Thanks to statmanhal for sharing this problem with me. And thanks to the smart "cat" who helped me understand it.
Quote:
Hero holds 33. The flop is Q93. Two villains saw the flop.
Given that they only play pocket pairs, what's the probability that at least one of them has a set?
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