Is this roughly the calculation for flopping set over set in a nine handed game of texas hold em assuming anyone with a pair sees the flop? Also, what is the probability of flopping an under set when set over set is flopped?

ASSUME: you hold 10-10 and your opponent holds Q-Q.
(We make no assumptions regarding the distribution of other cards to other
players at the table.)

This leave 48 cards unaccounted for. Therefore, based on our assumption,
there are C(48,3)=17296 equally likely flops possible. (Note: C(n,k) = the
number of possible combinations of k cards from n cards. The formula is
n!/(k!(n-k)!).)

In order for both players to flop sets, the flop must include one of the
two remaining queens, one of the two remaining tens, and one of the other
44 cards in the deck. The number of ways such a flop can occur is
2*2*44=176.

Therefore, the probability of set-over-set occurring, given our initial
assumption, 176 in 17296. This is about 1.02%, or roughly one in 98.

(Note: this rules out the possibility of the flop bringing quads vs a full
house. This is extremely unlikely - the probability of a TTQ flop, given
the above assumptions, is a very remote 2 in 17296, or about 0.0116%; the
probability of a QQT flop is, of course, the same.)

That is the odds for two opponents flopping sets (but not quads) when they both start with pocket pairs and both see the flop, at any size table. That is not the same as the odds of set-over-set when you simply assume all players with pocket pairs will see the flop, as you have to multiply this probability by the chance of two pocket pairs being dealt at an X-handed table. That's not as important a question anyway, since what you really want to know is the chance of being beat by another set when you have one. 98:1 is the answer if you know opponent started with a pocket pair.

Okay. On average should you flop an over set 50% of the time?

So you are asking that when set-over-set occurs, will your own set will be higher 50% of the time?

If you have top set, it will be best 100% of the time. If you have bottom set, it will lose 100% of the time. If you have middle set, it will be best when opponent has bottom set and worst when opponent has top set. So I suppose you could deduce that in set-over-set flops you will win them 50% of the time in the long run. But you won't play enough poker in your entire life for that to matter.

As mentioned before, you have to multiply the 98:1 odds by the chance that two players at an X-handed table are dealt pocket pairs AND by the chance that they both see the flop. The chance that two players at a 10-handed table are dealt pocket pairs of different ranks is about 9.5% (if same rank there is no set possibility). Then, for reasons I analysed in another thread, we can suppose that 24% of the time they will both see the flop (use any number you like better here).

Thus 98:1 x 24% x 9.5% = 4377:1
or once in 4378 deals at a 10-handed table, you might see set-over-set at the flop between two opponents at that table. If we assume you will be one of those players 1/10th of the time, then you should encounter this scenario once in about 44,000 hands played.

With a situation that happens that rarely, the long run odds don't matter to your winnings at all. Bottom line is if you flop a set, be willing to put all your money in. Even if you have bottom or middle set and the board is not straightening or flushing, still put all your money in. Don't get hung up on worrying about getting beat by set-over-set. It happens so rarely that it's practically irrelevant, and if you just shove your stack every time you have any set, you'll come out way ahead.

Happened twice to me in 1300 hands of six max. Ouch.

I have just experienced a run of under 100 hands where I flopped and underset 3x to another player at full ring. Talk about "tail" of a distribution.

OK, so we know the formula for calculating set over set...but what about set over set over set. It happened to me today on Stars and have been searching for the probability of that happening. Anybody know what the probability is or how to calculate it?

Here's the hand fwiw,

What were the poker Gods thinkin' when this went down?

PokerStars 25c/50c FR Limit

Pre Flop: (1.5sb) Hero is in BB with AA

UTG raises, 2 folds , MP 3!, 3 folds, SB calls, Hero CAPS, UTG calls, MP calls, SB calls

Flop: (16sb) JAK

SB checks, Hero bets, UTG calls, MP calls, SB folds

Turn: (19sb) 8

Hero bets, UTG raises, MP 3!, Hero CAPS, UTG calls, MP calls

River: (21.5bb) 6

Hero bets, UTG calls, MP calls


FINAL POT: 24.5bb

UTG mucks KK
MP mucks JJ
Hero shows AA (Three of a kind, Aces)
Hero wins pot.


Set over set over set! Never seen it before. Today turned out to be +75bb. Man, has this month turned around. June graphs thread?

Odds of flopping set over set over set, assuming you know that all 3 players have pocket pairs preflop and that each player does not know each others hands would be 1 to (50^3)/(2^3). Equaling, 1:15625.

If 3 players have pocket pairs, then there are 46 cards unaccounted for, and there are C(46,3) = 46*45*44/(3*2*1) = 15,180 possible flops. There are 2 cards of each of the 3 ranks that make sets for a total of 2*2*2 = 8 flops that make set over set over set, so the probability of these 3 players all flopping a set is 8/15,180 or odds of 1896.5-to-1.

