December 1st, 2010 at 9:50:54 PM
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Sorry, I cannot wrap my head around the workings of odds and probability to answer this question, so I thought I'd ask the question here.

Last night in a sit n' go I saw what I assume has to be an extremely rare occurrence. With 8 players left, on two consecutive hands someone was dealt Qc Qd (I was dealt it on the second occurrence). Qh came out on both flops and Qs came out on both turns to make quads on both hands.

What are the odds that this would happen, particularly at an 8-player table?

Last night in a sit n' go I saw what I assume has to be an extremely rare occurrence. With 8 players left, on two consecutive hands someone was dealt Qc Qd (I was dealt it on the second occurrence). Qh came out on both flops and Qs came out on both turns to make quads on both hands.

What are the odds that this would happen, particularly at an 8-player table?

December 2nd, 2010 at 5:08:04 AM
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The probability that any particular player will be dealt those two particular hole cards is 2/52/51 = 0.00754 = 1/1326.

The probability that someone at an 8 player table will be dealt those hole cards is 1-(1-0.00754)^8 = 0.0060173 = 1/166.

The probability that the flop will contain Qh (given that someone holds those pocket queens) is 3/50 = 0.06.

The probability that the turn will be Qs is 1/47 = 0.02128.

The probability that all these things happen at one time = 1/166 x 3/50 x 1/47 = 1 in 130,000.

The probability that this would occur on two consecutive hands is 1 in 16.9 billion.

The probability that someone at an 8 player table will be dealt those hole cards is 1-(1-0.00754)^8 = 0.0060173 = 1/166.

The probability that the flop will contain Qh (given that someone holds those pocket queens) is 3/50 = 0.06.

The probability that the turn will be Qs is 1/47 = 0.02128.

The probability that all these things happen at one time = 1/166 x 3/50 x 1/47 = 1 in 130,000.

The probability that this would occur on two consecutive hands is 1 in 16.9 billion.

December 2nd, 2010 at 7:49:00 AM
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Doesn't the 1/47 factor imply the assumption that the Qs did not show up on the flop along with the Qh? I think it might have to be 1/49 instead, in order to have the exact sequence stated in the original post. Let me know if/how I figured this incorrectly.Quote:PapaChubby... The probability that the turn will be Qs is 1/47 = 0.02128. ...