, and let $q_i$ denote the probability that the letter
will be found in some subsequent drawer (both $p_i$ and $q_i$ are conditional probabilities, since
they are based upon the assumption that the letter is not in the first $i$ drawers).  Show that
the $p_i$'s increase and the $q_i$'s decrease.  (This problem is from Falk et al.\footnote{R.\
Falk, A.\ Lipson, and C.\ Konold, ``The ups and downs of the hope function in a fruitless search,"
in \emx{Subjective Probability,} G.\ Wright and P.\ Ayton, (eds.) (Chichester: Wiley, 1994), pgs.
353-377.})\index{FALK, R.}\index{LIPSON, A.}\index{KONOLD, C.}
\item
The following data appeared in an article in the Wall Street Journal.\footnote{C. Crossen, ``Fright by the
numbers:  Alarming disease data are frequently flawed," \emx{Wall Street Journal,} 11 April 1996, p.
B1.}\index{Wall Street Journal}\index{CROSSEN, C.} For the ages 20, 30, 40, 50, and 60, the probability of a
woman in the U.S.\ developing cancer in the next ten years is 0.5\%, 1.2\%, 3.2\%, 6.4\%, and 10.8\%,
respectively.  At the same set of ages, the probability of a woman in the U.S.\ eventually developing cancer is
39.6\%, 39.5\%, 39.1\%, 37.5\%, and 34.2\%, respectively.  Do you think that the problem in part (a) gives an
explanation for these data?
\end{enumerate}

\i\label{exer 4.1.47}
Here are two variations of the Monty Hall problem that are discussed by Granberg.\footnote{D.
Granberg, ``To switch or not to switch," in
\emx{The power of logical thinking,} M. vos~Savant, (New York: St.\ Martin's
1996).}\index{GRANBERG, D.}\index{Monty Hall problem}
\begin{enumerate}
\item
Suppose that everything is the same except that Monty forgot to find out in advance which door has the car behind
it.  In the spirit of ``the show must go on," he makes a guess at which of the two doors to open and
gets lucky, opening a door behind which stands a goat.  Now should the contestant switch?
\item
You have observed the show for a long time and found that the car is put behind door A 45\% of the time,
behind door B 40\% of the time and behind door C 15\% of the time. Assume that everything else
about the show is the same.  Again you pick door A. Monty opens a door with a goat and
offers to let you switch. Should you?  Suppose you knew in advance that Monty was going to give you
a chance to switch.  Should you have initially chosen door A?
\end{enumerate}

\end{LJSItem}

%At one point, the following \choice command starting giving errors,
%so I commented it out (I also commented out the ending }).
%\choice{}{\section{Continuous Conditional Probability}\label{sec 4.2}
\section{Continuous Conditional Probability}\label{sec 4.2}

In situations where the sample space is continuous we will follow the same procedure as in the
previous section.  Thus, for example, if $X$ is a continuous random variable with density function
$f(x)$, and if $E$ is an event with positive probability, we define a conditional
density\index{conditional density}\index{density function!conditional} function by the formula
$$ f(x|E) = \left \{ \matrix{
 f(x)/P(E), &  \mbox{if} \,\,x \in E,       \cr
         0, & \mbox{if}\,\,x \not \in E.  \cr}\right.
$$
Then for any event $F$, we have
$$
P(F|E) = \int_F f(x|E)\,dx\ .
$$
The expression $P(F|E)$ is called the conditional probability of $F$ given $E$.  As in the previous
section, it is easy to obtain an alternative expression for this probability:
$$
P(F|E) = \int_F f(x|E)\,dx = \int_{E\cap F} \frac {f(x)}{P(E)}\,dx = \frac {P(E\cap F)}{P(E)}\ .
$$
\par
We can think of the conditional density function as being 0 except on $E$, and
normalized to have integral 1 over $E$.  Note that if the original density is a
uniform density corresponding to an experiment in which all events of equal
size are \emx {equally likely,} then the same will be true for the conditional
density.

\begin{example}\label{exam 4.12}
In the spinner\index{spinner} experiment (cf.\ Example~\ref{exam 2.1.1}),
suppose we know that the spinner has stopped with head in the upper half of the
circle, $0 \leq x \leq 1/2$.  What is the probability that $1/6 \leq x
\leq 1/3$?

Here $E = [0,1/2]$, $F = [1/6,1/3]$, and $F \cap E = F$.  Hence
\begin{eqnarray*}
P(F|E) &=& \frac {P(F \cap E)}{P(E)} \\
       &=& \frac {1/6}{1/2} \\
       &=& \frac 13\ ,
\end{eqnarray*}
which is reasonable, since $F$ is 1/3 the size of $E$.  The conditional density
function here is given by

$$
f(x|E) =  \left \{ \matrix{
               2, & \mbox{if}\,\,\, 0   \leq x < 1/2, \cr
               0, & \mbox{if}\,\,\, 1/2 \leq x < 1.\cr}\right.
$$
Thus the conditional density function is nonzero only on $[0,1/2]$, and is
uniform there.
\end{example}

\begin{example}\label{exam 4.13}
In the dart\index{darts} game (cf.\ Example~\ref{exam 2.2.2}),
suppose we know that the dart lands in the upper half of the target.  What is
the probability that its distance from the center is less than 1/2?

Here $E = \{\,(x,y) : y \geq 0\,\}$, and $F = \{\,(x,y) : x^2 + y^2 <
(1/2)^2\,\}$.  Hence,
\begin{eqnarray*}
P(F|E) & = & \frac {P(F \cap E)}{P(E)} = \frac {(1/\pi)[(1/2)(\pi/4)]}
{(1/\pi)(\pi/2)} \\
       & = & 1/4\ .
\end{eqnarray*}
Here again, the size of $F \cap E$ is 1/4 the size of $E$.  The conditional
density function is
$$
f((x,y)|E) =  \left \{ \matrix{
        f(x,y)/P(E) = 2/\pi, &\mbox{if}\,\,\,(x,y) \in E, \cr
        0,                   &\mbox{if}\,\,\,(x,y) \not \in E.\cr}\right.
$$
\end{example}

\begin{example}\label{exam 4.14}
We return to the exponential density\index{exponential density}\index{density
function!exponential} (cf.\ Example~\ref{exam 2.2.7.5}).  We suppose that we are observing a lump
of plutonium-239.  Our experiment consists of waiting for an emission, then starting a clock, and
recording the length of time $X$ that passes until the next emission.   Experience has shown that
$X$ has an exponential density with some parameter $\lambda$, which depends upon the size of the
lump.  Suppose that when we perform this experiment, we notice that the clock reads $r$ seconds,
and is still running.  What is the probability that there is no emission in a further $s$ seconds?

