View Full Version : 3 Flopped Sets...

09-08-2003, 01:53 AM
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...

09-08-2003, 06:41 AM
How many players at the table?

09-08-2003, 08:24 AM
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.

09-08-2003, 03:57 PM
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.

09-08-2003, 04:05 PM
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 /images/graemlins/smile.gif.

09-08-2003, 04:17 PM
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.

09-08-2003, 04:48 PM
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!

09-08-2003, 06:53 PM
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.


09-08-2003, 08:09 PM
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!

09-08-2003, 09:36 PM
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.

09-08-2003, 10:13 PM
Thanks for straightening me out.

Exact answer for my method is 1961.9:1.

You are right about the error of counting multiple queen flops, but this suggests my method would give an overestimate, so there must be an additional error since the actual probability is higher.

The answer is that if you makes the caveat that there is only one queen, the chance of you having a 7 is higher.

Call me an idiot, but I spent the time to figure out how to do this the hard way, correctly.

P(exactly one queen)=3(possible permutations)*2/46(a queen)*44/45(not the last queen)*43/44(not the last queen either)=12.46%

P(exactly one 7 given exactly one queen)=2*2/44(choosing from cards that are not qqqq7766)*42/43(any card that is not qqqq777766 from all cards not qqqq77766)=8.88%

P(other card is 6 given q7)=2/42(already disallowed it being a q or 7)=4.76%

So P(q76 on the flop)=P1*P2*P3=6/46*44/45*43/44*4/44*42/43*2/42=6*4*2/(46*45*44) check


09-09-2003, 06:37 AM
Thanks everyone!