# Texas Hold'em: Probabilities for Pocket Pairs

When playing pocket pairs in Texas Hold'em, you must be able to accurately estimate the percentage of times you don't flop a set but an overcard comes to your hand.

The below table is a good one to memorize and gives you an understanding of how strong a given pocket pair really is and how different pocket pairs compare.

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 Pocket Pair Probability of Not Hitting the Flop % AA 0 KK 20.67 QQ 37.84 JJ 51.83 TT 62.94 99 71.53 88 77.92 77 82.43 66 85.39 55 87.12 44 87.96 33 88.22 22 88.24

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Adrian Deacon 2012-05-05 23:00:07

I agree with Dean. However are discrepancies based on position negligable?

Dean 2010-09-07 20:38:55

My maths could be wrong, but I'm rather sure that's totally incorrect. I get 25% that you are the only one holding an ace:

If you have an ace, there are still 3 aces in the remaining 50 cards. So the remaining 9 people (18 cards) have to get cards other than aces.

For the first card (of the 18 cards your opponents have): 47 out of the 50 are not an ace, for the second one 46 out of 49, etc. This gives:
(47*46*...*30) / (50*49*...*33) = 0.253 = 25,3% chance that no-one else got an ace.

Sean Lind 2010-01-19 20:41:42

Paul,

The math is pretty much what you got going on, it works out to 77.4%.

7.7 X 10

That's not actually correct, but it's close enough to cheat.

-s

Paul 2010-01-19 08:10:46

These statistics are interesting and definitly good to know, but other than simply giving perspective, are there any practical applications (bet sizing, position in which to play the hands, etc...) to these numbers?

Also, I was watching the WSOP on tv and they dropped this stat on me (seen it a couple times) that I found interesting and wanted yall to confirm/deny it. It said that if, at a 10-handed table you are dealt an ace, that there's a 70% chance that you are the only player holding an ace. I tried to do the math on my own but had some problems coming up with an exact number....4/52 = 7.7%, 4/51, 4/50, etc...?

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