Strategy to win a card game that follows Wallenius' noncentral hypergeometric distribution Let's say that you have a two valued deck of N = 53 cards. Win and lose are the only values and we know that #win = 26 and #lose = 27. To play, we pay 1 unit of value. We draw one card at a time, but can't look at it. After a fix, but short, amount of time, the card is revealed to be either win or lose and then discarded. At any moment, before a card is revealed, you are allowed to "bet on it" and if it's a win card, you win 2 units of value and otherwise you lose. The game ends in one of two ways:
1) You actively bet on one card and either win/lose
2) The cards run out and you auto-bet on the last card
I've gathered that this asymmetric distribution of winning/losing cards makes the situation a Wallenius' noncentric hypergeometric distribution, or a "game with biased sampling and without replacement" but I can't seem to wrap my head around if it is winnable or not.
One naïve strategy, to just bet on the first card, is clearly at fault in the limit since your probability of winning is $P(win) = \frac{26}{26+27} < 0.5$
Now, one might instead suggest that you just wait until you have sampled two more loosing cards than winning ones, before you make a play on the next card, to naïvely get - in the best case scenario - $P(win) = \frac{26}{26+27-2} = \frac{26}{51} > 0.5$ but this doesn't really account for the fact that our playable deque is finite with 53 cards, and that this is only the best case of several possible ones with this strategy.
Now, is there a strategy to win this game, or a general way to calculate the probability of win per play as a function of earlier ones? I can't seem to find one, but I at least think that our ability to count the played cards could inform our "posterior" distribution of plays in order to always be able to choose one, under our conditions, that allow us to win in the limit. Is that true?
 A: Let there be $w$ win cards and $l$ loss cards.  The chance of a win, with optimal play, is still just $w/(w+l)$.  That's because at each stage you are making a choice between two (usually risky) options: to bet on the next card, or to pass and play the game with either $w-1$ win cards or $l-1$ loss cards.  You should take the option with the higher expected value. 
For simplicity, let's suppose you paid whatever you did to play the game and its payoff is one unit.  With this proviso, the chance of winning equals the expected payoff.  Let's optimize that value:


*

*Betting on the next card wins with probability $w/(w+l)$.  Its expectation therefore is $w/w+l$.

*Letting $p(w,l)$ be the value of this game, passing on the next card yields a game with value $p(w-1,l)$ with probability $w/(w+l)$ or value $p(w,l-1)$ with probability $l/(w+l)$.  The expectation of this option is therefore 
$$p(w-1,l) \frac{w}{w+l} + p(w,l-1) \frac{l}{w+l}.$$
By choosing the better of the two options, you will achieve a game with values determined by the recursion
$$p(w,l) = \max\left(\frac{w}{w+l},\quad p(w-1,l) \frac{w}{w+l} + p(w,l-1) \frac{l}{w+l}\right).\tag{*}$$
The initial conditions are $p(w,0) = 1$ when $w\gt 0$ and $p(0,l)=0$ when $l\gt 0$.
The unique solution is
$$p(w,l) = \frac{w}{w+l},$$
as you can readily check.  Indeed, this makes both arguments to $\max$ in $(*)$ equal, also showing that it doesn't matter how you play: the expectation will remain $w/(w+l)$.
