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Recently got asked this question and am not sure how to answer it. I'm not sure there is a singular concrete answer but any general thoughts would be appreciated.

-You are allowed to play ONE round of a game, who's rules are as follows.

-You start with a pot of 0\$

-You pay 1\$ to enter the game (so you start -1\$)

-Then, you flip a coin. If heads, 5\$ are added to your pot. If tails, you lose 1\$ and the game ends.

Now, if you received heads in the previous round, you have the option to flip the coin again. However, this time, the rewards and losses are multiplied by 5x, so you could have 25\$ added to your pot, or lose 5\$ (and have the game end).

This final step repeats indefinitely (until you lose or chose not to flip again).

Now, what is a strategy that has the highest expected winnings for playing this game? I'm confused because at any point, flipping the coin is a positive expected value choice, so theoretically, you should always flip the coin. And yet paradoxically, if you keep flipping the coin indefinitely, you will at some point lose all your money, so how to decide when to stop flipping? I initially thought about things like log utility, but realized that this doesn't apply a) by definition of the game (maximize the linear wealth), and also doesn't apply in a practical sense either since you are only playing 1 round and you aren't betting some portion of your wealth to enter the game.

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Yeah I would say this is a variation of, or at least has similar ideas to, the St Petersburg paradox.

More formally, you were asked about a strategy. What is a strategy? It is a function that tells you, given a game situation, what to do. In this game, the function consists of telling you, for every $n \ge 1$, if you have won $n$ flips in a row, whether to continue or to quit. I.e. a strategy is a function: $\mathbb{N}^+ \to \{Continue, Quit\}$.

Obviously, for this game it only matters the first time $N$ the function tells you to quit, i.e. $N(f) = \min_n \{n \in \mathbb{N}^+ \mid f(n) = Quit\}$. But any function with a finite $N(f)$ has lower expected value than another $f'$ with with a higher $N(f')$, e.g. $N(f') = N(f)+1$. This is easy to prove.

So no function with finite $N(f)$ can be optimal. OTOH the Never-Quit function has expected value $0$, so it's not optimal either. In short, there is no globally optimal function (just like there is no globally largest integer).

The point of the interview question (I'm guessing) must have been to see if you can explain why maximizing expected (linear) profit is not always the correct thing to do.

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  • $\begingroup$ agree with you on the likely point of the question. I guess i'm still a little confused with regards to how the never-quit function will be optimal, in the sense that say I've won the first round and have 5 wealth and I can choose between a) never quitting and b) ending the game. If I never quit, I am guaranteed to end up with 0 wealth since at some point I will lose, and I only take home my winnings if I quit. If I end the game, I take home 5, which is > 0. In spite of its per-round utility maximization, the practical result of never-quit makes me wonder how it can actually be optimal. $\endgroup$ – bopokippo Jan 7 at 21:34
  • $\begingroup$ Oh, good catch! the Never-Quit function has expected value $0$ (unlike the summation in the St Pete paradox). see my edits now. $\endgroup$ – antkam Jan 7 at 21:41

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