Lower pocket pair vs. higher pocket pair preflop in Hold 'Em
11-10-2011
, 10:30 PM
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.
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.
11-11-2011
, 02:01 AM
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?
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?
11-11-2011
, 02:32 AM
Quote:
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?
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?
1 - (1 - 72/1225)9
=~ 42.0%
Real close to the exact answer of about 41.9%.
11-12-2011
, 06:35 PM
Quote:
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%.
1 - (1 - 72/1225)9
=~ 42.0%
Real close to the exact answer of about 41.9%.