The probability of set over set over set at a 9-handed table, with all pairs seeing the flop, is exactly

C(13,3)*4³ / C(52,3) *
C(9,3) * 9/C(49,2) * 6/C(47,2) * 3/C(45,2)

=~ 1 in 111,666

The top line is the probability of 3 different ranks on the flop. The second line is the number of ways to choose the 3 players C(9,3), times the probability that each player has a different one of these 3 ranks. This is exact since only 3 players can have these 3 sets at one time.

The trick here is to consider the flop first. If you consider the players first, then the more pocket pairs that are out, the more likely it is that 3 players flop sets, so this becomes complicated. The above calculation avoids that.

Playing on Full Tilt Rush poker 3 tables at a time. I got it 11 times this week. And 3 times today. I won 2/11. I just started saving hand history the other day so i will dig through them to find the hands. I see about 600 hands per hour at 3 tables. And I've played about 18 hours this week. 600 X 18 = 10,800 hands. so I shouldve only saw it once. Yet i'm some how still up for the week.

Does that make me lucky in a way? for beating the odds?

Well I have no doubt that at least one was against me. I have lost to set over set at least 6 times in the last week playing Rush (maybe more - I don't feel like digging through the HH). I've been playing about 2500 hands per day and found this thread while searching for just how bad my luck has been lately. I'm on a sick downswing (10 buy ins or so), and have experienced some of the ugliest coolers of my life.

As soon as I make Ironman, I'm taking a break. I was destroying the Rush tables for the first 2-3 weeks, but I've given it all back now, so I guess it's time to reevaluate.

Cheers-

How do you get this 9.5%?

The first pocket pair at a 10-handed table is

1 - (1-(78/1326))^10 = 45.46%

The second pocket pair of a different rank, given that the first one happened, is

1 - (1 - (72/1225))^9 = 42.03%

The product of those is 19.1%, which is the correct answer. So I have no idea where I got 9.5% in the earlier post. It was wrong.

These are mutual-exclusive approximations but are very accurate. You could use inclusion exclusion to get the exact answer and it would be within about a tenth of a percent or so.

First of all, they are independent approximations, not mutually exclusive. Secondly, if you had used a mutually exclusive approximation, you would have quickly seen that the answer can be no bigger than 15.6%:

C(10,2)*78/C(52,2)*72/C(50,2) =~ 15.6%

Thirdly, if you were going to use an independence approximation, the following one would have given an even tighter upper bound of about 14.4%, though it takes some thought to realize that this is still an UPPER bound:

1 - [1 - 78/C(52,2)*72/C(50,2)]^C(10,2) =~ 14.4%

Fourthly, this turns out to be one case where the independence approximation is not very accurate as the exact answer is about 11.3%.

The exact calculation uses a variation of inclusion-exclusion where the adjustment factors are not just +/- 1, since we are starting with 2 players. Each line of inclusion-exclusion has multiple terms which each require their own adjustment factor which are shown in bold below. Getting these adjustment factors right takes some thought. The rest isn't difficult. Here are the first 4 terms:

C(10,2)*78*72 /C(52,2)/C(50,2) -

C(10,3)*(2*78*72*66 + 1*78*72*1*3) /C(52,2)/C(50,2)/C(48,2) +

C(10,4)*(3*78*72*66*60 + 2*78*72*66*1*C(4,2) + 1*78*1*72*1*3) /C(52,2)/C(50,2)/C(48,2)/C(46,2) -

C(10,5)*(4*78*72*66*60*54 + 3*78*72*66*60*1*C(5,2) + 2*78*72*1*66*1*15) /C(52,2)/C(50,2)/C(48,2)/C(46,2)/C(44,2)

=~ 11.3%


The first line is for 2 players, the second line is for 3 players, etc. The first summed term on the second line is for 3 different pairs. These are counted C(3,2) = 3 times by the first line, hence the adjustment factor subtracts 2. The second term is for 3 players holding 2 different pairs. These are counted twice by the first line, hence the adjustment factor subtracts 1.

The first term on each line is for N players having N different pairs, starting with N=2 on the first line, N=3 on the second line, etc. The second term on each line is for N players holding N-1 different pairs starting with N=3 on the second line. Starting with N=4 on the third line, the third term is for N players holding N-2 different pairs.

Finally, the question asked by the OP concerns set over set. For that problem, if you consider the flop first, then the above probability is not needed. It becomes an easy matter to make an exact calculation involving 2 or 3 players holding 2 or 3 specific pairs. See my earlier post in this thread with an example for 3 sets.

Bruce - I believe this is a difference between EXACTLY two players have a pair, and AT LEAST two players have a pair, which satisfies the reason for calculating it, and so that is what I gave although I should have been more clear. You gave the answer for EXACTLY two players, which doesn't really help us estimate how often two players will see the flop holding pairs, which was my original intent.

