Gambler $A$ has a dollars, gambler $B$ $- b$ dollars. $P(A \text{ wins}) = p$. When a player loses, he loses 1 dollar, when a player wins, his fortune does not change. Find Expected number of plays until $B$ has 0 dollars.
So if Ei – event that in first $(b+i)$ games, $B$ loses $b$ times (therefore $B$'s result is 0 dollars)
$A' – A$ wins
$$P(A') = \sum_{i=0}^{a-1} \binom{i+b-1}{b-1} p^bq^i$$
N = # of games until B ruined
then the Expected value is
$E(N) =\sum_{i=b}^{b+a-1} i \binom{i+b-1}{b-1} p^bq^{i-b}$.
how can I get this expectation in closed form? does this seem to be correct at all?
I have calculated the expected value based on A starting with 10 dollars and $B$ with 5, with probability of $A$ winning $=0.3$. So according to my formula above, the expected number of steps until $B$ is ruined is actually less than 5, which seems to be impossible. How can the expected number of steps be less than the minimum $(5)$ number of steps ?
Best Answer
We should be able to get a quick answer for the case when Player A has infinite capital. The number of steps until B's stake is gone will be a negative binomial random variable. The expected value is given by $$E[T]={b(1-p) \over p}$$ For the example you give we get $$E[T]={5(0.7) \over 0.3}= 11.6667$$
Now in reality Player A will sometimes lose before B runs out, so I am thinking the conditional mean is lower.
Follow-Up: Using Andre's definition of $N$ and his solution, the mean number of steps before B runs out, conditioned on B losing, is $$E[T] = 11.1584$$