FatBabyKate
FatBabyKate
Joined: Jul 6, 2015
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July 6th, 2015 at 7:30:38 PM permalink
Tonight at our holdem game one of the players offered me this bet.... He
wanted to bet $20 on every flop til one of us cried "UNCLE"!!

His bet was that a 2-4 or a J would come on the flop! THATS IT!

I didn't accept because I learned long ago not to take another mans bet.
I tried to do the math on this to see who had the best of it but couldnt do it.

I'm pretty certain that he has the best of this bet. But how much of an edge
does he have? Can someone show me the math on this?

I would also be interested to hear if anyone has any similar bets like this that
I might could try to pick up a few extra bucks on while playing poker.

Thank you for any help that anyone can give me.
beachbumbabs
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beachbumbabs
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July 6th, 2015 at 9:43:46 PM permalink
Quote: FatBabyKate

Tonight at our holdem game one of the players offered me this bet.... He
wanted to bet $20 on every flop til one of us cried "UNCLE"!!

His bet was that a 2-4 or a J would come on the flop! THATS IT!

I didn't accept because I learned long ago not to take another mans bet.
I tried to do the math on this to see who had the best of it but couldnt do it.

I'm pretty certain that he has the best of this bet. But how much of an edge
does he have? Can someone show me the math on this?

I would also be interested to hear if anyone has any similar bets like this that
I might could try to pick up a few extra bucks on while playing poker.

Thank you for any help that anyone can give me.



I'll take a stab at this. But the math guys can please check it.

Without regard to people's 2card hands, since there's no way of knowing what anybody holds any given hand, but they are dealt before the flop:

2's, 4's, and J's are 12 out of 52 cards. A bit confused about how you wrote 2-4; could mean 2 through 4, 2,4 or requiring both a 2 and a 4 OR a J.

Assuming you meant individually...

12/52 = .230769, or 23% of the time.
12/51 assuming you miss = .235294
12/50 assuming you miss = .24

so you have to add those together, since he gets 3 chances to hit.

.706063, or 70.61% of the time he will win.

Total sucker's bet IMO.
If the House lost every hand, they wouldn't deal the game.
MathExtremist
MathExtremist
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July 6th, 2015 at 9:50:22 PM permalink
I read it as "the flop will have at least one of 2, 3, 4, J; I'll bet you even money that I'm right." So in order to win, the other player has to see all 3 flop cards from 5, 6, 7, 8, 9, T, Q, K, A. 9 ranks, 36 cards, 36c3 = 7140. Total number of flops is 52c3 = 22100. 7140/22100 = 32.3% chance to win, 67.7% to lose. Total sucker bet.

(FWIW, adding isn't the right operation there.)
"In my own case, when it seemed to me after a long illness that death was close at hand, I found no little solace in playing constantly at dice." -- Girolamo Cardano, 1563
beachbumbabs
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beachbumbabs
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July 6th, 2015 at 9:51:16 PM permalink
Quote: MathExtremist

I read it as "the flop will have at least one of 2, 3, 4, J; I'll bet you even money that I'm right." So in order to win, the other player has to see all 3 flop cards from 5, 6, 7, 8, 9, T, Q, K, A. 9 ranks, 36 cards, 36c3 = 7140. Total number of flops is 52c3 = 22100. 7140/22100 = 32.3% chance to win, 67.7% to lose. Total sucker bet.

(FWIW, adding isn't the right operation there.)



Thanks, ME. Always learning, here.
If the House lost every hand, they wouldn't deal the game.
AlmondBread
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July 7th, 2015 at 7:14:23 AM permalink
Quote: beachbumbabs

12/52 = .230769, or 23% of the time.
12/51 assuming you miss = .235294
12/50 assuming you miss = .24

so you have to add those together, since he gets 3 chances to hit.

There are a couple ways to fix this.

a) Instead of adding the probabilities of hitting given that you missed the previous card, you can add the probabilities of missing and then hitting. You don't know yet if the first card missed, so you don't get to assume it beforehand. It has a probability which must be factored in.