Let $G(t)$ be the probability that the next particle is emitted after time $t$.  Then
\begin{eqnarray*}
G(t) & = & \int_t^\infty \lambda e^{-\lambda x}\,dx \\
     & = & \left.-e^{-\lambda x}\right|_t^\infty = e^{-\lambda t}\ .
\end{eqnarray*}

Let $E$ be the event ``the next particle is emitted after time $r$" and $F$ the event ``the
next particle is emitted after time $r + s$."  Then
\begin{eqnarray*}
P(F|E) & = & \frac {P(F \cap E)}{P(E)} \\
       & = & \frac {G(r + s)}{G(r)} \\
       & = & \frac {e^{-\lambda(r + s)}}{e^{-\lambda r}} \\
       & = & e^{-\lambda s}\ .
\end{eqnarray*}

This tells us the rather surprising fact that the probability that we have to
wait $s$ seconds more for an emission, given that there has been no emission in $r$ seconds,
is \emx {independent} of the time $r$.  This property (called the \emx {
memoryless}\index{memoryless property} property) was introduced in Example~\ref{exam 2.2.7.5}.
When trying to model various phenomena, this property is helpful in deciding whether the
exponential density is appropriate.
\par
The fact that the exponential density is memoryless means that it is reasonable to assume
if one comes upon a lump of a radioactive isotope at some random time, then the amount of time
until the next emission has an exponential density with the same parameter as the time between
emissions.  A well-known example, known as the ``bus paradox,"\index{bus paradox} replaces the
emissions by buses.  The apparent paradox arises from the following two facts:  1) If you know that,
on the average, the buses come by every 30 minutes, then if you come to the bus stop at a random
time, you should only have to wait, on the average, for 15 minutes for a bus, and 2) Since the buses
arrival times are being modelled by the exponential density, then no matter when you arrive,
you will have to wait, on the average, for 30 minutes for a bus.
\par
The reader can now see that in Exercises~\ref{sec 2.2}.\ref{exer 2.2.8.5},
\ref{sec 2.2}.\ref{exer 2.2.8.6}, and \ref{sec 2.2}.\ref{exer
2.2.11}, we were asking for simulations of conditional probabilities, under various assumptions
on the distribution of the interarrival times.  If one makes a reasonable assumption about this
distribution, such as the one in Exercise~\ref{sec 2.2}.\ref{exer 2.2.8.6}, then the average
waiting time is more nearly one-half the average interarrival time.
\end{example}

\subsection*{Independent Events}

If $E$ and $F$ are two events with positive probability in a continuous sample space, then, as in
the case of discrete sample spaces, we define $E$ and $F$ to be \emx {
independent}\index{events!independent}\index{independence of events} if
$P(E|F) = P(E)$ and $P(F|E) = P(F)$.  As before, each of the above equations imply the
other, so that to see whether two events are independent, only one of these equations must be
checked.  It is also the case that, if $E$ and $F$ are independent, then $P(E \cap F) = P(E)P(F)$.

\begin{example}(Example~\ref{exam 4.12} continued){\label{exam 4.15}}
In the dart\index{darts} game (see Example~\ref{exam 4.12}), let $E$ be the event that the
dart lands in the \emx {upper} half of the target ($y \geq 0$) and $F$ the
event that the dart lands in the \emx {right} half of the target ($x \geq
0$).  Then $P(E \cap F)$ is the probability that the dart lies in the first
quadrant of the target, and
\begin{eqnarray*}
P(E \cap F) & = & \frac 1\pi \int_{E \cap F} 1\,dxdy \\
            & = & \mbox{Area}\,(E\cap F)         \\
            & = & \mbox{Area}\,(E)\,\mbox{Area}\,(F)  \\
            & = & \left(\frac 1\pi \int_E 1\,dxdy\right) \left(\frac 1\pi \int_F
1\,dxdy\right) \\
            & = & P(E)P(F)
\end{eqnarray*}
so that $E$ and $F$ are independent.  What makes this work is that the events
$E$ and $F$ are described by restricting different coordinates.  This idea is made
more precise\linebreak[4] below.
\end{example}

\subsection*{Joint Density and Cumulative Distribution Functions}

In a manner analogous with discrete random variables, we can define joint density
functions and cumulative distribution functions for multi-dimensional continuous random
variables.

\begin{definition}\label{def 4.5}
Let $X_1,~X_2, \ldots,~X_n$ be continuous random variables associated with an experiment, and
let ${\bar X} = (X_1,~X_2, \ldots,~X_n)$.  Then the joint cumulative
distribution\index{cumulative distribution function!joint}\index{joint cumulative
distribution\\ function} function of
${\bar X}$ is defined by
$$F(x_1, x_2, \ldots, x_n) = P(X_1 \le x_1, X_2 \le x_2, \ldots, X_n \le x_n)\ .$$
The joint density function\index{joint density function}\index{density function!joint} of
${\bar X}$ satisfies the following equation:
$$F(x_1, x_2, \ldots, x_n) = \int_{-\infty}^{x_1} \int_{-\infty}^{x_2} \cdots
\int_{-\infty}^{x_n} f(t_1, t_2, \ldots t_n)\,dt_ndt_{n-1}\ldots dt_1\ .$$
\end{definition}

It is straightforward to show that, in the above notation,
\begin{equation}
f(x_1, x_2, \ldots, x_n) = {{\partial^n F(x_1, x_2, \ldots, x_n)}\over
{\partial x_1 \partial x_2 \cdots \partial x_n}}\ .\label{eq 4.4}
\end{equation}

\subsection*{Independent Random Variables}

As with discrete random variables, we can define mutual independence of continuous
random variables.

\begin{definition}\label{def 4.6} Let $X_1$,~$X_2$, \ldots,~$X_n$ be continuous random
variables with cumulative distribution functions $F_1(x),~F_2(x), \ldots,~F_n(x)$.  Then
these random variables are \emx {mutually independent}\index{independence of random
variables!mutual} if
$$F(x_1, x_2, \ldots, x_n) = F_1(x_1)F_2(x_2) \cdots  F_n(x_n)$$
for any choice of $x_1, x_2,
\ldots, x_n$.  Thus, if $X_1,~X_2, \ldots,~X_n$ are mutually independent, then the joint
cumulative distribution function of the random variable ${\bar X} = (X_1, X_2, \ldots,
X_n)$ is just the product of the individual cumulative distribution functions.  When two random
variables are mutually independent, we shall say more briefly that they are \emx {
independent.}\index{independence of random\\ variables}
\end{definition}

