Texas Hold'em Solutions

Since you have two hearts in your hand and their are two hearts in community cards there are nine remaining hearts among the 47 unknown cards. The probability of the turn card being a heart is then 9/47 &asymp 0.19. The probability of the turn card not being a heart but getting a heart on the river is (38/47) · (9/46) &asymp 0.16. Therefore the total probability of getting a flush is approximately 0.19 + 0.16 = 0.35. We can also do this problem by calculating the probability of not getting a flush. The chance of not drawing a flush on either the turn or the river is (38/47) · (37/46) &asymp 0.65, so the probability of getting a flush is 1 - 0.65 = 0.35, the same value calculated above.

In the second situation there are 10 remaining hearts among the 47 unknown cards. You need to draw a heart on both the turn and the river, so the probability of getting a flush is (10/47) · (9/46) &asymp 0.04.

The only way that you can end up with four of a kind is by drawing eights on the turn and river. There are only two eights left in the deck, so this occurs with probability (2/47) · (1/46) &asymp 0.0009.

There are many different types of full houses that you could draw in this situation. We calculate the probability of each type and sum the results to find the total chance of getting a full house.

• Three 8's, Two 1's. This requires getting an eight and a one as the last two cards. The chance of drawing an eight and then a one is (2/47) · (3/46) &asymp 0.0028, and the probability of drawing a one and then and eight is (3/47) · (2/46) &asymp 0.0028. So this full house occurs with probability 0.0056.
• Three 8's, Two 7's or Three 8's, Two 2's. By a calculation identical to the one above, these both occur with probability 0.0056.
• Three 1's, Two 8's. This requires getting a one on both the turn and the river, which occurs with probability (3/47) · (2/46) &asymp 0.0028.
• Three 7's, Two 8's or Three 2's, Two 8's. These are both analogous to the three 1's, two 8's situation and occur with probability 0.0028.

Therefore, the total probability of getting full house is 3 · (0.0056) + 3 · (0.0028) = 0.0252.

Only a 5 will complete your straight, so you have four outs and a (4/46) &asymp 0.087 probability of winning. You are considering a raise of $2 to a $20 pot giving a 1/10 ratio. Therefore pot odds say you should not call. However, if you hit your straight you will earn an additional $2 from each of your opponents for a total pot of $24. The ratio of the cost of your call to this potential pot is 2/24 &asymp 0.083 which is less than your 0.087 probability of winning, so implied odds say that you should make the call.

To calculate the frequency of four of a kind, first note that there are 13 different ranks in which you can get four of a kind. For any given rank, the possible hands that give four of a kind in that rank all include the four cards of that rank as well as any three additional cards. There are C_48,3 = 17,296 different ways of choosing these three additional cards, so we have a total of 13 · 17,296 = 224,848 different four of a kind hands. This gives a frequency of (224,848/133,784,560) = 0.0017.

To find the frequency of straight flushes, sort all straight flush hands by the high card of the highest straight flush in the hand. For ace high straight flushes in any of the four suits you need the A - K - Q - J - 10 of the given suit and then any 2 of the remaining 47 cards. This gives a total of C_47,2 = 1,081 distinct hands. For straight flushes that are not ace high the same argument holds except that one of the remaining 47 cards would give you higher straight flush if it were in your hand (for example, if you have 10 - 9 - 8 - 7 - 6 in hearts, if one of your two other cards was a jack of hearts you would have a jack high straight flush). Therefore, in these cases there are only C_46,2 = 1,035 distinct straight flush hands. So the total number of straight flush hands is (1,081 · 4) + (1,035 · 4 · 9) = 41,584 hands (the nine in the second parenthesis comes from the fact that there are nine different possible non-ace high cards for straights - a 2,3, or 4 high straight can not occur). The corresponding frequency is then (41,584/133,784,560) = 0.00031.

To count the number of full house hands, we divide up the types of full houses by looking at the two cards that are not used as part of the final hand. These two cards can either be a pair (but of a different rank than the triple or the pair you are using for the full house, or else you would have four of a kind), one of the two cards could be of the same rank as your pair (giving you two triples and one card of some different rank), or the two cards could be of different ranks from each other, the triple, and the pair.

• We first consider the case of the unused cards being a pair. We can choose the rank for the triple in 13 ways. Once a rank is chosen we can pick the three cards for the triple in C_4,3 = 4 ways. We can then choose the two ranks for the two pairs in C_12,2 = 66 ways. For each pair, once we have chosen the rank we can choose the cards for the pair in C_4,2 = 6 ways. So we have a total of 13 · 4 · 66 · 6² = 123,552 full house hands of this type.
• Now we consider the case of two triples. We can choose the ranks for the triples in C_13,2 = 78 ways, and for each triple we can then choose the cards for the triple in C_4,3 = 4 ways. There are then 44 remaining cards from which to choose the last card of the hand, so we have a total of 78 · 4² · 44 = 54,912 hands of this type.
• Finally we consider the case of two cards of different rank from each other, the triple, and the pair. As above, the cards for the triple can be chosen in 13 · C_4,3 = 52 ways and the cards for the pair can then be chosen in 12 · C_4,2 = 72 ways. We can choose the two ranks for the remaining two cards in C_11,2 = 55 ways, and for each rank we can choose any of the four cards of that rank. This gives a total of 52 · 72 · 55 · 4² = 3,294,720 hands of this type.

Therefore, we have a total of 3,473,184 full house hands. This gives a frequency of (3,473,184/133,784,560) = 0.02696.

For additional calculations, as well as the frequencies for 5-card poker hands (which tend to be significantly easier to calculate), see for example Wikipedia.