Expected value when the gambler leaves

Question

I have an interesting problem that I am not quite sure how to best proceed.

Suppose there is a man with a very large amount of money who is playing a game at a large casino.

The game has $3$ possible outcomes,

he can win $10$ with probability $0.25$, he can lose $2$ with probability $0.7$ or he can lose $50$ with probability $0.05$.

The man stops as soon as he has the event that he loses $50$.

What is his expected value of earning/loss when he stops?

I think it will need a sum, maybe a random sum.

I know that all the trials are independent, so that the probability that he loses 50 on the ith game is always 0.05. But I am not sure at all how to account for what happens before this.

Any ideas?

Answer 1

Let $Y_k$ be the payoff at step $k$. We have

$$ Y_k = \left\{\begin{aligned} &10 &&p = 1/4\\ &-2 && p = 7/10\\ &-50 && p = 1/20 \end{aligned} \right.$$ Let $\tau$ be the first step where $Y_k = -50$. We are interested in $$ \mathbb E \left[ \sum^\tau_{k=1} Y_k \right] $$ By conditioning on the stopping time, by the law of total expectation this satisfies $$ \mathbb E \left[ \sum^\tau_{k=1} Y_k \right] = \sum^\infty_{n=1} \mathbb E \left[ \sum^\tau_{k=1} Y_k \bigg| \tau = n \right] \mathbb P (\tau = n) $$ Now look at the expectation inside. This satisfies $$ \begin{align} \mathbb E \left[ \sum^\tau_{k=1} Y_k \bigg| \tau = n \right] &= \mathbb E \left[ \sum^{\tau-1}_{k=1} Y_k \bigg| \tau = n \right] + \mathbb E \left[ Y_\tau \bigg| \tau = n \right]\\ &= \sum^{\tau-1}_{k=1} \mathbb E [ Y_k \mid \tau = n ] - 50 \end{align} $$ by linearity and the fact that $Y_\tau = -50$ by definition of $\tau$. The expectation $\mathbb E [ Y_k \mid \tau = n ]$ is the expected payoff conditioned on that either $Y_k = 10$ or $Y_k = -2$. This happens to be $$ \mathbb E [ Y_k \mid \tau = n ] = \frac{\mathbb E[Y_k 1_{\lbrace Y_k \in \lbrace 10, -2 \rbrace \rbrace}]}{\mathbb P (\lbrace Y_k \in \lbrace 10, -2 \rbrace) } = \frac{10 * 0.25 - 2 *0.7}{0.95} = \frac{22}{19} $$ Thus we have $$ \mathbb E \left[ \sum^\tau_{k=1} Y_k \bigg| \tau = n \right] = (\tau - 1)\frac{22}{19} - 50 $$ As to $\mathbb P (\tau = n)$, this is the probability of having $n-1$ consecutive cases of $Y_k \neq -50$ and then $Y_n = 50$. This results in $$ \mathbb P(\tau = n) = p^{n-1}(1-p) $$ where $p = 19/20$ in our case. Putting it all together we have $$ \mathbb E \left[ \sum^\tau_{k=1} Y_k \right] = \sum^\infty_{\tau=1} \left( (\tau - 1)\frac{22}{19} - 50 \right) \left(\frac{19}{20}\right)^{n-1}\frac{1}{20} = -28 $$

Answer 2

What is the probability that our man exits on round $k$? Call it $p_k$ Well, we need round $k$ to be the bad event, probability $.05$, and the prior rounds to not be bad, probability $.95^{k-1}$. Hence $p_k=.95^{k-1}\times .05$

Conditioned on him losing in round $k$, what are his expected winnings? Call it $E_k$. Prior to the loss he loses $2$ with probability $\frac {.7}{.95}$ and wins $10$ with probability $\frac {.25}{.95}$ Thus on one of those rounds he expects to win $-2\times \frac {70}{95}+10\times \frac {25}{95}=\frac {110}{95}$. As there are $k-1$ pre-disaster rounds, and he loses $50$ on the last, we have $$E_k=\frac {110}{95}\times (k-1)-50$$ Thus the answer you want is $$\sum_{k=1}^{\infty}\left(\frac {110}{95}\times (k-1)-50\right)\times .95^{k-1}\times .05 = \fbox {-28}$$

Sanity check: He expects to lose on round $20$. If he loses on round $20$ he expects to net $$E_{20}=\frac {110}{95}\times 19-50=22-50=-28$$

Answer 3

Conditional on he does not lose 50 on the first game, the expected income in the first game is $w:=10 \times \frac{0.25}{0.95} -2 \times \frac{0.7}{0.95}$. So his total expected income is $-50+0.95 w +(0.95)^2 w +(0.95)^3 w +...$ which is just a geometric series.