• Poker Hand Probability Problems Solving
• Poker Hand Probability Problems Worksheets

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Find the probability of getting the following hands in a game of traditional 5 card poker.

(1)) P(Having exactly 2 Aces) (other 3 cards can be anything but aces)

(2)) P(Royal Flush) (aka 10, J, Q, K, A and all the same suit)

• After finding the number of ways to get four-of-a-kind for four cards in a poker hand, should you use: a. 48 choose 1 or b. 12 choose 1 x 4 choose 1.
• Probability of making a specific hand (7 out of 52) The following chart shows the probability of getting a certain hand. Whereas a pair floats by often enough, getting a straight or royal flush is less likely. 7 out of 52 means, that although you build your hand using 5 cards, you still have 7 cards from which to choose these 5.

(3)) P(Straight Flush) (aka same suit, in consecutive number order)(4)) P(Flush) (aka all cards in the same suit.)

Each of the 2,598,960 possible hands of poker is equally likely when dealt 5 cards from a standard poker deck. Because of this, one can use probability by outcomes to compute the probabilities of each classification of poker hand.

(5)) P(Straight) (aka all cards in consecutive number order.)(6)) P(Full house) (aka 3 of a kind and a pair)(7)) P(Four of a kind)(8)) P(Three of a kind)(9)) P(Two pair)(10)) P(One Pair)(11)) P(No pair)

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Poker Hand Probability Problems Solving

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Let _nC_k denote all the combinations of n objects in groups of k elements. It should be obvious that we need to work with combinations and not permutations as the order of the cards in our hand of cards is irrelevant.

(a) We have to select one suit out of 4 which can be done in ₄C₁ ways and, once the suit is chosen, we have to select 5 cards from that specific suit and this can be done in ₁₃C₅ ways. Finally, in a game of poker there are ₅₂C₅ possible hands. Hence, the probability of a flush is given by

(₄C₁ x ₁₃C₅)/ ₅₂C₅ = 0.001981 or 0.1981%

For part (b) we follow the same approach, but this is more complicated. In fact, in this case, we must recognize that

i) A pair can be formed in ₁₃C₁ ways. Once the card that forms the pair has been selected, we must choose 2 of the 4 cards with that particular number of them, i.e. ₄C₂.

ii) Then we have to choose other 3 cards from the other denominations, i.e. ₁₂C₃ and then for each of these cards we have ₄C₁ ways to select a card with that number of it.

Hence the probability of one pair is

(₁₃C₁ x ₄C₂ x ₁₂C₃ x ₄C₁ x ₄C₁ x ₄C₁)/ ₅₂C₅ = 0.4226 or 42.26%

For (c ) we have ₁₃C₂ ways to select what denominations will be used to form the 2 pairs, and once the two denominations are chosen, for each, we have ₄C₂ ways to select the two cards. The remaining card can be chosen from the remaining 11 denominations, i.e. ₁₁C₁ and then we have ₄C₁ ways to select the card from any of those denominations.

Poker Hand Probability Problems Worksheets

Thus the probability of exactly two pairs is

(₁₃C₂ x ₄C₂ x ₄C₂ x ₁₁C₁ x ₄C₁) / ₅₂C₅ = 0.0475 or 4.75%

We should now be sufficiently familiar with the arguments to use for part (d) and I will not provide all the details, but just the answer, i.e.:

(₁₃C₁ x ₄C₃ x ₁₂C₂ x ₄C₁ x ₄C₁) / ₅₂C₅ = 0.0213 or 2.13%

To find the probability of four of a kind, we have to select one denomination to use and then complete it with another card chosen among those the remaining 48 cards, i.e.

(₁₃C₁ x ₄C₄ x ₄₈C₁)/ ₅₂C₅ = 0.00024 or 0.024%