Probability Question

In Texas hold 'em I was playing heads-up and went all in pre-flop with Ace-2. My opponent called and he had King-Queen Suited. They were diamonds, I had one diamond.

I made a bet with a friend watching beside me that by the time all five cards were dealt my opponent would make, at the very least, Two Pair with his hand.

The first card was a King and there was a Queen on the turn, so indeed I lost the game but my prediction came true. What were the odds of me winning that bet?

Quote: JohnnyBlue17

In Texas hold 'em I was playing heads-up and went all in pre-flop with Ace-2. My opponent called and he had King-Queen Suited. They were diamonds, I had one diamond.

I made a bet with a friend watching beside me that by the time all five cards were dealt my opponent would make, at the very least, Two Pair with his hand.

The first card was a King and there was a Queen on the turn, so indeed I lost the game but my prediction came true. What were the odds of me winning that bet?

Okay, well congratulations on winning the Prop Bet, but I certainly hope there were some Odds involved because the bet sure as Hell wasn't 50/50.

I have to ask a couple quick questions before I decide whether or not I want to attempt this or leave it for someone with better skills than I have.

1.) It seems from the language you want the probability of any Two Pair or better, or do you just want Two Pair?

2.) Did he have to use zero, one or both of his cards? (ex. If the Flop would have come J-J-3 and a Three came on the Turn, do you still win the bet?)

odds of winning if the money went in preflop: you with Ace of diamonds 54.92% vs 44.47%, tie 0.61%, you with 2 of diamonds 54.37% vs 45.06%, tie 0.57%

well lets look at the math of making at least 2 pair by the river


48 x 47 x 46 x 45 x 44 = combos of boards knowing your 2 hands
205,476,480

42 x 41 x 40 x 39 x 38 = combos of boards with no kings or queens on them at all
102,080,160

205,476,480 - 102,080,160 = 103,396,320 hands with at least 1 king or 1 queen or better

pair of kings 3 x 42 x 41 x 40 x 39 = 8,058,960 x 5 (for each of the 5 card spots on board it could be) = 40,294,800 possible just pair of king hands
pair of queens 3 x 42 x 41 x 40 x 39 = 8,058,960 x 5 (for each of the 5 card spots on board it could be) = 40,294,800 possible just pair of queen hands

now add them together
40,294,800 + 40,294,800 = 80,589,600

now that we know this number we subtract it from our 103,396,320 (hands with at least 1 king or queen or better) which will tell us what hands are now better than just a pair of kings or queens

103,396,320 - 80,589,600 = 22,806,720

so 22,806,720 combos will give you 2 pair or better when holding KdQd knowing you hold A2d

22,806,720 / 205,476,480 = 11.1% of time

Hope you got at least 9 to 1 on your bet vs your buddy or you would be the "donkey" here.

pokercruncher gives the following values for the KdQd player (given As2d is out). (250k hand sim)

HighCard: 17.2%
1Pair: 42.6%
2Pair: 22.8%
3ofaKind: 4.49%
Straight: 4.47%
Flush: 5.48%
FullHouse: 2.39%
4ofakind: 0.14%
Straightflush: 0.11%

This is taking the best 5 out of 7 cards regardless of where they came from.

Quote: Mission146

It seems from the language you want the probability of any Two Pair or better.

Right.

Did he have to use zero, one or both of his cards?

He didn't have to use either of his cards.

Quote: nezbit

22,806,720 / 205,476,480 = 11.1% of time

Hope you got at least 9 to 1 on your bet vs your buddy or you would be the "donkey" here.

Thanks for that, fascinating stuff. Could I just ask if that 11% includes straights, flushes etc? He didn't need any Kings or Queens if a Jack, 10, 9 came out for example.

My friend would never have accepted 9/1, he was very cautious because of how lucky my opponent is especially against me. Another guy declined the bet altogether.

JohnnyBlue,

I believe that Socks posted the numbers most pertinent to your information, because those numbers also include all hands not involving your opponent's cards and just appearing on the board. In any event, you were roughly a 60/40 dog to win the bet, just shy of one, so getting paid 3:2 would have put you at a very slight advantage, but you're a huge dog at even money.

i guess my math is invisible to mission LOL.

11% to make 2 pair or better the math is all laid out in front of you...60/40 wtf...

Quote: JohnnyBlue17

Quote: Mission146

It seems from the language you want the probability of any Two Pair or better.

Right.

Did he have to use zero, one or both of his cards?

He didn't have to use either of his cards.

Thanks for that, fascinating stuff. Could I just ask if that 11% includes straights, flushes etc? He didn't need any Kings or Queens if a Jack, 10, 9 came out for example.

My friend would never have accepted 9/1, he was very cautious because of how lucky my opponent is especially against me. Another guy declined the bet altogether.

i worked at the problems backwards. I figured out hands that couldnt be in the sample, then just subtracted those from total possible hands. so yes all the flushes str8, 2 pairs, trips, everything is in there

Quote: nezbit

...
i worked at the problems backwards. I figured out hands that couldnt be in the sample, then just subtracted those from total possible hands. so yes all the flushes str8, 2 pairs, trips, everything is in there

You threw out lots of flushes/straights on each subtraction, and never too on-board pairs/trips/boats/quads into account.