Texas Hold 'Em Odds Question
February 18th, 2015 at 8:57:43 PM
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Hi Wizard and everyone else. Long time visitor to the site, but new to the forums.
Just playing some Texas hold em on Bovada right now and earlier I got dealt 3 different pocket pairs in a row. What are the odds of this? Thanks for the great site!
Just playing some Texas hold em on Bovada right now and earlier I got dealt 3 different pocket pairs in a row. What are the odds of this? Thanks for the great site!
February 18th, 2015 at 9:28:52 PM
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Quote: grambus12345Hi Wizard and everyone else. Long time visitor to the site, but new to the forums.
Just playing some Texas hold em on Bovada right now and earlier I got dealt 3 different pocket pairs in a row. What are the odds of this? Thanks for the great site!
This is an easy one, so I hope you don't mind me grabbing it.
Okay, the probability of being dealt any pair initially is:
(1 * 3/51) = 3/51 or 1 in 17
Now, you want to be dealt three DIFFERENT pairs on three consecutive hands, so now we need to figure out the probability of getting dealt a pair y while excluding pair x:
(48/52 * 3/51) = 0.05429864253
Okay, now you want to be dealt another pair that is not pair y or pair x:
(44/52 * 3/51) = 0.04977375565
(3/51) * (0.05429864253) * (0.04977375565) = 0.00015897925 or 1/0.00015897925 = 1 in 6290.12905772
The probability of getting three consecutive hands of inside pair, in general, is 1 in 4913, by the way.
https://wizardofvegas.com/forum/off-topic/gripes/11182-pet-peeves/120/#post815219
February 18th, 2015 at 9:36:57 PM
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Thank you very much!
February 18th, 2015 at 9:46:32 PM
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Quote: grambus12345Thank you very much!
No need to thank me, my work is its own reward.
Just being silly, you're welcome!
https://wizardofvegas.com/forum/off-topic/gripes/11182-pet-peeves/120/#post815219
April 29th, 2015 at 5:34:27 PM
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Here is a slight (perhaps picky) correction:
For a particular three hands in a row (say the first three hands you play), the probability of being dealt three different pocket pairs is exactly:
132/830297, which is 1 in 6290 17/132.
Calculating these numbers to eight significant digits results in
0.00015897926, which is one in 6290.1288.
This last number, when extended to more digits, is 6290.12878787..., with the 87 part repeating indefinitely.
But the OP's question was about the odds, not the probability. The odds against this happening are therefore 6289.12878787... to 1.
Of course if you play, say, 100 hands in a row, the odds against it happening in every possible set of three consecutive hands are much less.
For a particular three hands in a row (say the first three hands you play), the probability of being dealt three different pocket pairs is exactly:
132/830297, which is 1 in 6290 17/132.
Calculating these numbers to eight significant digits results in
0.00015897926, which is one in 6290.1288.
This last number, when extended to more digits, is 6290.12878787..., with the 87 part repeating indefinitely.
But the OP's question was about the odds, not the probability. The odds against this happening are therefore 6289.12878787... to 1.
Of course if you play, say, 100 hands in a row, the odds against it happening in every possible set of three consecutive hands are much less.
