Solution
The poker puzzle, #1175, was solved by Joseph DeVincentis, Rob Pratt, and Xie Jia (China). Three incorrect solutions were received.
I found my old source for this: Math and Logic, by Aaron Friedland, Dover, 1971.
The correct choice is any of the four full houses: AAA99, AAA88, AAA77, AAA66.
The first point is that any AAA full house will beat all other full houses; the choice of pair card is not relevant, since no one else can have three aces. All choices eliminate the same number of fours-of-a-kind, so the question is which choice will eliminate more straight flushes.
Since the aces eliminate all the royal flushes, the choice of 99 (or the others) is best as it eliminates more of the lower straight flushes, such as, say, 8-9-10-J-Q.
Here are some win probabilities for the case that there are only two players (thanks to Rob Pratt):
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AAA99
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This wins with probability 1533442/1533939, or 0.99967600
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AAAKK with all four suits represented (!)
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This wins with probability 511145/511313 or 0.99967143
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AAAKK with only three suits represented
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This is even worse because of the straight flush overlap, winning with probability 1533434/1533939, or 0.99967078.
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Joseph DeVincentis submitted a solution to the extra credit problem, which was the same as Problem 1175, except for 5-card draw instead of 5-card stud. He showed that the following is the optimum strategy, where ♣ ♦ ♥ ♠ refer to suits.
First five cards: 5♣ 6♠ 7♥ 8♦ 9♥ (a straight, but no matter).
For the exchange, discard these five and draw: A♦ A♥ A♠ 10♣ 10♠.
The final hand will lose to no full houses; to 228 fours-of-a-kind (four 2s, 3s, 4s, Js, Qs, or Ks, with 38 choices for the fifth card); and to four straight flushes:
2♥ 3♥ 4♥ 5♥ 6♥
2♦ 3♦ 4♦ 5♦ 6♦
3♦ 4♦ 5♦ 6♦ 7♦
9♦ 10♦ J♦ Q♦ K♦
Jeff Yoak submitted a nice variation, which goes back to a Martin Gardner column many years ago.
Alice and Bob are facing a deck of 52 cards, all face-up. The rules are that Alice can draw any five cards she wants, and then the same for Bob. Then Alice can exchange any of her five for new ones (the original five are not in play), and the same for Bob.
In the event of a tie, Bob wins.
There is a winning strategy for Alice. What is it?
In his Mathematical Puzzles and Diversions book, Gardner mentions that it was originally believed that there were only 48 solutions (4 10s, and then any other card to complete your initial hand), but that 40 more solutions exist: taking 3 10s and then in the final suit a combination like A-9 also works. This brings the number of solutions to 88, which appears to be the correct total number of solutions. The problem also appears in a book by Harry Anderson. More can be found at The New York Times, though some of the links for the solution are not working. A comment by "swells" has all the details of the solution. It appears that 88 is the correct answer to this nice poker problem.
And Rob Pratt asks: What is the largest set of cards from a standard deck that does not contain a straight?
These variations are not too hard, so I will not post the answers. You can ask me.
[Back to Problem 1175]
© Copyright 2014 Stan Wagon. Reproduced with permission.