# Statistic: Expected outcome of gambling

#### batman350z

##### New member
Studying for a stats final, and came across this problem that I'm not quite sure what to do, any help is appreciated.

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:

a) P(W > 0)
b) P(W < 0)
c) E(W)

For part (a), since the gambler only needs net win greater than 0, there is only one way that will happen is when he wins the first time. Hence, there P(W>0) = 1/2

For part (b), since the gambler can lose infinite times, and have just one win. Thus, the gambler can lose once and win once which gives probability of 1/4 (net winning = -1). But the gambler can lose twice and win once (net winnings = -2), which gives a probability of 1/8, and so on and so forth. So I would have, 1/4 + 1/8 + 1/16 + .... = 1/2. However, the textbook says it is 1/4. That is where I am confused.

Then for part (c), how do I account for the infinite combination of negative winnings?

#### tkhunny

##### Moderator
Staff member
On each round, there is a 50% chance of winning.

The expected number of turns, then is 1(1/2) + 2(1/4) + 3(1/8) + ... = $$\displaystyle \frac{\frac{1}{2}+\frac{\left(\frac{1}{2}\right)^{2}}{1-\frac{1}{2}}}{1-\frac{1}{2}}$$ = 2 = E[Turns]

That last thing looks a little scary, but it's just a little algebra.

The expected winnings, then is 1(1/2) + 0(1/4) + (-1)(1/8) + (-2)(1/16) + ... = 1/2 + 0 - (1/4)*[1(1/2) + 2(1/4) + 3(1/8) + ...] = 1/2 + 0 - (1/4)*E[Turns] = 0 = E[W]

This development suggests

P(W>0) = 1/2
P(W=0) = 1/4
There isn't much left
P(W<0) = 1/4

Very often, we do this thing called EXPLORATION. Play around with it. See what it does. Can you make sense of it? You can also pratice your algebra!

#### batman350z

##### New member
The expected number of turns, then is 1(1/2) + 2(1/4) + 3(1/8) + ... = $$\displaystyle \frac{\frac{1}{2}+\frac{\left(\frac{1}{2}\right)^{2}}{1-\frac{1}{2}}}{1-\frac{1}{2}}$$ = 2 = E[Turns]
I'm slightly confused by the last part. That is a geometric infinite sum?

In the end, I worked out E(W) = $$\displaystyle \frac{1}{2} * 1 - \sum_{k=1}^\infty k(\frac{1}{2})^{k+2}$$

Thanks for the explanation!

#### tkhunny

##### Moderator
Staff member
No, it is not Geometric. It has a linearly increasing component and takes a little more work. I rather deliberately left it in that awful form and didn't explain its development. It will be a worthwhile exercise for you.

As for your final result, it would be a little more instructive to include the second term, even though it is zero. The sum is good. Note how you can take the "+2" off the exponent and you will get the "1/4" outside and see that the answer is the same as the way I left it.

Good work.