Modeling for alternate gambling in an exercise of Karlin and Pinsky's book

Question

The following is an exercise from Pinsky and Karlin's An Introduction to Stochastic Modeling (4th edition):

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I'm trying to do (b) by analyzing the possible sequence of playing. Given that $A$ wins, denote $N$ as the number of plays. Denote $A0$ as $A$ losses and $A1$ as $A$ wins. When $N$ is odd, the possible situations are $$ A1;\\ A0B0A1;\\ A0B0A0B0A1;\\ \cdots $$ and when $N$ is even, $$ B0A1;\\ B0A0B0A1;\\ B0A0B0A0B0A1;\\ \cdots $$

I think that the distribution of $N$ is given by $$ P(N=2k+1)=x^kp,\quad P(N=2k)=(1-q)x^{k-1}p $$ where $x=(1-p)(1-q)$.

However, when I add these number up, I don't get the desired number $1$ unless $p=q=1/2$: $$ \sum_{k=0}^\infty x^kp+\sum_{k=1}^\infty(1-q)x^{k-1}p\neq 1. $$

Here are my questions:

• What is wrong with the reasoning? How would you write down the correct one?
• For the event in (a) be $(N\textrm{ is odd})$?

Answer 1

Your inaccuracy is $P(N=2k-1)=x^{k-1}p$ and $P(N=2k)=x^{k-1}(1-p)q$

So in order for $B$ to win in the $2k$th play, nobody should win in the first $2k-2$ plays, then $A$ should lose and $B$ should win.

Now it sums to $1$:

$$\sum_{k=1}^\infty x^{k-1}p+\sum_{k=1}^\infty x^{k-1}(1-p)q=\frac{p}{1-x}+\frac{(1-p)q}{1-x}=\frac{1-x}{1-x}=1$$

Your last sentence is correct:

$$P(A\text{ wins})=P(N\text{ odd})=\sum_{k=1}^\infty x^{k-1}p=\frac{p}{1-x}$$

Finally for part b):

$$E[N|A\text{ wins}]=\sum_{k=1}^\infty(2k-1)P(A\text{ wins in }2k-1\text{ plays}|A\text{ wins})\\ =\frac{1}{P(A\text{ wins})}\sum_{k=1}^\infty(2k-1)P(A\text{ wins in }2k-1\text{ plays})\\ =\frac{1-x}{p}p\sum_{k=1}^\infty(2k-1)x^{k-1}=(1-x)\frac{1+x}{(1-x)^2}=\frac{1+x}{1-x}$$

Answer 2

[Edited thanks to MOMO's comment]

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Bad notations (which appear in OP) would cause lots of confusion. I would try to be careful with the writing.

(a) What one wants to calculate is the conditional probability $$ P(\textrm{A wins}\mid \textrm{A plays first}). $$ Instead of working with the formula in the definition of conditional probability $$ P(C\mid D)=\frac{P(C\wedge D)}{P(D)}, $$ we work with the probability measure $P(\cdot\mid \textrm{A plays first})$ instead. The "conditional event" that "A wins" is the disjoint union of the following events:

A1;
A0B0A1;
A0B0A0B0A1; ...

the probability of which can thus be calculated as $$ \begin{align} &P(\textrm{A wins}\mid \textrm{A plays first})\\ =&\sum_{k=0}^\infty P(N=2k+1\mid\textrm{A plays first})\\ =&p(1+x+x^2+\cdots)=\frac{p}{1-x} \end{align} $$ where $x=(1-p)(1-q)$ and $N$ denotes the number of plays. Note that we explicitly assume that the event that "A (B) wins in a single play" is independent of the event "A plays first".

(b) This problem is ambiguous. It could possibly mean the following two cases:

• $E(N\mid\textrm{A wins})$
• $E(N\mid\textrm{A wins and A plays first})$

In the first case, without knowing the probability of who plays first, one does not have enough information to answer it. Suppose we are doing the second one. Then one should work on the conditional distribution of $N$, namely, find $P(N=k\mid\textrm{A wins and A plays first})$. Denote the event that A wins as $A_W$ and the event that A plays first as $A_F$.

Now we should work on the probability measure $P(\cdot\mid A_WA_F)$ instead of $P$. In the smaller world "given A wins and A plays first", "B loses a single play" is a sure event. (Note that "A wins" here means "A wins the game", not a single play.) On the other hand, $$ P(N=2k+1\mid A_WA_F)=\frac{P(N=2k+1, A_W\mid A_F)}{P(A_W\mid A_F)} $$ where the denominator is given by (a) and the numerator $$ P(N=2k+1, A_W\mid A_F)=x^{k}p. $$

Note in particular that $P(N=2k+1\mid A_WA_F)=0$ for any $k\in\mathbb{Z}_+$.

Now the conditional expectation is given by $$ E(N\mid A_WA_F)=\sum_{k=0}^\infty(2k+1)P(N=2k+1\mid A_WA_F) $$

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The original wrong answer is as follows $$ E(N\mid A_WA_F)=\sum_{k=0}^\infty(2k-1)(1-p)^kp. $$ An easy counter example is that when $q=p=1$, one should have $E(E\mid A_WA_F)=1$, but the result above gives $0$.