Gordon Pair Principle

[Mr. Orange]

http://sports.espn.go.com/espn/poker...amp;id=2665855

Saw this on the ESPN site today. The article just contains the chart but doesn't really say how he arrived at these numbers. Just wondering if someone can give me the math to back this up. Just one example is fine.

[Siegmund]

The 50 unseen cards can make 1225 different two-card hands. Of these 1225, there are 6 ways to make each overpair.

In the heads-up case, Phil has just rounded off 6/1225 to 1/200 or half a percent per rank.

When there are two or more players left to act, the expected number of overpairs goes up in proportion to the number of players. There is a small chance of two players both having an overpair on the same deal, and these are being double-counted by Phil's method of just multiplying by N. That is, his formula is slightly conservative in the 1-player case (1/200 instead of ~1/204) and slightly more conservative in the multiplayer cases.

A nice rule of thumb, I wonder why it hasn't ever made it into print before.

[jay_shark]

This isn't too hard to figure out if you're familiar with the principle of inclusion/exclusion .

For instance , suppose you have pocket queens utg and you're at a ten handed table . What is the probability one of your nine opponents has kings or aces ?

Here is the formula of inclusion/exclusion for two events A and B :

P(AUB) = PA + PB - P(AB)

AB is the event that both A and B occur

for three events :

P(AUBUC) = PA + PB + PC -PAB -PAC - PBC +PABC

In general

P(1u2u3u4u5u6u7u8u9)= P(i) {sum 1 to 9} - P(ij) {sum i<j} +P(ijk){sum over all i<j<k} - P(ijkl) {sum over i<j<k<l}

p(1) is the probability player 1 has aces or kings
P(12) is the probability player 1 AND 2 have aces or kings
P(123) is the probability player 1,2,3 have aces or kings
P(1234) is the probability player 1,2,3,4 have aces or kings .

For instance 12= {AcAd,AhAs},{AcAd,KhKs},{AcAd,KhKd},{AcAd,KhKc} ,{AcAd,KsKd},{AcAd,KsKc},{AcAd,KdKs}

The total number of ordered elements in (12) is 7 as i've listed them all above .

You can use a similar type of argument to figure out 123 or 1234 .

P1=P2=...P9 = 12/50C2
P(12) = 12*7/50C2*1/48C2
P(123)= 12(18)/50C2*1/48C2*1/46C2
P(1234) = 12*18/50C2*1/48C2*1/46C2*1/44C2

You must also remember that the number of ways of selecting 2,3,4 players from 9 is 9C2,9C3,9C4 respectively .

The final probability is the following :

9*12/50C2- 9C2*12*7/50C2*1/48C2 +9C3*12*18/50C2*1/48C2*1/46C2 - 9C4*12*18/50C2*1/48C2*1/46C2*1/44C2 = 8.6 %

Now try it yourself with pocket jacks :P

[jay_shark]

According to his chart , if you hold pocket deuces at a ten handed table then someone will get a higher pocket pair at least 54 % of the time . This is actually significantly off as the real probability should be around 38 % of the time .

This result comes from the inclusion/exclusion formula where he approximates P(1u2u3...ui) = i*p1

where i is the number of players at your table

[Copernicus]

[ QUOTE ]
According to his chart , if you hold pocket deuces at a ten handed table then someone will get a higher pocket pair at least 54 % of the time . This is actually significantly off as the real probability should be around 38 % of the time .

This result comes from the inclusion/exclusion formula where he approximates P(1u2u3...ui) = i*p1

where i is the number of players at your table

[/ QUOTE ]

some conservatism in the formula cant hurt since the really larger errors occur when more more bad things can happen when they dont have overpairs but still call

[jay_shark]

It turns out with pocket 2's there is a 42 % chance someone will hold a higher pocket pair .

[punter11235]

Wow. Very nice formula.

[SA125]

Good article. What is the 4 and 2 rule?

[jay_shark]

If you have x outs on the flop , then the probability you'll hit one of your outs with two cards to come is 4x%.

If you have x outs on the turn , then the probability you hit one of your outs is 2x % .

These are all close approximations .