Max Expected Payoff and Min Variance Strategy of a Card-Flipping Game

Question

I got the following question from https://jerryqin.com/posts/a-working-list-of-probability-problems (Day 7, Bonus Question).
You have $x$ red cards and $y$ blue cards. I flip them over one at a time. The probability of flipping a particular color is proportional to the amount of those colored cards left. You start with 1 and every flip you can bet some proportion of your money on red or blue. If you win the bet, you gain twice your bet, but if you lose the bet, you gain nothing. What is the strategy that maximizes expectancy and minimizes variance?
I think the question has something to do with Kelly's criterion: https://bjlkeng.github.io/posts/optimal-betting-and-the-kelly-criterion/ But one of the assumptions of Kelly's criterion is consistent probability of winning, which doesn't apply to this question.
Happy to hear different opinions.

Answer 1

As long as we only concern with maximizing the expectation or minimizing the variance, whether or not it produces a good strategy, here is the approach:

Let $X_n$ be the amount of money at step $n$. Then we can write

$$X_{n+1} = X_n + a_n\,X_n R_n,$$

where $a_n \in [0,1]$ is the proportion of the money the player bets at step $n$ and $R_n \in \{-1,1\}$ corresponds to winning $(R_n=1)$ or losing money $(R_n=-1)$ at that step. If $P_n$ is the probability of winning at step $n$ then, $R_n = 1$ or $-1$ with the probability $P_n$ or $(1-P_n)$ respectively. It is reasonable to think that at each turn the player chooses the color with higher frequency. For example if at some turns, $x_n$ red cards and $y_n$ blue cards are left with $x_n > y_n$, then the player bets on red cards. Simple calculations show that

$$E[X_{n+1}] = E[X_n]\, \left(1+a_n(2P_n-1)\right).$$

To maximize the expectation, player should look at the probability $P_n$. If $P_n > \frac12$ then $a_n = 1$, i.e. the player bets the whole money he/she has at that turn. Otherwise, $a_n=0$ and player doesn't bet money on that turn. Agian, the player has no difficulty finding or calculating $P_n$ at each step because as I mentioned above, the player bets on the color with higher frequency and therefore, the probability $P_n$ is simply $\frac{x_n}{x_n+y_n}$, if $x_n > y_n$ and $\frac{y_n}{x_n+y_n}$ otherwise.

To calculate the variance, first we have,

$X_{n+1}^2 = X_n^2 (1+a_n R_n)^2$

$\Rightarrow E[X_{n+1}^2 | X_n^2] = X_n^2\,(1+2a_n(2P_n-1)+a_n^2)$

$ \Rightarrow E[X_{n+1}^2] = E[X_n^2]\,(1+2a_n(2P_n-1)+a_n^2).$

So,

$$Var(X_{n+1}) = Var(X_n)\,(1+2a_n(2P_n-1)+a_n^2) + E[X_n]^2(4P_n-4P_n^2).$$

But maximizing the expectation ($a_n=1$ if $p_n > \frac12$, otherwise $a_n=0$), also maximizes the above variance. So, it not possible to maximize the expectation and minimize the variance simultaneously. This is kind of intuitive, because you can't maximize your profit without sacrificing the safe margin.

If you are asking for an offset between maximizing the expectation and maximizing the variance, that could be a good research.