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#1
03-25-2007, 01:47 AM
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math question
can you show the math for this question:
what is likelyhood if you have 74 that you will have 2 pair or better (not straight) by the turn? 
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#2
03-25-2007, 01:53 AM
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Re: math question
is probablity against
45/50 * 44/49 * 43/48 * 42/47 ???
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#3
03-25-2007, 02:01 PM
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Re: math question
[ QUOTE ]
is probablity against 45/50 * 44/49 * 43/48 * 42/47 ??? [/ QUOTE ] No, but if you want the probability of getting at least 2 of the six cards {7,7,7,4,4,4} on the board by the turn (which would give you 2-pair, trips, full house, or quads), this would be 1 minus the probability of NONE of these cards, minus the probability of exactly 1 of these cards. The probability of no 7 or 4 is (44/50 * 43/49 * 42/48 * 41/47). The probability of exactly one 7 or 4 is (6/50 * 44/49 * 43/48 * 42/47 * 4), where we multiply by 4 since the 7 or 4 can come on any of the 4 cards. All together, the probability of at least 2 of these cards on the board is 1 - (44/50 * 43/49 * 42/48 * 41/47) - (6/50 * 44/49 * 43/48 * 42/47 * 4) =~ 6.6% or 1 in 15.3 We can also do this using combinations. The number of 4 card boards with NO 7s or 4s is C(44,4), and the number of boards with exactly ONE 7 or 4 is 6*C(44,3) since there are 6 ways to get a 7 or 4, times C(44,3) ways to get the remaining 3 cards out of 44 non-sevens/fours. So all together this is 1 - C(44,4)/C(50,4) - 6*C(44,3)/C(50,4) =~ 6.6% or 1 in 15.3 Note that this does not include the hands 2-pair or better which do not contain at least 2 cards of {4,4,4,7,7,7} on the board.
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