Thanks to statmanhal for sharing this problem with me. And thanks to the smart "cat" who helped me understand it.

Easy, right? There are 66 pair combos remaining, 6 of which are sets.

P(at least one set) = P(villain A set) + P(villain A miss)*P(villain B set) = 6/66 + (60/66)(6/61) = 11/61 ≈ 18.033%

/thread...but first let's see what the solution looks like via the dumb long method of, "Count the combos and divide by all the combos"

Applying Bayes' Theorem: P(set | both have pocket pairs) = P(set AND both have pp's) / P(both have pp's)

P(set & both pp's) = P(both sets) + P(set & non-set pp) = (9 + 6*60) / C(47,4) / 3
P(both have pairs) = 369 + C(10,2)*6^2 + 10*3 / C(47,4) / 3

Plugging it in, you get 369/2019 = 123/673 ≈ 18.276%


Huh?? Where was the mistake in either of those solutions?

Let's see what PIE says: P(set) = P(villain A set | pocket pair) + P(villain B set | pp) - P(both set | both pp's)
P(villain A set | pp) = P(villain B set | pp), so their sum is 2*6/66 = 2/11

Notice that 2/11 < 123/673, so it looks like we can rule out the 2nd answer. Hopefully after subtraction, this 3rd solution will match the 1st answer.

We already calculated P(both set | pp) = 9/2019, so after subtracting that from 2/11 you get 1313/7403 ≈ 17.736%


Oh wow, so three different sensible solutions all get different answers! What's going on here? Is the answer none of the above?

I'll let you bang your heads on this for a while before spoiling it

funny thing is, when I tried to calculate this, I got one more answer that isn't either one of yours.

if hero holds 33 and they only play pocket pairs, this is what they can play:
10*6 for ranks 2,4,5,6,7,8,T,J,K,A = 60 combos of non-sets on the flop
2*3 for ranks 9,Q = 6 combos of sets on the flop
for rank 3, it is not possible that there is a pair out there

so the probability that someone has a set with this line of thinking would be
1 - 60C2/66C2 = 0.1748 = 17.48%

edit: this can't be correct, right now it came to my mind that they can't both have a set of 99 for example, which is counted with the above method.

Juk

@Juk - Your machine is correct. But what's wrong with the other solutions?

Juk

TC nailed it. The key when using the 1st or 3rd method is that we don't want P(A has set | A has pocket pair); we want P(A has set | both players have pp)

TC obtained that probability via what amounts to a shortcut of the 2nd method (ie, using the reduced sample space instead of the full sample space which would get canceled out anyway as it did in the calculation I showed).

There is another way to get that probability: solve p = p(3/63) + (1-p)(6/61)

See why?

The result is p = 63/673 which is the reduced form of TC's 378/4038. Once you have p, methods 1 & 3 get the right answer.

Of course the 2nd method was right because it assumed nothing, counted everything, and thoroughly applied Bayes (and for these reasons, sidestepped the trickiness of this problem). My other calculations in the OP both rested on the same false assumption. When "good" calculations get bad results, there's a false assumption somewhere.

What an interesting thread! Thank you heehaww and all.

Probability they both have the same pocket pair is 40/45 x 3/44 x 2/43 x 1/42

Probability they have two different non set pocket pairs is 40/45 x 3/44 x 36/43 x 3/42

Probability they both have sets is 6/45 x 2/44 x 3/43 x 2/42

Probability that one has a set while other one has a pocket pair is 6/45 x 2/44 x 40/43 x 3/42, times two, since it could be the second guy with a set.

Since all denominators are the same you just compare numerators.

No sets are 240 plus 12960 =13200

Sets are 72 plus 2880 =2952

At least one set is 2952 out of 16,152 (13200 +2952) = 18.276%

My working:

6 pair combos per card number
13 card numbers
= 78 pair combos

subtract pocket 3s = 72 pair combos
- one 9 = 3 combos of 99
- one Q = 3 combos of QQ
= 66 pair combos

Chance of set for 1 player given pair = 6/66

Chance neither player has set = 60/66 x 59/65
chance at least one has set = 1 - (60/66 x 59/65)
= 1 - (3540/4290)
= 750/4290
= 0.17 (2 sf) or 17%

EDIT: I didn't allow for overlapping ranges. ie. if player one has KhKd then there is only one KK combo left. So for when neither player has a set, whatever pair the first player has, it removes 5 hand combos, not 1.

So chance neither player has set = 60/66 x 55/65
chance at least one has set = 1 - (60/66 x 55/65)
= 1 - (3330/4290)
= 990/4290
= 0.23 (2 sf) or 23%

Nope, still forgot to subtract overlap in the denominator...

So chance neither player has set = 60/66 x 55/61
chance at least one has set = 1 - (60/66 x 55/65)
= 1 - (3330/4026)
= 696/4026
= 0.17 (2 sf) or 17%

... I may give up soon. Can anyone show me where I'm going wrong?

You meant for the red to be 61, which would have yielded an answer of 11/61, which still isn't quite right but is close. How you arrived at it is normally solid reasoning, but for this problem it rests on a false assumption.

Namely, you were using the probability that a player gets a set given that that player has a pocket pair. Instead, you need to use the probability given that both players have pocket pair. That's the subtle thing that makes this problem easy to get wrong.

I finally got back to looking at this seriously and came up with another way, though it is essentially what TC did.

Hero flops a set: Flop has two over-cards: If two villains see the flop with pairs, what is the probability at least one has an overset?

If both villains have the same pair, they can’t have a set. Let NC(E) be the number of combos for event E for the two villains:

NC(Vs have same pair) = 10*3*1 = 30

NC(Vs have different pairs) = NC(both Vs set by matching the two flop over-cards) + NC(only one V sets ) + NC(neither V sets)

= 3*3 + 2 *(3*10*6) + C(10,2)*6*6 = 9+ 360 + 1620 = 1989

NC(Total) = 30 + 1989 = 2019

Of the total combos, 369 are oversets, so P(overset) = 369/2019 = 0.1827