Using Equation~\ref{eq 4.4}, the following theorem can easily be shown to hold for mutually
independent continuous random variables.

\begin{theorem}\label{thm 4.2}
Let $X_1$,~$X_2$, \ldots,~$X_n$ be continuous random variables with density
functions $f_1(x),~f_2(x), \ldots,~f_n(x)$.  Then these random variables are \emx { mutually
independent} if and only if
$$f(x_1, x_2, \ldots, x_n) = f_1(x_1)f_2(x_2) \cdots  f_n(x_n)$$    for any choice of $x_1,
x_2, \ldots, x_n$.
\end{theorem}

Let's look at some examples.

\begin{example}
In this example, we define three random variables, $X_1,\ X_2$, and $X_3$.  We will show that
$X_1$ and $X_2$ are independent, and that $X_1$ and $X_3$ are not independent.  Choose a point
$\omega = (\omega_1,\omega_2)$ at random from the unit square.  Set $X_1 =
\omega_1^2$, $X_2 = \omega_2^2$, and $X_3 =
\omega_1 + \omega_2$.  Find the joint distributions $F_{12}(r_1,r_2)$ and $F_{23}(r_2,r_3)$.

We have already seen (see Example~\ref{exam 2.2.7.1}) that
\begin{eqnarray*}
 F_1(r_1) & = & P(-\infty < X_1 \leq r_1) \\
         & = & \sqrt{r_1}, \qquad \mbox{if} \,\,0 \leq r_1 \leq 1\ ,
\end{eqnarray*}
and similarly,
$$ F_2(r_2) = \sqrt{r_2}\ ,$$
if $0 \leq r_2 \leq 1$.
Now we have (see Figure~\ref{fig 5.15})
\putfig{3truein}{PSfig5-15}{$X_1$ and $X_2$ are independent.}{fig 5.15}
\begin{eqnarray*}
 F_{12}(r_1,r_2) & = & P(X_1 \leq r_1 \,\, \mbox{and}\,\, X_2 \leq r_2) \\
                & = & P(\omega_1 \leq \sqrt{r_1} \,\,\mbox{and}\,\, \omega_2 \leq
\sqrt{r_2}) \\
                & = & \mbox{Area}\,(E_1)\\
                & = & \sqrt{r_1} \sqrt{r_2} \\
                & = &F_1(r_1)F_2(r_2)\ .
\end{eqnarray*}
In this case $F_{12}(r_1,r_2) = F_1(r_1)F_2(r_2)$ so that $X_1$ and $X_2$ are
independent.  On the other hand, if $r_1 = 1/4$ and $r_3 = 1$, then (see Figure~\ref{fig
5.16})
\putfig{3truein}{PSfig5-16}{$X_1$ and $X_3$ are not independent.}{fig 5.16}
\begin{eqnarray*}
F_{13}(1/4,1) & = & P(X_1 \leq 1/4,\ X_3 \leq 1) \\
              & = & P(\omega_1 \leq 1/2,\ \omega_1 + \omega_2 \leq 1) \\
              & = & \mbox{Area}\,(E_2) \\
              & = & \frac 12 - \frac 18 = \frac 38\ .
\end{eqnarray*}
Now recalling that
$$F_3(r_3) = \left \{ \matrix{
                  0, & \mbox{if} \,\,r_3 < 0, \cr
         (1/2)r_3^2, & \mbox{if} \,\,0 \leq r_3 \leq 1, \cr
   1-(1/2)(2-r_3)^2, & \mbox{if} \,\,1 \leq r_3 \leq 2, \cr
                  1, & \mbox{if} \,\,2 < r_3,\cr}\right.
$$ (see Example~\ref{exam 2.2.7.2}), we have $F_1(1/4)F_3(1) = (1/2)(1/2) = 1/4$.  Hence, $X_1$
and $X_3$ are not independent random variables.  A similar calculation shows that $X_2$ and
$X_3$ are not independent either.
\end{example}

Although we shall not prove it here, the following theorem is a useful one.  The
statement also holds for mutually independent discrete random variables.  A proof may be
found in R\'enyi.\index{R\'ENYI, A.}\footnote{A. R\'enyi, \emx {Probability Theory} (Budapest:
Akad\'emiai Kiad\'o, 1970), p.~183.}

\begin{theorem}\label{thm 4.3}
Let $X_1, X_2, \ldots, X_n$ be mutually independent continuous random variables and let
$\phi_1(x), \phi_2(x), \ldots, \phi_n(x)$ be continuous functions.  Then $\phi_1(X_1),$
\newline
$\phi_2(X_2), \ldots, \phi_n(X_n)$ are mutually independent.
\end{theorem}

\subsection*{Independent Trials}

Using the notion of independence, we can now formulate for continuous sample spaces the notion
of independent trials (see Definition~\ref{def 5.5}).

\begin{definition}\label{def 5.12} A sequence $X_1$,~$X_2$, \dots,~$X_n$ of random variables
$X_i$ that are mutually independent and have the same density is called an \emx {independent
trials\linebreak[4] process.}\index{independent trials process}
\end{definition}
\par
As in the case of discrete random variables, these independent trials processes arise
naturally in situations where an experiment described by a single random variable is repeated
$n$ times.

\subsection*{Beta Density}

We consider next an example which involves a sample space with both discrete
and continuous coordinates.  For this example we shall need a new density
function called the \emx {beta density.}\index{density function!beta}\index{beta density}  This
density has two parameters
$\alpha$,~$\beta$ and is defined by
$$
B(\alpha,\beta,x) =  \left \{ \matrix{
(1/B(\alpha,\beta))x^{\alpha - 1}(1 - x)^{\beta - 1}, & {\mbox{if}}\,\, 0 \leq x \leq 1, \cr
                                                   0, & {\mbox{otherwise}}.\cr}\right.
$$
Here $\alpha$ and $\beta$ are any positive numbers, and the beta function
$B(\alpha,\beta)$ is given by the area under the graph of $x^{\alpha - 1}(1 -
x)^{\beta - 1}$ between 0 and 1:
$$
B(\alpha,\beta) = \int_0^1 x^{\alpha - 1}(1 - x)^{\beta - 1}\,dx\ .
$$
Note that when $\alpha = \beta = 1$ the beta density if the uniform density.
When $\alpha$ and $\beta$ are greater than 1 the density is bell-shaped, but
when they are less than 1 it is U-shaped as suggested by the examples in
Figure~\ref{fig 4.6}.

