Background:
$$\text{Show}: \quad E[R\mid\text{win},S=4] = E[R\mid S=4]$$
-
where $R$ is the number of rolls of the dice in a game of craps
-
$S$ is the initial sum and for this question let's say it's $4$
-
You keep rolling two dice until you get either a sum of $4$ (and win) or a sum of $7$ (and lose)
It is true that $E[R\mid\text{win},S=4] = E[R\mid S=4]$ and the explanation in words from Lord Sheldon Ross is in the screenshot below.
I now want to show the same conditional independence but using equations.
$$\text{Show}: \quad E[R\mid\text{win},S=4] = E[R\mid S=4]$$
$$ \sum_r r P(R=r \mid\text{win}, S=4) = \sum_r r P(R=r \mid S=4) $$
$$ \sum_r r \frac{P(R=r, \text{win}, S=4)}{P(\text{win},S=4)} = \sum_r r \frac{P(R=r, S=4)}{P(S=4)}$$
$$ \sum_r r \frac{P(R=r, S=4) P(\text{win})}{P(S=4)P(\text{win})} = \sum_r r \frac{P(R=r, S=4)}{P(S=4)}$$
My specific question is about the LHS of the final equality above. I don't believe the probability of winning a craps game is independent of the initial sum being $4$. That is I don't think you can say $P(\text{win}, S=4) = P(\text{win}) P(S=4)$. For example, if your initial sum is $7$ then your probability of winning is $1$. But I know the LHS equals the RHS through the verbal reasoning in the screenshot below. Therefore, to get the LHS to equal the RHS I'd like to take "win" out using conditional independence, that way it will cancel out and the LHS will equal the RHS. Where am I going wrong here?
Book answer below
Best Answer
Both sides can be calculated directly. Preliminary: $\sum_{n=1}^\infty nx^{n-1} = \frac{1}{(1-x)^2}$
$$P(R=r\mid\text{win},S=4)=\frac{P(R=r,\text{win}\mid S=4)}{P(\text{win}\mid S=4)}$$
Let $p_k=\operatorname{prob}(\text{roll} = k).$ Let $u=p_4+p_7$ and $v=1-u$, so termination at any roll has a probability $u$. Therefore $P(R=r,\text{win} \mid S=4) = p_4v^{r-1}$, while $P(R=r\mid S=4)=uv^{r-1}$. Forming the expectations, $E(\text{win stop}\mid S=4)=p_4\sum_{r=1}^\infty rv^{r-1} = \frac{p_4}{(1-v)^2} = \frac{p_4}{(p_4+p_7)^2}$ and $P(\text{win}\mid S=4)=\frac{p_4}{p_4+p_7}$. Therefore $E(\text{stop}\mid\text{win},S=4)=\frac{1}{p_4+p_7}$ Next $E(\text{stop}\mid S=4) = u \sum_{r=1}^\infty rv^{r-1} = \frac{u}{(1-v)^2)^2}=\frac{1}{p_4+p_7}$, so the two expectations are equal.