There's no doubt this is a close correct approximation for some player getting a PP at 10-handed:
1 - (1-(78/1326))^10 = 45.46%

For someone else having a pair (excluding same rank) I used the same approximation method and got:
1 - (1 - (72/1225))^9 = 42.03%

where Alspach calculated this the (very) long way as 42.06%. So that's about where this approximation should fall relatively and I believe it is correct for another pair at the table.
http://www.math.sfu.ca/~alspach/comp35/

So this gives us the chance that AT LEAST one other player has a pair, not that EXACTLY one other player has a pair, and I'm thinking that difference is the entire difference between my answer and yours.

All that leaves is the combined probability of the two events. It seems we should just multiply to get that answer, which gives 19.1% for at LEAST two players holding a pair. Agree?

There's no reason the independent approximation should be much off at all in this case, and I don't think it is as long as we define what we are calculating.

No, I computed the probability of AT LEAST 1 pair.

I showed you trivially that this probability can be no greater than 15.4%, and this is also for AT LEAST 1 pair.

I see no reason why your independence approximation should be accurate at all.

EDIT: Removed calculation of exactly 1 pair.

But yet it is accurate for another player having a pair. Something is wrong with my combined probability then.

Is there some reason we would not simply multiply the two probabilities together in this case? I'm confused. Apparently multiplying the chance one player has a pair, times the chance another player at the table has a pair, is wrong in this case but I'm not sure why. I know that given I have a pair, the chance another player has one is the number I show.

So once I have the chance some player has a pair, and the chance for another player to have one given the first, what is the correct way to combine those here? Is that just a wrong approach? Is there a shortcut?

PS - I know your ~11.3% is the correct answer. I just can't figure out why mine is so far off.

You want:

P(1 or more pairs)*P(2 or more pairs | 1 or more pairs)

but you are computing:

P(1 or more pairs)*P(2 or more pairs | player 1 has a pair)


These are not the same because the second probabilities are very different, even though they may sound similar. In the first case, the second probability is

(# hands with 2 or more pairs) / (# hands with 1 or more pairs)

In the second case, the second probability is

(# hands with 2 or more pairs AND player 1 has pair) / (# hands where player 1 has pair)


Notice that the numerator in both cases is dominated by hands with exactly 2 pairs, and the numerator in the second case contains exactly 1/5 of these. Now what about the denominators? Most of these are hands with only 1 or 2 pairs, and there will be more hands with 1 pair than with 2 pairs. The denominator in the second case contains exactly 1/10 of the hands with only 1 pair, and exactly 1/5 of the hands with exactly 2 pairs. So the denominator in the second case is not 1/5 of the denominator in the first case; it is between 1/10 and 1/5 and closer to 1/10. So comparing the 2 fractions, the denominator scales from something between 1/10 and 3/20, and the numerator scales by about 1/5, so the second fraction is larger by a factor of between 1.33 and 2.

By the inclusion-exclusion calculation, the second probability should be about 11.3% / 45.46% =~24.9%. Compare this to the 42.03% which you used. This is different by a factor of about 1.7, in agreement with the above analysis. Offhand I know of no shortcuts to getting the 11.3% other than the inclusion-exclusion calculation that I've shown.

I recomputed the probability of EXACTLY 2 pairs by changing the adjustment factors in the inclusion-exclusion method, and this gave about 9.5%, which coincidentally is the number you had posted originally.

I don't think it was coincidence, I just don't remember how I got it. I probably looked up someone else's calc on it.

Thanks for the explanation, makes sense.

How about the odds of set over set (I had the higher one) with us both beaten by the flush draw that gets there on the goddamn Barry Greenstein Ace On The River©?

I'd say the odds are slim, but... this kinda crap happens to me daily.

Yeah yeah, so he had correct odds to chase his draw, but the friggin board coulda paired (prob giving my other opponent quads LOL).

Poker is a stupid friggin game.

Here's a somewhat decent approximation to the 11.3% figure using independence only to compute the probability that I get a pair and someone else gets a different pair (accurate), and multiplying this by 5 since I take part in 1/5 of the 2 pair cases. This will be slightly high since the cases of more than 2 pairs mean I take part in more than 1/5 of the 2+ pair cases.

P(I get pair) * P(someone else has different pair) * 5

1/17 * [1 - (1 - 72/1225)^9] * 5

=~ 12.4%

this just happened to me in my final table of a 6 handed tornament 4 people. gaaaaay

Was playing donkey limit poker with friends at the AC Showboat around Pres Day '09 and saw set over set over set on the flop...it was like a spoon in a microwave. I'd have laid bottom set down by the turn, but all three went capped to the river.

Think I figured the flop at like 2200-1.

If you know that your opponent started with a pocket pair then these odds are way off. Assuming that I flopped a set, then the opponent, who we know has a pocket pair, will have two cards to also flop a set

2/49+(47/49*2/48)= 8.08%

8.08% ~ 11.5:1

i.e. not nearly as uncommon as your fuzzy math suggests