\putfig{4truein}{PSfig4-6}{Beta density for $\alpha = \beta = .5,1,2.$}{fig 4.6}

We shall need the values of the beta function only for integer values of
$\alpha$ and $\beta$, and in this case
$$
B(\alpha,\beta) = \frac{(\alpha - 1)!\,(\beta - 1)!}{(\alpha + \beta - 1)!}\ .
$$

\begin{example}\label{exam 4.16}
In medical problems it is often assumed that a drug is effective with a
probability~$x$ each time it is used and the various trials are
independent, so that one is, in effect, tossing a biased coin with probability
$x$ for heads.
Before further experimentation, you do not know the value $x$ but past experience might give
some information about its possible values. It is natural to represent this information
by sketching a density function
to determine a distribution for $x$.  Thus, we are considering $x$ to be a continuous random
variable, which takes on values between 0 and 1.  If you have no knowledge at all, you would
sketch the uniform density.  If past experience suggests that
$x$ is very likely to be near 2/3 you would sketch a density with maximum at 2/3 and a
spread reflecting your uncertainly in the estimate of 2/3. You would then want to find a
density function that reasonably fits your sketch.  The beta densities provide a class of
densities that can be fit to most sketches you might make. For example, for
$\alpha > 1$ and $\beta > 1$ it is bell-shaped with the parameters $\alpha$ and $\beta$
determining its peak and its spread.
\par
Assume that the experimenter has chosen a beta density to describe the state of
his knowledge about $x$ before the experiment.  Then he gives the drug to $n$
subjects and records the number $i$ of successes.   The number $i$ is a discrete random
variable, so we may conveniently describe the set of possible outcomes of this experiment
by referring to the ordered pair $(x, i)$.
\par
We let $m(i|x)$ denote the probability that we observe $i$ successes given the value of $x$.  By our
assumptions, $m(i|x)$ is the binomial distribution with probability~$x$ for success:
$$
m(i|x) = b(n,x,i) = {n \choose i} x^i(1 - x)^j\ ,
$$
where $j = n - i$.
\par
If $x$ is chosen at random from $[0,1]$ with a beta
density $B(\alpha,\beta,x)$, then the density function for the outcome of the
pair $(x,i)$ is
\begin{eqnarray*}
f(x,i) & = & m(i|x)B(\alpha,\beta,x) \\
       & = & {n \choose i} x^i(1 - x)^j \frac 1{B(\alpha,\beta)} x^{\alpha - 1}(1 -
x)^{\beta - 1} \\
       & = & {n \choose i}  \frac 1{B(\alpha,\beta)} x^{\alpha + i - 1}(1 - x)^{\beta +
j - 1}\ .
\end{eqnarray*}
Now let $m(i)$ be the probability that we observe $i$ successes \emx {not}
knowing the value of $x$.  Then
\begin{eqnarray*}
m(i) & = & \int_0^1 m(i|x) B(\alpha,\beta,x)\,dx \\
     & = & {n \choose i} \frac 1{B(\alpha,\beta)} \int_0^1 x^{\alpha + i - 1}(1 -
x)^{\beta + j - 1}\,dx \\
     & = & {n \choose i} \frac {B(\alpha + i,\beta + j)}{B(\alpha,\beta)}\ .
\end{eqnarray*}
Hence, the probability density $f(x|i)$ for $x$, given that $i$ successes were
observed, is
$$
f(x|i)  =  \frac {f(x,i)}{m(i)}
$$
\begin{equation}
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =
\frac {x^{\alpha + i - 1}(1 - x)^{\beta + j - 1}}{B(\alpha + i,\beta
+ j)}\ ,\label{eq 4.5}
\end{equation}
that is, $f(x|i)$ is another beta density.  This says that if we observe $i$
successes and $j$ failures in $n$ subjects, then the new density for the
probability that the drug is effective is again a beta density but with
parameters $\alpha + i$, $\beta + j$.
\par
Now we assume that before the experiment we choose a beta density with
parameters $\alpha$ and $\beta$, and that in the experiment
we obtain $i$ successes in $n$ trials.  We have just seen that in this case, the new density for $x$
is a beta density with parameters $\alpha + i$ and $\beta + j$.
\par
Now we wish to calculate the probability that the drug is effective on the next subject.  For any
particular real number $t$ between 0 and 1, the probability that $x$ has the value $t$ is given by
the expression in Equation~\ref{eq 4.5}.  Given that $x$ has the value $t$, the probability that the drug
is effective on the next subject is just $t$.  Thus, to obtain the probability that the drug is effective
on the next subject, we integrate the product of the expression in Equation~\ref{eq 4.5} and $t$ over all
possible values of $t$.  We obtain:
\begin{eqnarray*}
\lefteqn{\frac 1{B(\alpha + i,\beta + j)}
\int_0^1 t\cdot t^{\alpha + i - 1}(1 - t)^{\beta + j - 1}\,dt}
\\
  & = & \frac {B(\alpha + i + 1,\beta + j)}{B(\alpha + i,\beta + j)} \\
  & = & \frac {(\alpha + i)!\,(\beta + j - 1)!}{(\alpha + \beta + i + j)!} \cdot
\frac {(\alpha + \beta + i + j - 1)!}{(\alpha + i - 1)!\,(\beta + j - 1)!} \\
  & = & \frac {\alpha + i}{\alpha + \beta + n}\ .
\end{eqnarray*}
If $n$ is large, then our estimate for the probability of success after the experiment is
approximately the proportion of successes observed in the experiment, which is
certainly a reasonable conclusion.
\end{example}

The next example is another in which the true probabilities are unknown and must be
estimated based upon experimental data.

\begin{example}(Two-armed bandit problem)\label{exam 4.17}\index{two-armed bandit}
You are in a casino and confronted by two slot machines.  Each machine pays off
either 1 dollar or nothing.  The probability that the first machine pays off a
dollar is $x$ and that the second machine pays off a dollar is $y$.  We assume
that $x$ and $y$ are random numbers chosen independently from the interval
$[0,1]$ and unknown to you.  You are permitted to make a series of ten plays,
each time choosing one machine or the other.  How should you choose to maximize
the number of times that you win?

