K6 Prop Bet

I was playing 1-3 NLTHE at Aria and chatting with a fellow player who said there is a mysterious link between Kings and Sixes. He said whenever there is a K on the flop, a 6 is sure to come on the turn or river. Similarly he proposed that when a 6 flops, there will be a K on the turn or river. He wanted to bet on it so I took him up on it for fun. The terms were:

If there's a K and 6 BOTH on the flop, he wins $15.
If there's a K on the flop and a 6 on the turn or river, he wins $15.
If there's a 6 on the flop and a K on the turn or river, he wins $15.
If there's a K or 6 on the flop and the other card doesn't show up on the turn or river, I win $5.
I can't win the bet unless we see the river card. So if a K flops but everyone folds to a flop-bet, our prop bet is off because we didn't get to see the turn & river.

Knowing a little about the math of Hold'Em, I figured when a 6 or K shows up on the flop, he only has 4 outs so I'm a 5-to-1 favorite to win the bet. And since I'm only paying him 3-to-1, I should make money on this prop bet & be entertained while patiently waiting for good hands. Not so!

I finally cried UNCLE! when I was down $65!

What are the true odds of this prop bet?

Thanks,
Brandon

You can only get paid if hand goes to the river..ouch. I'll leave the math to the experts. I'm assuming he gets paid on his end regardless of if hand goes to river?

Quote: slyther

You can only get paid if hand goes to the river..ouch. I'll leave the math to the experts. I'm assuming he gets paid on his end regardless of if hand goes to river?

Prop bet is off unless hand goes to the river (according to the post).

If I may simplify this, let's say you lose $15 if both a K and 6 appear on the board, and you win $5 if one or the other appears. Also assume that every hand goes to the river. In that case I show your expected win per bet resolved is $1.56. I know that isn't quite what you asked, but let's see if we can agree on the simple case first.

Yikes. Given a K or 6 flops, he actually has 4 outs for the other one on four streets (the two flop cards, turn and river)! The odds of the other one not appearing is 71.4%. You are already behind 72 cents on this alone (for an edge of 4.8% on your $15).

Then, he can win the bet if they both flop or turn, but the river doesn't show - while you can't win! Depending on how loose or tight the table is, this will hurt even more. That is, given a K or 6 appeared, you won't even win 71.4% of the time, because you won't always see the river to prove you would have won. So your actual win % is less, and edge higher.

The fair thing to do would be NO ACTION either way unless you see the river. Then the fair bet is very close to his $10 to your $25, but that is still 0.1% in his favour.

EDIT: I seem to disagree with the Wiz on the 1st part. I get OP wins 47/51*46/50*45/49*44/48 given that one card on the board is a K or 6. Am I cutting corners?

Quote: Wizard

If I may simplify this, let's say you lose $15 if both a K and 6 appear on the board, and you win $5 if one or the other appears. Also assume that every hand goes to the river. In that case I show your expected win per bet resolved is $1.56. I know that isn't quite what you asked, but let's see if we can agree on the simple case first.

The original question doesn't say what happens if you get a K and a 6 on the turn and the river only. Your simplified game calls this a $15 loss. So, is it a $15 loss, a push, or a $5 win?

Quote: thecesspit

Prop bet is off unless hand goes to the river (according to the post).

I'm not so sure. OP said turn/river legs of bet is off if all fold on the flop. I think we need clarification from OP.

The way I read it he has to pay if there is K-6-x on the flop regardless if the other streets are seen. Likewise I would think he pays if the board runs K-x-x-6 then all fold on turn.

That's how I read it. That arrangement is bad for the OP.

Clarification from "OP" :)

K x 6 flop only, I lose
K x x 6 no river, I lose
K x x no turn, bet is off
K x x x no river, bet is off
K x x x x, I win

I hope that clears it up. Thanks for all the replies!

Brandon

What am I missing? I get an advantage for the OP, assuming all hands go out to the river.

