Flopping a boat, losing to quads, 2 hands in a row

[_Towelie_]

Is there a way of determining the probability of this happening? This happened to a friend of mine yesterday and he was wondering. I don't have the suits. Hero got it all-in good on the flop in hand 1 and the turn in hand 2.

Hand 1

Hero: 77
Villian: 33

Board: JJ7 3 3

Hand 2:

Hero: KK
Villain: JQ

Board: KQQ 3 Q

[jay_shark]

1) The probability that villain hits two perfect cards is
1/45c2

2) The probability villain hits the queen on either the turn or the river is 1/45+ 1/45 =2/45

[_Towelie_]

I appreciate the response but I'm a math donk and the answer doesn't make sense to me. Is it possible to express this in a single percentage value?

[jay_shark]

maybe this will help .

1) The turn can be any of the 2 three's . The probability of this is 2/45 . If the turn is one of the three's , then the river must contain the last 3 available . This is just 1/44 . Now just multiply to give you 2/45*1/44 = 0.0010101 or 0.101%.

2) Here is an exercise . Someone gives you two cards face down from a pile of 52 cards . You are to determine the probability that either of the two cards is the ace of spades .
The answer should be 1/52+1/52 =2/52 . Does that make any sense to you ? Now apply this to part 2) and use the queen of some fixed suit to give you 2/45 =0.0444 or 4.44%

[_Towelie_]

Thanks a lot, this helps.

[bigpooch]

I think the replies only gave you the probabilities of the
villain making quads, which doesn't reflect the true equity.

In Hand 1, the villain can also "chop" on running jacks
with a probability of 1/C(45,2) = 1/990.

In Hand 2, the villain was all-in on the TURN, so the odds
of the villain winning were 1/44. If the probability
calculation is made postflop, there is the unlikely event of
a "bad beat" where the hero also makes quads: after the
flop, the chances are just 1/C(45,2) = 1/990.