One strategy that sounds reasonable is to calculate, at every stage, the
probability that each machine will pay off and choose the machine with the
higher probability.  Let win($i$), for~$i = 1$ or~2, be the number of times
that you have won on the $i$th machine.  Similarly, let lose($i$) be the number
of times you have lost on the $i$th machine.  Then, from Example~\ref{exam
4.16}, the probability $p(i)$ that you win if you choose the $i$th machine is
$$
p(i) = \frac {{\mbox{win}}(i) + 1} {{\mbox{win}}(i) + {\mbox{lose}}(i) + 2}\ .
$$
Thus, if $p(1) > p(2)$ you would play machine 1 and otherwise you would play
machine 2.  We have written a program {\bf TwoArm}\index{TwoArm (program)} to simulate this
experiment.  In the program, the user specifies the initial values for $x$ and
$y$ (but these are unknown to the experimenter).  The program calculates at
each stage the two conditional densities for $x$ and $y$, given the outcomes of
the previous trials, and then computes $p(i)$, for~$i = 1$,~2.  It then chooses
the machine with the highest value for the probability of winning for the next
play.  The program prints the machine chosen on each play and the outcome of
this play.  It also plots the new densities for~$x$ (solid line) and $y$
(dotted line), showing only the current densities.  We have run the program for
ten plays for the case $x = .6$ and $y = .7$.  The result is shown in
Figure~\ref{fig 4.7}.

\putfig{4.5truein}{PSfig4-7}{Play the best machine.}{fig 4.7}

The run of the program shows the weakness of this strategy.  Our initial
probability for winning on the better of the two machines is .7.  We start with
the poorer machine and our outcomes are such that we always have a probability
greater than .6 of winning and so we just keep playing this machine even though
the other machine is better.  If we had lost on the first play we would have
switched machines.  Our final density for $y$ is the same as our initial
density, namely, the uniform density.  Our final density for $x$ is different
and reflects a much more accurate knowledge about $x$.  The computer did pretty
well with this strategy, winning seven out of the ten trials, but ten trials
are not enough to judge whether this is a good strategy in the long run.

\putfig{4.5truein}{PSfig4-8}{Play the winner.}{fig 4.8}

Another popular strategy  is the \emx {play-the-winner strategy.}  As the name
suggests, for this strategy we choose the same machine when we win and switch
machines when we lose.  The program {\bf TwoArm} will simulate this
strategy as well.  In Figure~\ref{fig 4.8}, we show the results of running this program
with the play-the-winner strategy and the same true probabilities of .6 and .7
for the two machines.  After ten plays our densities for the unknown
probabilities of winning suggest to us that the second machine is indeed the
better of the two.  We again won seven out of the ten trials.

Neither of the strategies that we simulated is the best one in terms of
maximizing our average winnings.  This best strategy is very complicated but is
reasonably approximated by the play-the-winner strategy.  Variations on this
example have played an important role in the problem of clinical tests of drugs
where experimenters face a similar situation.
\end{example}

\exercises
\begin{LJSItem}

\i\label{exer 4.2.1} Pick a point $x$ at random (with uniform density) in the interval
$[0,1]$.  Find the probability that $x > 1/2$, given that
\begin{enumerate}
\item $x > 1/4$.

\item $x < 3/4$.

\item $|x - 1/2| < 1/4$.

\item $x^2 - x + 2/9 < 0$.
\end{enumerate}

\i\label{exer 4.2.2} A radioactive material emits $\alpha$-particles at a rate described by
the density function
$$
f(t) = .1e^{-.1t}\ .
$$
Find the probability
that a particle is emitted in the first 10~seconds, given that
\begin{enumerate}
\item no particle is emitted in the first second.

\item no particle is emitted in the first 5 seconds.

\item a particle is emitted in the first 3 seconds.

\item a particle is emitted in the first 20 seconds.
\end{enumerate}

\i\label{exer 4.2.3} The Acme Super light bulb\index{light bulb} is known to have a useful life
described by  the density function
$$
f(t) = .01e^{-.01t}\ ,
$$
where time $t$ is measured in hours.
\begin{enumerate}
\item Find the \emx {failure rate} of this bulb
(see Exercise~\ref{sec 2.2}.\ref{exer 2.2.6}).

\item Find the \emx {reliability} of this bulb after 20 hours.

\item Given that it lasts 20 hours, find the probability that the bulb lasts another 20
hours.

\item Find the probability that the bulb burns out in the forty-first hour, given
that it lasts 40 hours.
\end{enumerate}

\i\label{exer 4.2.4} Suppose you toss a dart at a circular target of radius 10 inches.  Given
that the dart lands in the upper half of the target, find the probability that
\begin{enumerate}
\item it lands in the right half of the target.

\item its distance from the center is less than 5 inches.

\item its distance from the center is greater than 5 inches.

\item it lands within 5 inches of the point $(0,5)$.
\end{enumerate}

\i\label{exer 4.2.5} Suppose you choose two numbers $x$ and $y$, independently at random from
the interval $[0,1]$.  Given that their sum lies in the interval $[0,1]$,
find the probability that
\begin{enumerate}
\item $|x - y| < 1$.

\item $xy < 1/2$.

\item $\max\{x,y\} < 1/2$.

\item $x^2 + y^2 < 1/4$.

\item $x > y$.
\end{enumerate}

\i\label{exer 4.2.6} Find the conditional density functions for the following experiments.
\begin{enumerate}
\item A number $x$ is chosen at random in the interval $[0,1]$, given that
$x > 1/4$.

\item A number $t$ is chosen at random in the interval $[0,\infty)$ with
exponential density $e^{-t}$, given that $1 < t < 10$.

\item A dart is thrown at a circular target of radius 10 inches, given that
it falls in the upper half of the target.

\item Two numbers $x$ and $y$ are chosen at random in the interval
$[0,1]$, given that $x > y$.
\end{enumerate}

\i\label{exer 4.2.7} Let $x$ and $y$ be chosen at random from the interval $[0,1]$.  Show
that the events $x > 1/3$ and $y > 2/3$ are independent events.

\i\label{exer 4.2.8} Let $x$ and $y$ be chosen at random from the interval $[0,1]$.  Which
pairs of the following events are independent?
\begin{enumerate}
\item $x > 1/3$.

\item $y > 2/3$.

\item $x > y$.

