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Here is a braintest:

We draw from a standard 100 card deck, 50 red and 50 black. You can stop at any round with payoff the ratio of red card in total draws. What's the payoff of this game?

We denote $f(r,b)$ the expected value of r red and b black left (i.e. 50-r and 50-b has been opened). if you stop, you will been paid: $$\dfrac{50-r}{100-r-b};$$ if you continue, the value will be: $$\dfrac{r}{r+b}f(r-1,b) + \dfrac{b}{r+b}f(r,b-1).$$ Therefore we have $$f(r,b) = \max\left(\dfrac{50-r}{100-r-b},\quad \dfrac{r}{r+b}f(r-1,b) + \dfrac{b}{r+b}f(r,b-1)\right)$$ with $$f(r,0) = \dfrac{1}{2},\ f(0,b) = \dfrac{r}{100-b}.$$

Am I correct? And do we have the closed form of $f(26,26)?$

user6703592
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  • Should note: your payout is not clear. If you get Red initially, is the payout infinite? This game, which is very well discussed online, is usually set to payout $1$ for red and $-1$ for black. Of course, the backwards induction model described in the duplicate can easily accommodate alternate payout structures, but you ought to clarify. – lulu Jan 01 '23 at 16:19
  • @LULU if your first draw is red, you will be paid $1/1 = 1?$ Here the ratio is the percentage of total drawing. – user6703592 Jan 01 '23 at 16:22
  • It's hard to parse your formula. Where do $50$ and $100$ come in? Are you assuming a $100$ card deck? In any case, study the duplicate. It gives a numerical framework which you should be able to modify to accommodate your payout structure. – lulu Jan 01 '23 at 16:27
  • @lulu for r: +1, b: -1 case, I don't understand why the update will be $v(r,d) = \max(0, f(r,b))$ but not $v(r,d) = \max((26-r)-(26-b), f(r,b)),$ i.e. if you stop you will been paid related to the red/black you already opend – user6703592 Jan 01 '23 at 16:29
  • @lulu see my update – user6703592 Jan 01 '23 at 16:29
  • Their $f$ is not the same as yours (well, I don't understand what yours is, but I assume it is different). Their $f$ is the option valued payout...it could certainly be negative: if, say, you drew $26$ red cards in a row then you know for certain that the next draw will produce a loss. Of course, in that case, you stop. You do not exercise your option to continue. – lulu Jan 01 '23 at 16:36

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