3 Flopped Sets...

[RockLobster]

This happened to me tonite, so I have to ask... what are the odds?

Three of us see the [Q-7-6] flop. One of us had QQ, another 66, and I had 77. First time I can recall seeing 3 flopped sets...

[Copernicus]

How many players at the table?

[RockLobster]

[ QUOTE ]
How many players at the table?

[/ QUOTE ]

A full Party 2/4 table (10 players). Lots of betting (as you can imagine), no one improved. I can't ever recall having seen this before.

[Copernicus]

very roughly I get about 4400/1, although that seems a bet more likely than I would have thought.

The probability of any pair being dealt to a player is .0588. The probability of 3 players being dealt a pair is about

(.0588^3)*C(10,3)=14.66%. (I didnt force the other 7 to not have pairs, since there could have been more that folded to the pre-flop action. Also this is a little low because as pairs are dealt it becomes a little easier to deal another pair). So say 3 or more pairs will be dealt 15% of the time. A single pair has an 11.5% chance of flopping a set, so to flop 3 sets its roughly .115^3 (again thats a little low because of the existence of the other pairs/sets) or .15%. .15*.0015=.0000225 or 4443/1.

[RockLobster]

Hi C--

The probability of 3 players being dealt a pair is about (.0588^3)*C(10,3)=14.66%.

I must be misunderstanding this... 14 out of every 100 hold'em hands will include 3 pairs? That sounds high. Then assuming that 3 pairs are dealt, for each flop card to match one of the pairs sounds a lot less likely than 4400:1. But of course I haven't taken the time to try and figure this out, so I thank you for doing so [img]/images/graemlins/smile.gif[/img].

[ramjam]

I make the first approximation about 2.44%, not 14.66% [I think you might be out by a factor of 6 - perhaps you took P(10,3) rather than C(10,3)?].

That gives a slightly more plausible final estimate of 26,919 to 1.

[RockLobster]

That gives a slightly more plausible final estimate of 26,919 to 1.

Aah, that makes more sense to me. And when you add in the fact that there are times (maybe 50% or more) that people fold pairs preflop, you'd SEE this even less often.

Thanks both!

[Bozeman]

But the chance of the second (& third) person flopping a set is not independent of the other player flopping his set, so even your flop analysis is an overestimate.

This is probably wrong, but 1-44/46*43/45*42/44=12.8% chance that a q flops, 1-43/45*42/44=8.8% that a 7 flops given a q, and 1-42/44=4.5% chance that a 6 flops given the 7,q. Thus if q,7,6's see the flop, they will all get sets ~1 in 2000.

Craig

[ramjam]

I think your answer is very clever but a bit too clever for me! Also, I think there might be a slight problem in that your formula for a Q flopping is not for just one Q flopping but for one or more Qs flopping - and the same for the 7 formula. As you point out, Copernicus's estimate has a similar problem but to much a larger extent.

I think an easy way to calculate this piece of the puzzle accurately is as follows:
-first card on flop must be a Q, 7 or 6 (6 cards out of 46 unknown)
-second card must match one of the other two pocket pairs (4 out of 45)
- third card must match the final pair (2 cards out of 44).

This gives 6.4.2/(46.45.44) = roughly 0.05% or 1,896 to 1 (very close to your answer - not sure why the odds my way are slightly shorter - there must be some other factor at play).

Anyway if you combine this with the 2.44% estimate of 3 pocket pairs seeing the flop it takes you up to massive 77,000+ to 1!

[BruceZ]

Here's the exact solution.

It's much easier if you first consider the odds of getting any 3 different cards on the flop, that's
52*48*44/(52*51*50). Then multiply this by the probability that the 3 pairs are out that match the 3 cards on the flop. Note that no more than 1 player can hold each of the pairs and there are 3 ways to hold each one. The probability of 3 specific players holding these pairs is 9/C(49,2) * 6/C(47,2) * 3/C(45,2). Finally, there are C(10,3) ways to choose the 3 players. Putting it all together we have:

52*48*44/(52*51*50)*9/C(49,2)*6/C(47,2)*3/C(45,2)*C(10,3) = 0.0013% or 1 in 78,166.