\item $x + y < 1$.
\end{enumerate}

\i\label{exer 4.2.8.5}  Suppose that $X$ and $Y$ are continuous random variables with
density functions $f_X(x)$ and $f_Y(y)$, respectively.  Let $f(x, y)$ denote the joint
density function of $(X, Y)$.  Show that
$$
\int_{-\infty}^\infty f(x, y)\, dy = f_X(x)\ ,
$$ and
$$
\int_{-\infty}^\infty f(x, y)\, dx = f_Y(y)\ .
$$

\istar\label{exer 4.2.9} In Exercise~\ref{sec 2.2}.\ref{exer 2.2.13} you
proved the following: If you take a stick of unit length and break it into three pieces,
choosing the breaks at random (i.e., choosing two real numbers independently
and uniformly from [0, 1]), then the probability that the three pieces form
a triangle is 1/4.  Consider now a similar experiment: First break the stick at
random, then break the longer piece at random.  Show that the two experiments
are actually quite different, as follows:
\begin{enumerate}
\item Write a program which simulates both cases for a run of 1000 trials,
prints out the proportion of successes for each run, and repeats this process
ten times.  (Call a trial a success if the three pieces do form a triangle.)
Have your program pick $(x,y)$ at random in the unit square, and in each case
use $x$ and $y$ to find the two breaks.  For each experiment, have it plot
$(x,y)$ if $(x,y)$ gives a success.

\item Show that in the second experiment the theoretical probability of
success is actually $2\log 2 - 1$.
\end{enumerate}

\i\label{exer 4.2.10} A coin has an unknown bias $p$ that is assumed to be uniformly
distributed between 0 and 1.  The coin is tossed $n$ times and heads turns up
$j$ times and tails turns up $k$ times.  We have seen that the probability that
heads turns up next time is
$$
\frac {j + 1}{n + 2}\ .
$$
Show that this is the same as the probability that the next ball is black for
the Polya urn model\index{Polya urn model} of Exercise~\ref{sec 4.1}.\ref{exer 4.1.18}.
Use this result to explain why, in the Polya urn model, the proportion of black
balls does not tend to 0~or~1 as one might expect but rather to a uniform
distribution on the interval $[0,1]$.

\i\label{exer 4.2.11} Previous experience with a drug suggests that the probability~$p$ that
the drug is effective is a random quantity having a beta density with
parameters $\alpha = 2$ and $\beta = 3$.  The drug is used on ten subjects and
found to be successful in four out of the ten patients.  What density should we
now assign to the probability~$p$?  What is the probability that the drug will
be successful the next time it is used?

\i\label{exer 4.2.12} Write a program to allow you to compare the strategies play-the-winner
and play-the-best-machine for the two-armed bandit problem of Example~\ref{exam
4.17}.  Have your program determine the initial payoff probabilities for
each machine by choosing a pair of random numbers between 0 and 1.  Have your
program carry out 20 plays and keep track of the number of wins for each of the
two strategies.  Finally, have your program make 1000 repetitions of the 20
plays and compute the average winning per 20 plays.  Which strategy seems to be
the best?  Repeat these simulations with 20 replaced by 100.  Does your answer to the
above question change?

\i\label{exer 4.2.13} Consider the two-armed bandit problem of Example~\ref{exam 4.17}.
Bruce Barnes\index{BARNES, B.} proposed the following strategy, which is a variation on the
play-the-best-machine strategy.  The machine with the greatest probability of
winning is played \emx {unless} the following two conditions hold: (a)~the
difference in the probabilities for winning is less than .08, and (b)~the ratio
of the number of times played on the more often played machine to the number of
times played on the less often played machine is greater than 1.4.  If the
above two conditions hold, then the machine with the smaller probability of
winning is played.  Write a program to simulate this strategy.  Have your
program choose the initial payoff probabilities at random from the unit
interval $[0,1]$, make 20 plays, and keep track of the number of wins.
Repeat this experiment 1000 times and obtain the average number of wins per 20
plays.  Implement a second strategy---for example, play-the-best-machine or one
of your own choice, and see how this second strategy compares with Bruce's on
average wins.

%\end{LJSItem}}
\end{LJSItem}

\section{Paradoxes}\label{sec 4.3}

Much of this section is based on an article by Snell and Vanderbei.\index{SNELL, J.
L.}\index{VANDERBEI, R.}\footnote{J. L. Snell and R. Vanderbei, ``Three Bewitching
Paradoxes," in \emx {Topics in Contemporary Probability and Its Applications}, CRC Press, Boca
Raton, 1995.}
\par
One must be very careful in dealing with problems involving conditional probability.
The reader will recall that in the Monty Hall problem (Example~\ref{exam 4.5}), if the contestant
chooses the door with the car behind it, then Monty has a choice of doors to open.  We made an
assumption that in this case, he will choose each door with probability 1/2.  We then noted
that if this assumption is changed, the answer to the original question changes.   In this
section, we will study other examples of the same phenomenon.

\begin{example}\label{exam 4.3.1}
Consider a family with two children.  Given that one of the children is a boy,
what is the probability that both children are boys?
\par
One way to approach this problem is to say that the other child is equally likely to
be a boy or a girl, so the probability that both children are boys is 1/2.  The ``text-book"
solution would be to draw the tree diagram and then form the
conditional tree by deleting paths to leave only those paths that are consistent
with the given information.  The result is shown in Figure~\ref{fig 4.3}. We see that the
probability of two boys given a boy in the family is not 1/2 but rather 1/3.
\putfig{3.5truein}{PSfig4-3}{Tree for Example \protect\ref{exam 4.3.1}\protect.}{fig 4.3}
\end{example}

This problem and others like it are discussed in Bar-Hillel and
Falk.\index{BAR-HILLEL, M.}\index{FALK, R.}\footnote{M. Bar-Hillel and R. Falk, ``Some teasers
concerning conditional probabilities," \emx {Cognition}, vol. 11 (1982), pgs. 109-122.} These
authors stress that the answer to conditional probabilities of this  kind can change depending
upon how the information given was actually obtained.   For example, they show that 1/2 is the
correct answer for the following scenario.

\begin{example}\label{exam 4.3.2}
Mr.\ Smith is the father of two.  We meet him walking along the street with a young boy whom
he proudly introduces as his son.  What is the probability that Mr.\ Smith's other child is also
a boy?
\par
As usual we have to make some additional assumptions.  For example, we will
assume that if Mr.\ Smith has a boy and a girl, he is equally likely to choose
either one to accompany him on his walk. In Figure~\ref{fig 4.3.5} we show the tree analysis of this
problem and we see that 1/2 is, indeed, the correct answer.
\putfig{5truein}{PSfig4-3-5}{Tree for Example \protect\ref{exam 4.3.2}\protect.}{fig 4.3.5}
\end{example}

\begin{example}\label{exam 4.3.3}
It is not so easy to think of reasonable scenarios that would lead to the classical 1/3 answer.
An attempt was made by Stephen Geller\index{GELLER, S.} in proposing this problem to Marilyn vos
Savant.\index{vos SAVANT, M.}\footnote{M. vos Savant, ``Ask Marilyn," \emx {Parade
Magazine}, 9  September;  2 December;  17  February 1990, reprinted in Marilyn
vos Savant, \emx {Ask Marilyn}, St. Martins, New York, 1992.}
Geller's problem is as follows:  A shopkeeper says she has two new baby beagles to show you, but she
doesn't know whether they're both male, both female, or one of each sex.  You tell her that
you want only a male, and she telephones the fellow who's giving them a bath.  ``Is at least
one a male?" she asks. ``Yes," she informs you with a smile.  What is the probability that
the \emx {other} one is male?
\par
The reader is asked to decide whether the model which gives an answer of 1/3 is a reasonable one
to use in this case.
\end{example}

