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!
Best Answer
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.