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You have $x$ red cards and $y$ black cards. I flip them over one at a time. The probability of flipping a particular colour is proportional to the amount of those coloured cards left. You start with $1$ and for 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 correct strategy to use is the Kelly Criterion, but I honestly do not know how to set the formula and how to find the expected value and variance of the game. The fact that the probability is dynamic really confuses me.

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  • $\begingroup$ Why don't you study the first flip in detail? Formulate the optimization problem and try to solve it. Then worry about the following flips. $\endgroup$ Oct 4, 2021 at 10:11
  • $\begingroup$ Do you know the value of $x$ and $y$ or just $n = x + y$? $\endgroup$
    – Bob Jansen
    Oct 4, 2021 at 12:57
  • $\begingroup$ I know just x + y = n $\endgroup$
    – Mining
    Oct 4, 2021 at 12:58

2 Answers 2

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This is what I would do without access to pen and paper. I don't know whether this strategy is optimal but it is easy to execute and I invite others to do better :)

The problem in this setup is that I don't know my probability of winning as $x$ and $y$ are unknown and I can't apply the Kelly criterion. To wit (Wikipedia):

Where losing the bet involves losing the entire wager, the Kelly bet is: $$f^* = p-\frac{q}{b} = p + \frac{p-1}{b}$$ where:

  • $f^{*}$ is the fraction of the current bankroll to wager.
  • $p$ is the probability of a win.
  • $q$ is the probability of a loss ($ q = 1 - p$).
  • $b$ is the amount gained with a win. E.g. If betting \$10 on a 2-to-1 odds bet, (upon win you are returned \$30, winning you \$20), then $b = 20 / 10 = 2$. As an example, if a gamble has a 60% chance of winning ($p = 0.6, q = 0.4$), and the gambler receives 1-to-1 odds on a winning bet ($b=1$), then the gambler should bet 20% of the bankroll at each opportunity ($f^{*} = 0.6-\frac{0.4}{1} = 0.2$), in order to maximize the long-run growth rate of the bankroll.

I have $b = 1$ and $f^* = 2p - 1 $ but don't know $p$. I can only estimate it from what I have seen. I have seen nothing and therefore in the first round I shouldn't bet at all! After the first round, I have a small bit of information about the distribution which I can use. Assume the first card was red, in that case I set $p = \frac{1}{2} + \frac{1}{n}$.

If the second card is blue I set $p = \frac{1}{2}$ and bet nothing. If the second card is red I bet $p = \frac{1}{2} + \frac{2}{n}$. That is, my formula for $p$ when betting on red is $$p = \frac{1}{2} + \frac{r - b}{n}$$ where $r$ is the number of red cards drawn and $b$ the formula for betting on blue is analogous.

If I was risk averse I would limit the fraction regardless of the observed proportion.

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Here is my attempt at some formal analysis: Denote by $V(x,n,W)$ the value function where, $n\equiv x+y$ We have:

\begin{equation} \begin{aligned} V\left(x,n,W\right) = \max_{\left\{0\leq \lambda_x, \lambda_y \leq 1\right\}} & \frac{x}{n}V\left(x-1,n-1,W\left(1+\lambda_x-\lambda_y\right)\right) \\ & \quad + \frac{n-x}{n}V\left(x,n-1,W\left(1+\lambda_y-\lambda_x\right)\right) \end{aligned} \end{equation} FOC are given by: \begin{align} \frac{x}{n}V_3\left(x-1,n-1,W\left(1+\lambda_x-\lambda_y\right)\right)-\frac{n-x}{n}V_3\left(x,n-1,W\left(1+\lambda_y-\lambda_x\right)\right)) & = 0 \end{align} So: $$ x V_3\left(x-1,n-1,W\left(1+\lambda_x-\lambda_y\right)\right) = (n-x) V_3\left(x,n-1,W\left(1+\lambda_y-\lambda_x\right)\right)) $$ Conjecture: $V(*,*,W)=A(*,*)+\log W$ Then: $$ \lambda_x-\lambda_y=\frac{2x-n}{n} $$ Verify: \begin{equation} \begin{aligned} n A(x,n) & = xA(x-1,n-1)+x\log\left(x\right) \\ & \quad +(n-x)A(x,n-1) \\ & \quad +(n-x)\log\left(n-x\right) -\log(2n) \end{aligned} \end{equation} Finally, if we can solve that equation, we want to use all our wealth, so that $\lambda_x+\lambda_y=1$, hence: $$ \lambda_x = \frac{1}{2} + \frac{2x-n}{2n} = \frac{x}{n} $$

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