# Flipping Cards Probability

You have a deck of cards, 26 red, 26 black. These are turned over, and at any point you may stop and exclaim "The next card is red.". If the next card is red you win £10.

What's the optimal strategy? Prove this is the optimal strategy.

I feel like the optimal strategy is whenever you are in a state that you flip more black than red cards you call out red card, but I do not know how to prove it, nor can I think of why it must be the absolute best strategy.

Thanks for any help.

-

1
The following seems persuasive to me. Let Alice play the optimal strategy for red, and let Bob simultaneously play the optimal strategy for black. If random variables $X$ and $Y$ represent their winnings, then $E(X+Y)=10$, so by symmetry $E(X)=5$. – André Nicolas Jun 16 '12 at 19:24
Indeed. It doesn't seem there's anything you can do to increase your odds beyond 50/50, so the optimal strategy is to predict the first card red and be done with it. – Théophile Jun 16 '12 at 19:32
I have replaced [brainteaser] by [puzzle], but I'm not sure either is needed. – Asaf Karagila Jun 16 '12 at 19:40

## 2 Answers

There is no optimal strategy (or rather, every strategy is optimal). Assuming your strategy involves exclaiming before the last card is drawn, your expected winnings are 5 pounds.

If you don't exclaim, your winnings are obviously zero.

Here's a reference: "Games People Don't Play" (Peter Winkler)

-

This is a really fun paper. Thanks! – MJD Jun 17 '12 at 17:45

We deal first with deterministic strategies, adding the obvious rule that a call must be made on or before the $52$nd card.

The first step is an existence proof: there are optimal deterministic strategies. For among the finitely many deterministic strategies, there must exist one or more that maximizes expectation.

Suppose that Alice chooses an optimal strategy, and plays it for red, and independently Bob chooses and plays an optimal strategy for black. Let random variable $X$ denote Alice's winnings, and let $Y$ denote Bob's winnings. (The sample space is the set of permutations of the cards, assumed all equally likely.)

Then $E(X+Y)=10$. So by linearity of expectation, $E(X)+E(Y)=10$, and by symmetry $E(X)=E(Y)$, so $E(X)=5$.

Suppose that under the same conditions, we selflessly try to minimize the expected return. The above argument shows that again the expectation is $5$. So all deterministic "strategies" lead to the same expected return.

Now we briefly consider probabilistic strategies, for which the probability of what one does next (call or not call) is a function of the sequence of cards already seen, but nothing else. It is not hard to show that again in this case there are optimal strategies, and the rest of the argument is the same.

There may be more general notions of strategy. The above arguments are silent about these.

-
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 12, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9313902258872986, "perplexity_flag": "middle"}