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Topic: Worst winning hand (Read 542 times) |
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John_Thomas
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Worst winning hand
« on: Dec 7th, 2008, 12:45pm » |
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A standard 52-card deck is divided into 4 piles. Each pile must contain at least 5 cards but the piles need not be equal in size. The best 5 card poker hand is chosen from each pile, and then the best of these 4 hands is selected as the winner. What is the worst poker hand that can be the winner?
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rmsgrey
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Re: Worst winning hand
« Reply #1 on: Dec 8th, 2008, 12:51pm » |
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Initial thoughts:
If you have no pairs, then each pile contains 13 cards, one of each value, so you have an ace-high straight in each pile (and no clear winner).
To avoid an ace-high straight, you need each pile to lack at least one of AKQJT, so there must be at least one pile with at least a pair of jacks (at least one pile must have at least one ten, so that pile must be missing a jack or better. Having a pair of jacks each in two of the piles, and a pair of tens each in the other two breaks up the ace-high straights without having any pairs higher than a jack.
That leaves 9-high (or ten-high) straights to worry about, so similar logic gives you four two-pair hands, JJ66A, JJ66A, TT55A, TT55A. There are now no straights, no three-of-a-kind, and still no clear winner. Since each pile still contains 13 cards, it should be possible to arrange matters so there are no more than 4 cards of any given suit in any given pile, avoiding flushes.
(if 5-high straights are allowed, then the piles with pairs of fives must have a hole lower, meaning the piles without fives must have a third pair, but that doesn't improve the 5-card hands any)
It seems that the worst hand that could be an outright winner is two-pair, Jacks and Sevens, with an Ace kicker, and the worst hand that can be tied for winner is two-pair, Jacks and Sixes, Ace kicker
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