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View Full Version : flush over flush or set over set - which is rarer?

Big Bend
08-26-2005, 01:24 AM
Which am I less likely to see? Each player has 2 suited cards of the same suit and 3 appear on the board.. or each player has pocket pair and both get a set? Assuming everybody stays to the river. Just curious... I'm always amazed when I see either occurance.

allen

kiddj
08-26-2005, 09:56 AM
Although both are relatively rare, set over set is much less likely to happen. My calculations show that flush over flush is about 4 times more likely, but this could be reduced slightly by the fact that people are more likely to play a pocket pair than any 2 suited.

aloiz
08-26-2005, 10:20 AM
Given that two players have suited hole cards of the same suit and they both see the river, the odds that they both make the flush:

C(9,3)*C(39,2)/C(48,5) + C(9,4)*39/C(48,5) + C(9,5)/C(48,5) =~ 0.0393 or about 24.4:1

Given that two players have different pocket pairs, and they both stay to the river, the odds that both make at least a set including the possiblity that one or both hit quads:

2*2*C(44,3)/C(48,5) + 2*C(44,2)/C(48,5) + 44/C(48,5) =~ 0.0321 or about 30.2:1 against.

Please note that this is not a good way of comparing the frequency that you will observe these types of hands at a table, as you are starting off by assuming that you have two players with pocket pairs or two players with suited cards. The odds that you have two players with two pocket pairs is much less than the odds that you have two with suited cards. (Of course this is somewhat offset by the fact that people will play a lower percentage of suited hands compared to pocket pairs). Also you are assuming that the both players stay to the river. The best way to get accurate real-world results would be to query a large poker tracker database.

aloiz

kiddj
08-26-2005, 10:44 AM
[ QUOTE ]
Given that two players have suited hole cards of the same suit and they both see the river, the odds that they both make the flush:

C(9,3)*C(39,2)/C(48,5) + C(9,4)*39/C(48,5) + C(9,5)/C(48,5) =~ 0.0393 or about 24.4:1

Given that two players have different pocket pairs, and they both stay to the river, the odds that both make at least a set including the possiblity that one or both hit quads:

2*2*C(44,3)/C(48,5) + 2*C(44,2)/C(48,5) + 44/C(48,5) =~ 0.0321 or about 30.2:1 against.

Please note that this is not a good way of comparing the frequency that you will observe these types of hands at a table, as you are starting off by assuming that you have two players with pocket pairs or two players with suited cards. The odds that you have two players with two pocket pairs is much less than the odds that you have two with suited cards. (Of course this is somewhat offset by the fact that people will play a lower percentage of suited hands compared to pocket pairs). Also you are assuming that the both players stay to the river. The best way to get accurate real-world results would be to query a large poker tracker database.

aloiz

[/ QUOTE ]
In my calculations (that i'm too lazy to provide), I assumed heads up from the start of the deal. So my results included the begining probabilities of 2 people being dealt all the same suit or each getting a different pocket pair.

just wanted to clarify. /images/graemlins/smile.gif

AaronBrown
08-26-2005, 01:31 PM
I agree with kiddj for the heads up version if 3.6 counts as "about 4." If you just deal two hold'em hands and the board, there are 2,781,381,002,400 possible combinations of cards. This is C(52,2)*C(50,2)*C(48,5).

Flush over flush can be done in four suits. Hand 1 can be set up C(13,2) ways, hand 2 C(11,2) and the board flush cards C(9,3). We have to divide by 2, because the two hands are indistinguishable. The other two board cards can be chosen in C(39,2) ways. This doesn't count cases with four and five flushes on the board. It does count cases where one or both players have straight flushes, but that's the only had better than a flush either one can have. There are 1,068,107,040 possibilities, so the chance is 0.000192 or 1 in 5,208.

Now with 10 players, you can multiply this by about 45 and say you would see this about every 116 hands if every suited hand stayed in until showdown. As kiddj says, a lot of suited hands are folded preflop. Even postflop, unless at least two of the flush cards are there, and sometimes when they are. Since both flushes can't hold the high outstanding card of the suit, and usually neither will, there could be a fold after making the flush but before showdown (although most players are so overjoyed at filling that they underweight the possibility of a higher flush). Two pair or three of a kind on the board will also scare some flushes. Given all of this, I'd expect to see this at showdown about one in 1,000 hands or so.

For three of a kind, there are C(13,2) ways to choose the two ranks, and 6 ways to pick suits for each player. There are then four ways to choose the suits of the matching cards for the board. The remaining three cards can be any of the 44 non-matching cards, C(44,3). Some of these could form hands higher than three of a kind. Multiply these all out and you get 148,756,608 ways for this to happen. Divide by 2,781,381,002,400 as before to get 0.000053 or 1 chance in 18,698 making the double flush about 3.6 times more likely.

If you multiply by 45, you get once in 416 hands. Given that most pairs will see the flop, and that sets are likely to stay in, I think that you'll see about the same 1 in 1,000 as I estimate for double flushes.