Expected Value with stopping rule.

Question

GAME 1) Suppose you play game where you throw a coin 100 times. If it's heads you win 2 if it is tails you lose 1.

a) What is the expected value of this game? This one is easy and the answer is E[game1] = 50

GAME 2) Suppose you start with $50 and you stop when you run out of money. Otherwise you stop after 100 throws.

b) What is the expected value of game 2 (not counting the initial amount you started with)? Actually the question only asks if E[game2] > E[game1] or E[game2] < E[game1] or E[game2] = E[game1].

If they are different is the difference significant?

I am puzzled by part b. I am not sure how to go about it? Any suggestions/ answers will be most welcome.

Answer 1

There doesn't seem to be any difference, really. In the worst case scenario you run out of money at exactly 100 throws, so no matter what, you get 100 throws, which is exactly game 1.

For starting with \$50, it's rather harder; since you might stop early, it's possible to have what would be a winning run of 100 items that nonetheless fails - if your sequence would be 50 losses and 50 wins, since you run out of money on the 50th throw, you stop there. I suspect a more calculation-intensive approach is required, which I'll show for starting at \$3 and limiting to just 6 throws:

• After 1 throw, you may have \$2 (1/2 the time) or \$5 (1/2)
• After 2 throws, you may have \$1 (1/4) or \$4 (2/4 = 1/2) or \$7 (1/4)
• After 3 throws, you may have \$0 (1/8) or \$3 (3/8) or \$6 (3/8) or \$9 (1/8)
• After 4 throws, you may have quit at \$0 (2/16), or \$2 (3/16), or \$5 (6/16), or \$8 (4/16), or \$11 (1/16)
• After 5 throws, you may have quit at \$0 (4/32), or \$1 (3/32), or \$4 (9/32), or \$7 (10/32), or \$10 (5/32), or \$13 (1/32)
• After 6 throws, you may have quit at \$0 (8/64), or just arrived at \$0 (3/64), or \$3 (12/64), or \$6 (19/64), or \$9 (15/64), or \$12 (6/64), or \$15 (1/64)

The expectation is then $$\frac{0\cdot 8 + 0\cdot 3 + 3\cdot 12 + 6\cdot 19 + 9\cdot 15 + 12\cdot 6 + 15 \cdot 1}{64}-3 = 2.8125$$

Which is actually slighty lower than the unrestricted expectation of $\frac{2-1}{2}\cdot6=3$. This is because even when it's going bad, the remaining expectation is still positive: in the situation where you give up, the remaining expectation is still +1.5, just like the others.

Answer 2

A way to (rigorously) compare the expectations without doing any calculations is to use Wald's equation. Let $X_i$ be a random variable taking values $2$ and $-1$, each with probability $1/2$. Then we may write:

\begin{equation} G_1 = \sum_{i=1}^{100} X_i, \\ G_2 = \sum_{i=1}^{N} X_i, \end{equation} where $N = \text{inf} \{ n : X_n = -50 \} \wedge 100$, which is a well-defined stopping time satisfying $N \leq 100$.

Clearly $\mathbb{E}[G_1] = 100\mathbb{E}[X_i] = 50$, whereas by Wald's identity, $\mathbb{E}[G_2] = \mathbb{E}[N] \mathbb{E}[X_i] < 100\mathbb{E}[X_i] = 50$, since there is a non-zero (yet small) probability of $N < 100$ corresponding to going bankrupt. Note that the smaller $\mathbb{P}[N<100]$ gets, the closer these expectations. This is dependent on the level of starting capital.