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#2
03-24-2005, 08:33 PM
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Re: What are the odds of 2 people flopping a flushin Hold\'em?
The probability of a monotone flop is 22/425 ~ 5.1764%.
Given that there is a monotone flop, the following is the distribution of the number of people with flushes (or straight-flushes, henceforth counted as flushes): 0 66.017% 1 29.877% 2 3.932% 3 0.172% 4 0.002% 5 0.000003% Given that you flop a flush (~ 1/505) the following is the distribution of the number of people with flushes, including you: 1 78.079% (no other flopped flushes) 2 20.552% 3 1.347% 4 0.022% 5 0.00004% The probability that you flop a flush is about 1/505. The probability at least one person flops a flush is about 1/56.8, so the probability that you flop a flush that is high is about 1/568. 1/505-1/568 ~ 1/4511 of the time, you flop a flush that is not high. 
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#3
03-24-2005, 08:43 PM
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Re: What are the odds of 2 people flopping a flushin Hold\'em?
[ QUOTE ]
The probability of a monotone flop is 22/425 ~ 5.1764%. Given that there is a monotone flop, the following is the distribution of the number of people with flushes (or straight-flushes, henceforth counted as flushes): 0 66.017% 1 29.877% 2 3.932% 3 0.172% 4 0.002% 5 0.000003% Given that you flop a flush (~ 1/505) the following is the distribution of the number of people with flushes, including you: 1 78.079% (no other flopped flushes) 2 20.552% 3 1.347% 4 0.022% 5 0.00004% The probability that you flop a flush is about 1/505. The probability at least one person flops a flush is about 1/56.8, so the probability that you flop a flush that is high is about 1/568. 1/505-1/568 ~ 1/4511 of the time, you flop a flush that is not high. [/ QUOTE ] nice answer, pzhon. i'm just going to point out that i believed you assumed random cards on the part of your opponent. this offers a useful baseline estimate, but more accurate approximations can be obtained by assuming more reasonable starting hand selection.
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#4
03-25-2005, 03:32 AM
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Re: What are the odds of 2 people flopping a flushin Hold\'em?
This calculation is made from outside view (by not taking into account own hole cards).
The flop is suited. The odds of one player (a specific one) to hold two cards of that suit are C(10,2)/C(49,2) = 45/1176= 3.826%. The odds of two players (two specific players) to hold two cards of same suit are C(10,2)*C(8,2)/(C(49,2)*C(47,2))= 0.099%. If you want the odds of two players (any) to hold that suit you have to multiply 0.099% by C(n,2)= n*(n-1)/2, where n is the number of players. If you want the odds for at least two players to hold that suit, the formula is much more heavy (maximum 5 players could simultaneously hold that suit).
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