UTH Quad Kings x 3 within 30 hands... Odds
April 25th, 2012 at 8:48:01 PM
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Just got back from vegas trip .. Very unusual
Occurance at UTH table at Red Rock this last
Saturday
Within two hours, a player at same seat made
Quad kings, wirh pair kings in her hole card
THREE times......
I'm guessing this was a one in five million (or more )
Occurance.... (maybe 30 hands, 15/ hr full table)
Dose anyone have a simple calculation?
Occurance at UTH table at Red Rock this last
Saturday
Within two hours, a player at same seat made
Quad kings, wirh pair kings in her hole card
THREE times......
I'm guessing this was a one in five million (or more )
Occurance.... (maybe 30 hands, 15/ hr full table)
Dose anyone have a simple calculation?
May 12th, 2012 at 10:52:33 PM
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why no interest in this math question?
May 12th, 2012 at 11:25:27 PM
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I'll take a stab at it. I'm going to assume you want the probability of it happening regardless of whether the hand wins or loses the showdown.
If the rank matters (i.e. must be kings), then:
The probability of being dealt pocket kings = combin(4,2)/combin(52,2) = 0.004525
The probability of a pocket pair turning into quads = combin(2,2)*combin(48,3)/combin(50,5) = 0.008163
The combined probability (p) = 0.004525 * 0.008163 = 0.0000369379
The probability of this happening 3 times in 30 hands is (p3) * (1-p)27 * combin(30,3) = 0.000000000204413 = 1 in 4,892,064,710
If the rank doesn't matter (i.e. first quads were aces, second quads were nines, third quads were fours), then:
The probability of being dealt a pocket pair = combin(13,1)*combin(4,2)/combin(52,2) = 0.058824
The probability of a pocket pair turning into quads = combin(2,2)*combin(48,3)/combin(50,5) = 0.008163
The combined probability (p) = 0.004525 * 0.008163 = 0.000480
The probability of this happening 3 times in 30 hands is (p3) * (1-p)27 * combin(30,3) = 0.000000443751 = 1 in 2,253,518
If the first quad can be of any rank, but the two subsequent quads must match the first quad's rank, then:
pa = the p value from the first paragraph above
pb = the p value from the second paragraph above
The probability of it happening once and then happening twice more with the same rank, all within 30 hands, is:
pb * pa2 * (1-pa)27 * combin(30,1) * combin(29,2) = 0.000000103637 = 1 in 9,649,043
I'm not 100% certain about that last figure; something in there doesn't sit well with me, so take it with a grain of salt. I'm pretty sure the first two figures are accurate though.
If the rank matters (i.e. must be kings), then:
The probability of being dealt pocket kings = combin(4,2)/combin(52,2) = 0.004525
The probability of a pocket pair turning into quads = combin(2,2)*combin(48,3)/combin(50,5) = 0.008163
The combined probability (p) = 0.004525 * 0.008163 = 0.0000369379
The probability of this happening 3 times in 30 hands is (p3) * (1-p)27 * combin(30,3) = 0.000000000204413 = 1 in 4,892,064,710
If the rank doesn't matter (i.e. first quads were aces, second quads were nines, third quads were fours), then:
The probability of being dealt a pocket pair = combin(13,1)*combin(4,2)/combin(52,2) = 0.058824
The probability of a pocket pair turning into quads = combin(2,2)*combin(48,3)/combin(50,5) = 0.008163
The combined probability (p) = 0.004525 * 0.008163 = 0.000480
The probability of this happening 3 times in 30 hands is (p3) * (1-p)27 * combin(30,3) = 0.000000443751 = 1 in 2,253,518
If the first quad can be of any rank, but the two subsequent quads must match the first quad's rank, then:
pa = the p value from the first paragraph above
pb = the p value from the second paragraph above
The probability of it happening once and then happening twice more with the same rank, all within 30 hands, is:
pb * pa2 * (1-pa)27 * combin(30,1) * combin(29,2) = 0.000000103637 = 1 in 9,649,043
I'm not 100% certain about that last figure; something in there doesn't sit well with me, so take it with a grain of salt. I'm pretty sure the first two figures are accurate though.
May 12th, 2012 at 11:32:58 PM
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tnxs JB.......I quoted a prob of 1 in five to ten million when it happened ....it was just guess....... but it “sits right" with me.......ur calculations are awesome.
