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Old 11-10-2011, 11:46 AM   #1
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Lower pocket pair vs. higher pocket pair preflop in Hold 'Em

Let's take a worst case scenario: I'm playing a 10-handed Hold 'Em game. I'm UTG with 22. There's 9 opponents and 12 higher pockets to worry about.

What are the odds someone has a higher pocket?

I'm no math whiz, so is there a simple calculation I could use at the table? I'd like to do a bunch of scenarios and see what it would be with like 66 vs. 5 opponents, JJ vs. 7 opponents, 99 vs. 4 opponents, etc. Also, let's ignore the odds of being against lower pockets for now. If I had say 66, I just want to know what are the odds someone has 77+, don't worry about 55 and down.
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Old 11-10-2011, 09:30 PM   #2
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Re: Lower pocket pair vs. higher pocket pair preflop in Hold 'Em

Wikepedia has a table but it uses an approximation that has been shown to be not very good for the higher number of opponents. Brian Alspach has calculated this exactly on his web page but I have found errors for even number of opponents.

Phil Gordon had a simple approximation formula on one of his web sites some time back but I haven't been able to find it now. It wasn't too bad for many cases, as I recall. I may have copied the formula somewhere and will post it if I can locate it.

The following table is based on a simulation program I wrote. One million runs for each case were done and the results track very well with other published data.

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Old 11-10-2011, 11:38 PM   #3
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Re: Lower pocket pair vs. higher pocket pair preflop in Hold 'Em

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Originally Posted by ShortyTheFish View Post
Let's take a worst case scenario: I'm playing a 10-handed Hold 'Em game. I'm UTG with 22. There's 9 opponents and 12 higher pockets to worry about.

What are the odds someone has a higher pocket?
Here's my exact calculation of that very problem, written on the talk page of wiki. As you can see, it agrees with statmanhal's 41.9%. It also has the general equation. I will be generating a complete table soon.

http://en.wikipedia.org/wiki/Talk:Po...iple_opponents
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Old 11-11-2011, 01:01 AM   #4
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Re: Lower pocket pair vs. higher pocket pair preflop in Hold 'Em

So basically, even in a worst case scenario of 22 UTG at a full table, there's still roughly a 60% chance no one else has a pocket. Easy game!

I have another question though: I understand there's 1225 preflop combos and 72 ways for an opponent to make a higher pocket than 22 (12 higher pockets with 6 combos each). So that's a 5.88% chance for one guy to have a higher pocket. So why not just multiply that by the number of opponents? I mean, they all have an equal chance to get a pocket, so why not just multiply 5.88% by the number of opponents (9 in this case, making it 52.9% as opposed to 41.9%).

What am I missing from the equation?
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Old 11-11-2011, 01:32 AM   #5
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Re: Lower pocket pair vs. higher pocket pair preflop in Hold 'Em

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Originally Posted by ShortyTheFish View Post
So basically, even in a worst case scenario of 22 UTG at a full table, there's still roughly a 60% chance no one else has a pocket. Easy game!

I have another question though: I understand there's 1225 preflop combos and 72 ways for an opponent to make a higher pocket than 22 (12 higher pockets with 6 combos each). So that's a 5.88% chance for one guy to have a higher pocket. So why not just multiply that by the number of opponents? I mean, they all have an equal chance to get a pocket, so why not just multiply 5.88% by the number of opponents (9 in this case, making it 52.9% as opposed to 41.9%).

What am I missing from the equation?
That over counts the times multiple opponents have a higher pair. That's why we use inclusion-exclusion for the exact calculation. If you want a good approximation, use independence. It will generally get you to within a few tenths of a percent for higher pairs.

1 - (1 - 72/1225)9

=~ 42.0%

Real close to the exact answer of about 41.9%.
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Old 11-12-2011, 05:27 PM   #6
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Re: Lower pocket pair vs. higher pocket pair preflop in Hold 'Em

I found the Phil Gordon approximation. The following is on one of his blogs:

"The Gordon Pair Principle

Let C = percent chance someone left to act has a bigger pocket pair Let N = number of players left to act Let R = number of higher ranks than your pocket pair (i.e., if you have Q-Q, there are two ranks higher. If you have 8-8, there are six ranks higher)

Then, C = (N x R) / 2"

Gordon's formula is okay for higher pairs (say 9 or higher) and 5 or fewer players. It will usually get you to within 1%. It can be way off otherwise. For example, 22 with 9 opponents, the formula gives 54% vs. correct 42%. But it is very simple and easy to remember and apply.
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Old 11-12-2011, 05:35 PM   #7
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Re: Lower pocket pair vs. higher pocket pair preflop in Hold 'Em

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That over counts the times multiple opponents have a higher pair. That's why we use inclusion-exclusion for the exact calculation. If you want a good approximation, use independence. It will generally get you to within a few tenths of a percent for higher pairs.

1 - (1 - 72/1225)9

=~ 42.0%

Real close to the exact answer of about 41.9%.
By "for higher pairs", I mean for the probability that someone has a higher pair, not just that it only works well for high ranking pairs. It works well for any pairs and any number of players 2-10.
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Old 11-12-2011, 11:45 PM   #8
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Re: Lower pocket pair vs. higher pocket pair preflop in Hold 'Em

wow i was just gonna ask something along these lines. I have had a brutal 2 days Fri and Sat. lost with pocket 9's twice to pocket 10's and pocket aces, and lost with pocket kings to pocket aces, all three games were tourneys. Know I'm wondering how much of a fish i am for not folding these hands pre flop. And i would really appreciate some insight on how to figure these out at the live table. I'm just tired now and will check back later, just got back from the last tourney when my pocket kings lost to pocket aces, how big of a fish am i?
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