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Poker Solutions

Poker Solutions. How Many Hands are Possible?. All of these problems will have a numerator which is the number of ways to form the hand of interest. The denominator will always be the number of ways to form any five card hand Order does not matter, so we use combination

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Poker Solutions

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  1. Poker Solutions

  2. How Many Hands are Possible? • All of these problems will have a numerator which is the number of ways to form the hand of interest. • The denominator will always be the number of ways to form any five card hand • Order does not matter, so we use combination • We are choosing 5 things from a universe of 52, so evaluate 52C5 to get the denominator which turns out to be 2,598,960

  3. One Pair • How many ways can you form exactly one pair from a deck of cards? • Choose one of 13 denominations to pair up • Within that chosen denomination, choose 2 of the 4 available cards • Of the 12 remaining denominations, choose 3 to get the other cards in the hand • This ensures that we don’t accidentally get two pairs or some higher hand • Within each of those 3 chosen denominations, choose one of the 4 available cards • So the numerator is: 13C1 x 4C2 x 12C3 x 4C1 x 4C1 x 4C1 • Dividing by 52C5 gives P(one pair) = .423

  4. Two Pairs • Choose two of 13 denominations that will have the pairs. • Choose two of the four card in each chosen denomination to form the pairs. • Choose one of 11 denominations and one of its four cards to finish the hand. • Numerator: 13C2 x 4C2 x 4C2 x 11C1 x 4C1 • Divide to get P(two pairs) = .047

  5. Three of a Kind • Choose one of 13 denominations and choose 3 of its 4 cards • Choose two of the remaining 12 denominations and choose one of four cards from each • Numerator: 13C1 x 4C3 x 12C2 x 4C1 x 4C1 • Divide to get P(3 of a kind) = .021

  6. Straight • The order of the cards (low to high) is 2-3-4-5-6-7-8-9-10-J-Q-K-A • The low card of any straight cannot be higher than a 10 • So there are 9 denominations possible to start a straight • Choose one of the 9 denominations and choose one of its 4 cards • Choose one of 4 cards from the next higher denomination • Choose one of 4 cards from each of the next 3 higher denominations • Numerator: 9C1 x 4C1 x 4C1 x 4C1 x 4C1 x 4C1 • Divide to get P(straight) = .0035

  7. Flush • Choose one of four suits for the flush • Choose 5 of its 13 cards • Numerator: 4C1 x 13C5 • Divide to get P(flush) = .0020

  8. Full House • Is a full house of three 5’s and two 6’s the same as two 5’s and three 6’s? • No! The hand with the higher three of a kind would win. • That means order matters, so we start with permutation • Choose two of the 13 denominations to form the full house. • Within the denomination that will use 3 of a kind, choose 3 of its 4 cards (and order does not matter) • Within the denomination that will use 2 of a kind, choose 2 of its 4 cards. • Numerator: 13P2 x 4C3 x 4C2 • Divide to get P(full house) = .0014

  9. Four of a Kind • Choose one of 13 denominations and use all 4 cards • Choose one of 12 remaining denominations and choose one of its 4 cards • Numerator: 13C1 x 4C4 x 12C1 x 4C1 • Divide to get P(4 of a kind) = .0002

  10. Straight Flush • Choose one of the 4 suits for the flush • Choose one of the 9 possible bottom cards to start the straight • Numerator: 4C1 x 9C1 • Divide to get P(straight flush) = .00001 • Note: Some people distinguish between a straight flush and a “royal flush”, which is just a straight flush that goes 10-J-Q-K-A • There are only 4 ways to choose the suit, so the probability is 4 / 52C5 = 1.54 x 10-6 • This was actually included in the calculation of P(straight flush) above

  11. What About None of the Above? • To find the probability of getting a 5 card hand that contains none of the named groupings, just subtract from 1 all of the probabilities we have calculated. • The result is that there are 1,179,936 ways to get nothing, and the probability is .454

  12. Summary

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