I came across a question in a book that I am stumped on. There is a chance there is a typo, but I could simply be confused.
There is a can with $50$ red and $50$ green chips. You cannot see the chips in the can, but you can see them when you pull them out. You pull chips out one by one. If the last chip you pull out (meaning the can is now empty) is red you win $10$. If the last chip you pull out is green, you lose $1$. However, you have the option at any time to change the game to a new game where if the next chip you pull out is red, you win $9$ and if the next chip you pull out is green, you lose $1$ (for those of you familiar with finance, this is like an American option that you can exercise early). What is the optimal strategy (the one that maximizes expected value)?
I tried using rollback:
At all green left, you lose anyway.
At all red left, you win guaranteed so do not exercise the option.
If $1R$ vs $1G$, the regular game is $10\times \frac{1}{2} - 1 \times \frac{1}{2}$ and the option is $9\times \frac{1}{2} - 1 \times \frac{1}{2}$ so do not exercise
If $2R$ vs $1G$, the regular game is $10\times (\frac{1}{3} + \frac{2}{3} \times \frac{1}{2}) - 1\times (1-(\frac{1}{3} + \frac{2}{3} \times \frac{1}{2})) = 6.33$ and the option is $9\times \frac{2}{3} - \frac{1}{3} = 5.67$ (for the probabilities in the regular game, there is a 1/3 chance you win after the $1^{st}$ draw and $\frac{2}{3} \times \frac{1}{2}$ chance you win on the second) so do not exercise
If $1R$ vs $2G$, the regular game is $10\times ( \frac{2}{3} \times \frac{1}{2} ) - (1 - \frac{2}{3} \times \frac{1}{2} ) ) = 2.67$ and the option is $9\times \frac{1}{3} - \frac{2}{3} = 2.33$ so you do not exercise again.
Given these examples, I intuitively assume that you never exercise. Is this correct? If so, how do you formally prove it (I assume there is a way to do it without going through every single case, but maybe there isn't)?
Thanks!
Presumably the chips are being drawn randomly, i.e. at every draw, every chip currently in the can has equal probability of being drawn.
The probabilities for the next chip being red or green are always the same as the probabilities for the last chip being red or green (namely $r/(r+g)$ and $g/(r+g)$, where $r$ and $g$ are the numbers of red and green chips currently in the can). Your expected payout for the new game is never better than your expected payout for the old game. Therefore there is never any advantage to switching.