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#1
05-04-2004, 03:04 AM
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Two Boats on the Flop?
As a random exercise to kill time, I decided to try to calculate the probability that, in hold'em, given that two players see a flop, one of which has a pocket pair and the other has an unpaired hand dominated by the other player (e.g., 88 vs. 87), both players will flop a boat. My statistics are rusty, so I wanted to see whether my calculations are correct.
Given my example, in which one player holds 87 vs. his opponent's 88, the flop necessary to satisfy my condition is 8 7 7. There are only 4 cards left in the deck that can make the necessary 3-card flop: the 1 remaining 8 and the 3 remaining 7's. The probability of 3 of these 4 cards hitting on the flop, given 48 unseen cards, is: (4/48)*(3/47)*(2/46). Given these four cards, and letting the 3 remaining 7's be 7a, 7b, and 7c, respectively, there are four possible 3-card combinations: 7a 7b 7c, 8 7a 7b, 7a 8 7c, and 7a 7b 8. Thus the necessary flop will arise 3/4 times that 3 of the 4 necessary cards hit on the flop (i.e., when the flop isn't three 7's). Thus, the probability that, given the aforementioned hands, both players will hit a boat on the flop is as follows: (4/48)*(3/47)*(2/46)*(3/4) = 72/415104 = 3/17296 ~ 1/5765 ~ 0.0001735, or 0.01735%. I'm only posting this to see whether I'm calculating the probabilities correctly in the hopes that some of the posters here who have a better command of statistics than I will be courteous enough to respond. Thanks in advance. 
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#3
05-04-2004, 08:10 AM
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Re: Two Boats on the Flop?
If both players have a pocket pair and the flop is 3 of a rank then both players have made a full house.
Lost Wages
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#4
05-04-2004, 08:19 AM
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Re: Two Boats on the Flop?
One player has 88 and and the other 87. There are 48 cards available for the flop leading to C(48,3)= (48*47*46)/6 = 17,296 possible flops.
There are 1*C(3,2) = 3 flops of 877. 3/17,296 = 0.01735% So, yes you are correct. Lost Wages
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