# Expectation of a gamblers winnings

Gambles are independent, and each one results in the player being equally likely to win or lose 1 unit. Let W denote the net winnings of a gambler whose strategy is to stop gambling immediately after his first win.

Find P(W < 0).

I am not sure how to calculate the probability for this. I know that P(W > 0) is 1/2.

Find E[W] How does it equal 0?

• What is $Pr(W=0)$? I.e. What is the probability that the sequence you observe is $lw$? What of $Pr(W=-1)$? I.e. what is the probability that the sequence you observe is $llw$? $Pr(W=-n)$ corresponds to the sequence being $\underbrace{lll\cdots l}_{n+1~\text{times}}w$. Finally, notice, what is the outcome of the series: $\sum\limits_{i=0}^\infty -\frac{1}{2^i}$? – JMoravitz Apr 28 '16 at 2:55

All other combinations of gambles result in a net loss, so P(W < 0) = $1 -$P(W > 0)$-$P(W = 0)
The expected value can be obtained by summing the products of each profit and the probability of that profit. In this case, you have $1 * 0.5 + 0 * 0.25 + (-1) * 0.125 + (-2) * 0.0625$, etc. So your E(W) is the sum $$\sum_{n=0}^{\infty}{\frac{1-n}{2^{n+1}}}=0$$
He only wins if he wins the first gamble. He breaks even if he loses the first and wins the second. Every other sequence of results has at least two losses, so $\bf {W} \lt 0$
If we assume that the expectation $E(W)$ exists, we can calculate it by conditioning on the result of the first toss. If it is a win, we have won $1$ dollar. If it is a loss, we are down $1$ dollar, and the game starts again, so our conditional expectation is $E(W)-1$. Thus $$E(W)=\frac{1}{2}+\frac{1}{2}(E(W)-1).$$ Solve. We get $E(W)=0$.