(In each case, * represents a card that is neither K nor 6)

The flops with both K and 6 are:
K K 6 (24 different flops (4 choices for the K, 6 for the pair))
K 6 6 (again, 24 different flops)
K 6 * (704 different flops (4 choices for the K, 4 for the 6, 44 for the other card))
That is 752 different flops where the player loses 15

The flops with one or more Ks but no 6s are:
K K K (4 different flops)
K K * (264 different flops (6 choices for the pair; 44 for the other card))
K * * (3784 different flops (4 choices for the K; (44 x 43) / 2 = 946 choices for the other two cards))
6 6 6 (4 different flops)
6 6 * (264 different flops (6 choices for the pair; 44 for the other card))
6 * * (3784 different flops (4 choices for the 6; (44 x 43) / 2 = 946 choices for the other two cards))
That is 8104 different flops with one card but not the other
In each of these cases, the probability of not seeing the other card (i.e. not getting a 6 with a K in the flop, or a K with a 6 in the flop) in the turn or river = 45/49 x 44/48, and the probability of seeing the other card = 1 - (45/49 x 44/48).
The EV for each flop = (45/49 x 44/48) x 5 - (1 - (45/49 x 44/48)) x 15 = 1.8367347.
The total EV over all flops = 1.8367347 x 8104 - 15 x 752 = about 3605. Since there are 22,100 distinct possible flops, the EV per flop that reaches the river = about 0.163.

Aren't there more distinct possible flops? 52 x 51 x 50 = 132,600

Also, is your math correct only assuming every hand goes to the river? Many times my "buddy" won the bet when the K or 6 fell on the turn & we never saw the river.

Quote: bbdsan

Aren't there more distinct possible flops? 52 x 51 x 50 = 132,600

Also, is your math correct only assuming every hand goes to the river? Many times my "buddy" won the bet when the K or 6 fell on the turn & we never saw the river.

I treat, for example, As 2h 3d the same as 3d 2h As - there are (52 x 51 x 50) / 6 = 22,100 distinct flops.

As for the assumption that every hand goes to the river, I think I misread an earlier post that said that the other player's advantage was 0.1% "even if the bet was 25 against 10 instead of 15 against 5".

If we assume 90% of flops reach the turn, and 90% of these (i.e. 81% overall) reach the river:
Chance of losing on the turn = 9/10 x 4/49 = 36/490
Chance of losing on the river = 9/10 x 45/49 x 9/10 x 4/48
Chance of winning = 9/10 x 45/49 x 9/10 x 44/48
The total EV over all 22,100 flops is now -116.33.

I don't follow poker that closely, but I am under the impression that the percentage of hands where everybody but one player folds pretty much right after the flop is greater than 10%; the higher the percentages go, the lower the player's EV gets.

Thank you for explaining the number of distinct flops. I've been trying to figure that out. I saw the formula online (n!/k!(n-k)!) but I just wasn't getting it. You have to divide by 6 because there are 6 possible orders for the same 3 cards, and of course the order doesn't matter. 6 different flops are actually all the same flop.

As for the other part of your post, this confirms what I suspected all along. The "trick" to this prop bet is that we have to reach the river for me to win. I don't know how accurate it is to say that 90% of flops reach the turn and 90% of turns reach the river. But it doesn't seem unreasonable. And look at that EV! Seems like it was a sucker bet for me & I should have called it off much sooner!

Thanks again,
Brandon

If/when someone sells this as a prop bet, I suggest the name "Royal Six."

And a cut of the take. Thank you.

LOL so far I'm the only sucker I know that's taken the bad end of the bet. Now that I know the math I'll push it a lot harder on my fellow players. "Royal Six" I like it!

Perhaps there really is a "strange" connection between the two cards... Was the dealer' nickname, 'wrench"?

Quote: bbdsan

LOL so far I'm the only sucker I know that's taken the bad end of the bet. Now that I know the math I'll push it a lot harder on my fellow players. "Royal Six" I like it!

I expect 10% as royalty for the name. I'll sue otherwise :P

Glad you like it. It seems both obvious and misleading, which should fulfill the requirements of any sucker/side bet ;)