odds of set over set

[muckyouraces]

anyone know the odds of getting a set in holdem and getting beat by a higher set?

I'm not talking about when the board pairs and you have trips and get beat by someone with a higher kicker, I am only talking about when you have a pocket pair and someone else has a pocket pair, and you both hit a set. this has happened 3 days in a row and seems really odd

[therockofgibraltar]

I have had 7 set over sets in the past 20000 hands (VPIP ~22). Normal or unlucky?

[AaronBrown]

If you have a pair of Kings in a ten player game, there is a 4.4% chance that another player has a pair of Aces. If you have a pair of Queens the odds are almost twice as high that another player has a higher pair (it's not quite twice as high due to the probability of more than one player having higher pocket pairs, but for this problem it's a good approximation to assume the odds are proportional to the number of pairs that beat your pair).

Given that two players have different pairs, the odds that the board will contain exactly one of each card is 3.1%. Some of these boards will make straights or flushes, or have three of a kind on the board.

Multiplying the two together gives 0.14% times the number of cards that beat your pair.

[Overdrive]

Well, check this paragraph out from Phil Gordon's Little Green Book, page 82:

"When I flop a set, I never worry about my opponent having a bigger set. If both players start with a pocket pair, set over set will happen after the flop only about one out of one hundred times. Against those odds, I'm willing to risk going broke."

[MrWizard]

get this i had 3 set over sets in 2 hours yesterday and another today. I am on the worst run of my life.

[Riorin]

[ QUOTE ]
Well, check this paragraph out from Phil Gordon's Little Green Book, page 82:

"When I flop a set, I never worry about my opponent having a bigger set. If both players start with a pocket pair, set over set will happen after the flop only about one out of one hundred times. Against those odds, I'm willing to risk going broke."

[/ QUOTE ]

That's what I read also, and I think Phil may want to rewrite that chapter if he ever played on Crypto as I get sucked out 1-2 times per day (me having lower set) on average (in 1-2hours of play) at their site!

My top reason for believing they are rigged!

[Alcibadies]

Hey aaron. Will you explain where I went wrong? Thanks.

The probability of just exactly 2 players geting distinct PPs is:
13(4,2)/(52,2)*12(4,2)/(48,2) *100 =0.346%

At a 9 player table you can estimate the probability of 2 players out of 9 getting a distinct PP by [C(9,2)*0.00346]*100= 12.456%

The probability of a player flopping a set is 11.76%. The probability of 2 players who start with a PP flopping a set is (0.1176^2)*100= 1.38%


The probability on any give hand that 2 players will flop a set is ~ (0.12456*0.0138)*100= 0.17%

[BruceZ]

[ QUOTE ]
Hey aaron. Will you explain where I went wrong? Thanks.

The probability of just exactly 2 players geting distinct PPs is:
13(4,2)/(52,2)*12(4,2)/(48,2) *100 =0.346%

[/ QUOTE ]

The (48,2) should be (50,2), but your number is right.

[ QUOTE ]
At a 9 player table you can estimate the probability of 2 players out of 9 getting a distinct PP by [C(9,2)*0.00346]*100= 12.456%

[/ QUOTE ]

Note that Aaron's 4.4% = 9*6/C(50,2) is the approximate probability that one of 9 opponents at a 10-player table has a particular higher pocket pair when YOU have a pocket pair. You are computing the probabilty that ANY 2 players at a 9-player table both have distinct pocket pairs, which is a very different thing.


[ QUOTE ]
The probability of a player flopping a set is 11.76%. The probability of 2 players who start with a PP flopping a set is (0.1176^2)*100= 1.38%

[/ QUOTE ]

You can't just square 11.76% because the 2 sets are not independent. When one player flops a set, the probability that the other player flops a set will be smaller than 11.76% because there are only 2 remaining flop cards. The probability that both pairs flop sets is:

2*2*44/C(48,3) =~ 1.02%

OR

4/48 * 2/47 * 44/46 * 3 =~ 1.02%

where we multiply by 3 since the non-set card can come in any of 3 positions. This does not include full houses/quads.

Aaron did this for the whole board, not just the flop. This gives:

2*2*C(44,3)/C(48,5) =~ 3.1%.

This includes boards that pair to make full houses for both players, but not boards that pair to make a full house for one player and quads for the other. These would add another 2*2*1*C(44,2)/C(48,5) =~ 0.2%, half of which would allow the smaller pair to regain the pot.


[ QUOTE ]
The probability on any give hand that 2 players will flop a set is ~ (0.12456*0.0138)*100= 0.17%

[/ QUOTE ]

~(0.12456*0.0102)*100= 0.13%

[Alcibadies]

Thanks Bruce. At least I'm getting closer.

[BruceZ]

Note that Aaron is computing the probability that another player at a 10-player table beats you with set over set by the RIVER when YOU have a pocket pair, and for this he approximates the probability that another player has a particular pocket pair higher than yours as 9*6/C(50,2) =~ 4.4%, and he multiplies this by the 3.1% probability that you and the other player both make a set by the river to get 0.14%, and this number is meant to be multiplied by the number of denominations higher than your pair. You are computing the probability that ANY 2 players at a 9-player table FLOP a set, which is a completely different thing that just happens to come out to almost exactly the same number (~0.13%).