In the preceding examples, the apparent paradoxes could easily be resolved by clearly stating
the model that is being used and the assumptions that are being made.  We now turn to some
examples in which the paradoxes are not so easily resolved.

\begin{example}\label{exam 4.3.4}
Two envelopes each contain a certain amount of money.  One envelope is given to Ali\index{Ali}
and the other to Baba\index{Baba} and they are told that one envelope contains twice as much money
as the other.  However, neither knows who has the larger prize.  Before anyone has
opened their envelope, Ali is asked if she would like to trade her envelope with Baba.
She reasons as follows:  Assume that the amount in my envelope is $x$.  If I switch,
I will end up with $x/2$ with probability 1/2, and $2x$ with probability 1/2.  If I were
given the opportunity to play this game many times, and if I were to switch each time,
I would, on average, get
$$
\frac 12 \frac x2 + \frac 12 2x = \frac 54 x\ .
$$
This is greater than my average winnings if I didn't switch.
\par
Of course, Baba is presented with the same opportunity and reasons in the same way
to conclude that he too would like to switch.  So they switch and each thinks that
his/her net worth just went up by 25\%.
\par
Since neither has yet opened any envelope, this process can be
repeated and so again they switch.  Now they are back with their original
envelopes and yet they think that their fortune has increased 25\% twice.
By this reasoning, they could convince themselves that by repeatedly switching the
envelopes, they could become arbitrarily wealthy.  Clearly, something is wrong with
the above reasoning, but where is the mistake?
\par
One of the tricks of making paradoxes is to make them slightly more difficult
than is necessary to further befuddle us.  As John Finn\index{FINN, J.} has suggested, in this
paradox  we could just have well started with a simpler problem.  Suppose Ali and Baba know that
I am going to give then either an envelope with \$5 or one with \$10
and I am going to toss a coin to decide which to give to Ali, and then give the other to
Baba.  Then Ali can argue that Baba has $2x$ with probability $1/2$ and $x/2$ with probability
$1/2$.  This leads Ali to the same conclusion as before.   But now it is clear that this is
nonsense, since if Ali has the envelope containing \$5, Baba cannot possibly have half of this,
namely \$2.50, since that was not even one of the choices.  Similarly, if Ali has \$10,
Baba cannot have twice as much, namely \$20. In fact, in this simpler problem the possibly
outcomes are given by the tree diagram in Figure~\ref{fig 4.9}.
\putfig{3truein}{PSfig4-9}{John Finn's version of Example \protect\ref{exam 4.3.4}\protect.}{fig 4.9}%%4.5truein
From the diagram, it is clear that neither is made better off by switching.
\end{example}

In the above example, Ali's reasoning is incorrect because he infers that if the amount
in his envelope is $x$, then the probability that his envelope contains the smaller amount
is 1/2, and the probability that her envelope contains the larger amount is also 1/2.  In
fact, these conditional probabilities depend upon the distribution of the amounts that
are placed in the envelopes.
\par
For definiteness, let $X$ denote the positive integer-valued random variable which represents
the smaller of the two amounts in the envelopes.  Suppose, in addition, that we are given
the distribution of $X$, i.e., for each positive integer $x$, we are given the value of
$$p_x = P(X = x)\ .$$
(In Finn's example, $p_5 = 1$, and $p_{n} = 0$ for all other values of $n$.)
Then it is easy to calculate the conditional probability that an envelope contains the
smaller amount, given that it contains $x$ dollars.  The two possible sample points are
$(x, x/2)$ and $(x, 2x)$.  If $x$ is odd, then the first sample point has probability 0,
since $x/2$ is not an integer, so the desired conditional probability is 1 that $x$ is
the smaller amount.  If $x$ is even, then the two sample points have probabilities $p_{x/2}$
and $p_x$, respectively, so the conditional probability that $x$ is the smaller amount is
$$\frac{p_x}{p_{x/2} + p_x}\ ,$$
which is not necessarily equal to 1/2.
\par
Steven Brams\index{BRAMS, S.} and D. Marc Kilgour\index{KILGOUR, D. M.}\footnote{S. J. Brams
and D. M. Kilgour, ``The Box Problem:  To Switch or Not to Switch," \emx {Mathematics Magazine},
vol. 68, no. 1 (1995),  p. 29.} study the problem, for different distributions, of whether or
not one should  switch envelopes, if one's objective is to maximize the long-term average
winnings.  Let
$x$ be the amount in your envelope.  They show that for any distribution of $X$, there
is at least one value of $x$ such that you should switch.  They give an example of a
distribution for which there is exactly one value of $x$ such that you should switch (see
Exercise~\ref{exer 4.3.5}).  Perhaps the most interesting case is a distribution in which
you should always switch.  We now give this example.

\begin{example}\label{exam 4.3.5}
Suppose that we have two envelopes\index{envelopes} in front of us, and that one envelope contains
twice the amount of money as the other (both amounts are positive integers).  We are given
one of the envelopes, and asked if we would like to switch.
\par
As above, we let $X$ denote the smaller of the two amounts in the envelopes, and let
$$p_x = P(X = x)\ .$$
We are now in a position where we can calculate the long-term average winnings, if we switch.
(This long-term average is an example of a probabilistic concept known as expectation, and
will be discussed in Chapter~\ref{chp 6}.)  Given that one of the two sample points has
occurred, the probability that it is the point $(x, x/2)$ is
$$\frac{p_{x/2}}{p_{x/2} + p_x}\ ,$$
and the probability that it is the point $(x, 2x)$ is
$$\frac{p_x}{p_{x/2} + p_x}\ .$$
Thus, if we switch, our long-term average winnings are
$$\frac{p_{x/2}}{p_{x/2} + p_x}\frac x2 + \frac{p_x}{p_{x/2} + p_x} 2x\ .$$
If this is greater than $x$, then it pays in the long run for us to switch.
Some routine algebra shows that the above expression is greater than $x$ if and only
if
\begin{equation}
\frac{p_{x/2}}{p_{x/2} + p_x} < \frac 23\ .
\label{eq 4.3.1}
\end{equation}
\par
It is interesting to consider whether there is a distribution on the positive integers
such that the inequality~\ref{eq 4.3.1} is true for all even values of $x$.  Brams and
Kilgour\footnote{ibid.} give the following example.
\par
We define $p_x$ as follows:
$$
p_x = \left \{ \matrix{
               \frac 13 \Bigl(\frac 23\Bigr)^{k-1}, & \mbox{if}\,\, x = 2^k, \cr
               0, & \mbox{otherwise.}\cr
}\right.
$$
It is easy to calculate (see Exercise~\ref{exer 4.3.4}) that for all relevant values of $x$,
we have
$$
\frac{p_{x/2}}{p_{x/2} + p_x} = \frac 35\ ,
$$
which means that the inequality~\ref{eq 4.3.1} is always true.
\end{example}

