Puzzle time - Next Card Red
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Suppose I shuffle a deck of 52 cards thoroughly, then play cards face up one at a time, from the top of the deck. At any time, you can interrupt me and bet $1 that the next card will be red. You bet once and only once; if you never interrupt me, you're deemed to have bet on the last card.
What's your best strategy? How much better than even can you do? (Assume there are 26 red and 26 black cards in the deck.)
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I guess you'd develop a formula for maximum expected ratio of red to black in the remaining deck, as the rest of the cards were played one by one, given a current number of red and black cards. Then you'd make a bet if the current ratio was >= maximum expected ratio. Now the issue would be developing that formula.
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I think my above answer is coherent, but I think it results in betting on the first card. The reductions to two- or four-card decks indicate this. The counter-intuitive corollary is that there is no strategy whatsoever that changes expected results. You can neither increase nor decrease your expected results by employing any a-priori strategy. It's tempting to say "just wait till there are more reds than blacks in the deck", but the small risk of that never happening exactly balances the small gain to be had by immediately betting when it happens. There's always a symmetry to what happens between betting on a card and not betting, no matter what the rest of the deck holds.
Or to put it another way, betting that the next card is red is identical to betting that the last card is red, no matter what the deck holds. Therefore all strategies are identical to no strategy.
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@jon-nyc said in Puzzle time - Next Card Red:
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I have an inductive proof that no strategy beats betting on the first card. I’ll post it later. Official solution comes out Saturday.
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I'll bet it looks a lot like this:
Or to put it another way, betting that the next card is red is identical to betting that the last card is red, no matter what the deck holds. Therefore all strategies are identical to no strategy.
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