Conditional Expectation and Craps Bet

Question

I was hoping someone could help be understand a question regarding the conditional expectation of a 'pass' line craps bet and by extension conditional expectation a bit better.

For context, in a game of Craps there's three possible outcomes when you bet the 'pass' line: $1.$ you win if you roll a sum of $7$ or $11$, $2.$ you lose if you roll a sum of $2,3$ or $12$. $3.$ you roll any other sum, that is $4,5,6,8,9,10$, in which case establishes a 'point-number' that you must roll again before a $7$.

In my question, $i=1,2$ and $N_i$ is the number of trials needed for outcome $i$ to occur and $N := N_1 \wedge N2$. Also in my book the author defined $x \wedge y$ as $min(x,y)$, so $N = min(N_1, N_2)$

$So$, I'm trying to evaluate $E[N_1| N_1 < N_2]$. As far as I understand the theorem that's used for actually evaluating these expectations is $$E[X] = \sum_{i=1}^n E[X|A_i]P(A_i) $$ where r.v. $Y$ takes outcomes $A_1,..A_n$

So, my main issue comes from actually rewriting $E[N_1| N_1 < N_2]$ and solving for $E[N]$. I tried doing $$E[N] = E[N|N_1]P(N1) + E[N|N2]P(N_2) \\= 1(p_1) + (1+E[N])(p_2)$$

but solving doesn't give me $E[N]=\frac{1}{p_1+p_2}$ which is what I'm supposed to get.

Anyway, sorry this post is so long-winded, but any help would be appreciated. Thanks!

Answer 1

For both of the events $\ \left\{N_1=n \right\}\ $ and $ \left\{N_1<N_2\right\}\ $ to occur, neither outcome $1$ nor outcome $2$ can occur on any of the first $\ n-1\ $ trials, and outcome $1$ must occur on the $\ n^\text{th}\ $. The probability of neither outcome $1$'s nor $2$'s occurring at any trial is $\ 1-p_1-p_2\ $, and the probability of outcome $1$'s occurring on the $\ n^\text{th}\ $ is $\ p_1\ $. Since the trials are independent, we have $$ P\left(\left\{N_1=n\right\} \wedge \left\{N_1< N_2\right\}\right)=(1-p_1-p_2)^{n-1}p_1 $$ and \begin{align} P\left( \left\{N_1< N_2\right\}\right)&=\sum_\limits{n=1}^\infty P\left(\left\{N_1=n\right\} \wedge \left\{N_1< N_2\right\}\right)\\ &=\frac{p_1}{p_1+p_2}\ . \end{align} Therefore \begin{align} P\left(\left\{N_1=n\right\} \left| \left\{N_1< N_2\right.\right\}\right)&=\frac{P\left(\left\{N_1=n\right\} \wedge \left\{N_1< N_2\right\}\right)}{P\left( \left\{N_1< N_2\right\}\right)}\\ &=(1-p_1-p_2)^{n-1}(p_1+p_2) \end{align} and \begin{align} E\left(N_1 |N_1<N_2\right)&=\sum_\limits{n=1}^\infty n P\left(\left\{N_1=n\right\} \left| \left\{N_1< N_2\right.\right\}\right)\\ &=\frac{p_1+p_2}{\left(p_1+p_2\right)^2}\\ &=\frac{1}{p_1+p_2}\ . \end{align}

Answer 2

Well, to start, the expression you need is:

$$\mathsf E(N_1\wedge N_2) =\mathsf E(N_1\mid N_1<N_2)~\mathsf P(N_1< N_2)+\mathsf E(N_2\mid N_1>N_2)~\mathsf P(N_1> N_2)$$

Well, $\mathsf P(N_1<N_2)$ is the probability that the first roll that is not-outcome 3 is outcome 1. So this equals $p_1/(p_1+p_2)$, where $p_k$ is the probability that a given roll is outcome $k$.

Now $\mathsf E(N_1\mid N_1<N_2)~$ is the expected count of rolls until you roll the first outcome 1 when given that you roll the first outcome 2 is rolled after that.