So far, we have been able to resolve paradoxes by clearly stating the assumptions being made
and by precisely stating the models being used.  We end this section by describing a paradox
which we cannot resolve.

\begin{example}\label{exam 4.3.6}
Suppose that we have two envelopes\index{envelopes} in front of us, and we are told that the
envelopes contain
$X$ and $Y$ dollars, respectively, where $X$ and $Y$ are different positive integers.  We randomly
choose one of the envelopes, and we open it, revealing $X$, say.  Is it possible to determine, with
probability greater than 1/2, whether $X$ is the smaller of the two dollar amounts?
\par
Even if we have no knowledge of the joint distribution of $X$ and $Y$, the surprising answer
is yes!  Here's how to do it.  Toss a fair coin until the first time that heads turns up.  Let $Z$
denote the number of tosses required plus 1/2.  If $Z > X$, then we say that $X$ is the smaller of the
two amounts, and if $Z < X$, then we say that $X$ is the larger of the two amounts.
\par
First, if $Z$ lies between $X$ and $Y$, then we are sure to be correct.  Since $X$ and $Y$ are
unequal, $Z$ lies between them with positive probability.  Second, if $Z$ is not between $X$ and $Y$,
then $Z$ is either greater than both $X$ and $Y$, or is less than both $X$ and $Y$.  In either
case, $X$ is the smaller of the two amounts with probability 1/2, by symmetry considerations (remember,
we chose the envelope at random).  Thus, the probability that we are correct is greater than 1/2.
\end{example}

\exercises
\begin{LJSItem}

\i\label{exer 4.3.1}
One of the first conditional probability paradoxes was
provided by Bertrand.\index{BERTRAND, J.}\footnote{J. Bertrand, \emx {Calcul des Probabilit\'{e}s},
Gauthier-Uillars, 1888.}  It is called the \emx {Box Paradox}\index{Box paradox}.  A cabinet has
three drawers. In the first drawer there  are two gold balls, in the second drawer  there are two
silver balls, and in the third drawer there  is one silver and one gold ball. A drawer is picked
at random and a ball chosen at random from the  two balls in the drawer.  Given that a  gold ball
was drawn, what is the probability that the drawer with the two gold balls was chosen?

\i\label{exer 4.3.2}
The following problem is called the \emx {two aces problem}\index{Two aces problem}. This problem,
dating back to 1936, has been attributed to the English mathematician J. H. C.
Whitehead\index{WHITEHEAD, J. H. C.} (see Gridgeman\index{GRIDGEMAN, N. T.}\footnote{N. T.
Gridgeman, Letter, \emx {American Statistician}, 21 (1967), pgs. 38-39.}). This problem was also
submitted to Marilyn vos Savant\index{vos SAVANT, M.} by the master of mathematical puzzles Martin
Gardner\index{GARDNER, M.}, who remarks that it is one of his favorites.
\par
A bridge\index{bridge} hand has been dealt, i.\ e. thirteen cards are dealt to each player.  Given that
your partner has at least one ace, what is the probability that he has at least two aces?  Given that
your partner has the ace of hearts, what is the probability that he has at least two aces?  Answer these
questions for a version of bridge in which there are eight cards, namely four aces and four kings, and
each player is dealt two cards.  (The reader may wish to solve the problem with a 52-card deck.)

\i\label{exer 4.3.3}
In the preceding exercise, it is natural to ask ``How do we get the information that the given hand
has an ace?" Gridgeman considers two different ways that we might get this information.  (Again, assume
the deck consists of eight cards.)
\begin{enumerate}
\item
Assume that the person holding the hand is asked to ``Name an ace in your hand" and answers ``The ace
of hearts."  What is the probability that he has a second ace?
\item
Suppose the person holding the hand is asked the more direct question ``Do you have the ace of
hearts?" and the answer is yes.  What is the probability that he has a second ace?
\end{enumerate}

\i\label{exer 4.3.4}  Using the notation introduced in Example~\ref{exam 4.3.5}, show that in the
example of Brams\index{BRAMS, S.} and Kilgour\index{KILGOUR, D. M.}, if $x$ is a positive power of
2, then
$$
\frac{p_{x/2}}{p_{x/2} + p_x} = \frac 35\ .
$$

\i\label{exer 4.3.5} Using the notation introduced in Example~\ref{exam 4.3.5}, let
$$
p_x = \left \{ \matrix{
               \frac 23 \Bigl(\frac 13\Bigr)^k, & \mbox{if}\,\, x = 2^k, \cr
               0, & \mbox{otherwise.}\cr
}\right.
$$
Show that there is exactly one value of $x$ such that if your envelope contains $x$,
then you should switch.

\istar\label{exer 4.3.6} (For bridge\index{bridge} players only.  From
Sutherland.\index{SUTHERLAND, E.}\footnote{E. Sutherland, ``Restricted Choice\index{restricted
choice, principle of} --- Fact or Fiction?", \emx {Canadian Master Point}, November 1, 1993.})
Suppose that we are the declarer in a hand of bridge, and we have the king, 9, 8, 7, and 2  of a
certain suit, while the dummy has the ace, 10, 5, and 4 of the same suit.  Suppose that  we want
to play this suit in such a way as to maximize the probability of having no losers  in the suit.
We begin by leading the 2 to the ace, and we note that the queen drops on our  left.  We then lead
the 10 from the dummy, and our right-hand opponent plays the six (after playing the three on the
first round).  Should we finesse or play for the drop?

\end{LJSItem}