12/52 + (40/52)(12/51) + (40/52)(39/51)(11/50) = correct answer (for the problem you answered)

b) Add 12/52 three times, then use inclusion-exclusion since that overcounts the ways two or three of the cards can hit. Draw 3 overlapping circles. One represents P(J hits), another P(2 hits) and the third P(4 hits). When you add all 3 circles, how many times have you counted each of the regions where 2 circles intersect? So then subtract those intersections as needed to make them only counted once. But once that step is done, how many times is the triple-intersection counted?
ThatDonGuy
ThatDonGuy 
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July 7th, 2015 at 7:48:29 AM permalink
Quote: FatBabyKate

Tonight at our holdem game one of the players offered me this bet.... He
wanted to bet $20 on every flop til one of us cried "UNCLE"!!

His bet was that a 2-4 or a J would come on the flop! THATS IT!

I didn't accept because I learned long ago not to take another mans bet.
I tried to do the math on this to see who had the best of it but couldnt do it.

I'm pretty certain that he has the best of this bet. But how much of an edge
does he have? Can someone show me the math on this?


In a case like this, it's easier to figure out the probability that none of those cards would come up on the flop, then subtract the result from 1.
There are C(52,3) cards in a flop; there are C(36,3) different flops that do not contain any of those cards.
The probability of one of these flops showing up is C(36,3) / C(52,3) = ((36 x 35 x 34) / 6) / ((52 x 51 x 50) / 6) = 99/455.
The probability that a flop has at least one of the four cards is 1 - 99/455 = 356/455, or about 78% of the time.

Even if it is limited to a 2-4 showing up, that is 1 - C(40,3) / C(52,3) = 47/85, or about 55.3% of the time.
studmuffn
studmuffn
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July 7th, 2015 at 10:48:16 AM permalink
Quote: ThatDonGuy

There are C(52,3) cards in a flop; there are C(36,3) different flops that do not contain any of those cards.
The probability of one of these flops showing up is C(36,3) / C(52,3) = ((36 x 35 x 34) / 6) / ((52 x 51 x 50) / 6) = 99/455.
The probability that a flop has at least one of the four cards is 1 - 99/455 = 356/455, or about 78% of the time.



Something went awry with your fractions. The factorial equation is right, but it simplifies down to 147/455, or 21/65, which agrees with ME's 32.3%.

In any event, the Hold Em player did not have much respect for the others at the table. Reminds me of a coworker who stood a Coke on a desk and laid one on the ground, and measured the height between the two. He then switched the two and asked if I would bet that they had the same difference in height, with the one laid on the desk and the other upright on the ground. He said he's won hundreds at the bar from people betting that the height difference was the same.
MathExtremist
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July 7th, 2015 at 11:09:28 AM permalink
Quote: studmuffn

Something went awry with your fractions. The factorial equation is right, but it simplifies down to 147/455, or 21/65, which agrees with ME's 32.3%.

In any event, the Hold Em player did not have much respect for the others at the table. Reminds me of a coworker who stood a Coke on a desk and laid one on the ground, and measured the height between the two. He then switched the two and asked if I would bet that they had the same difference in height, with the one laid on the desk and the other upright on the ground. He said he's won hundreds at the bar from people betting that the height difference was the same.


Please clarify: let the height of the desk be D, height of the can be H, width of the can be W, where H > W. Are you saying that the bet was D+H-W = D+W-H? If so, that's not even a math problem, just an exploit of people who are bad at spatial reasoning. I suppose it would help if the mark is drunk, but that's still just mean.
"In my own case, when it seemed to me after a long illness that death was close at hand, I found no little solace in playing constantly at dice." -- Girolamo Cardano, 1563
studmuffn
studmuffn
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July 7th, 2015 at 12:05:45 PM permalink
Quote: MathExtremist

Please clarify: let the height of the desk be D, height of the can be H, width of the can be W, where H > W. Are you saying that the bet was D+H-W = D+W-H? If so, that's not even a math problem, just an exploit of people who are bad at spatial reasoning. I suppose it would help if the mark is drunk, but that's still just mean.



You've got it exactly right. The kicker is that he wasn't being mean at all. He had lost money at the bar to this bet, and was so impressed that he used it to trick his friends. But even while showing me, he was still impressed at the 3-4" change and my efforts to explain proved fruitless. Very talented welder and rig operator, not much of a logician.
MathExtremist
MathExtremist
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July 7th, 2015 at 2:41:21 PM permalink
I prefer the "flip four coins" bet, but you usually can't pull it off more than once. The mark bets that they'll see 2 heads and 2 tails in any order and you fade them at even money. It goes down one of two ways:

a) heads, tails, (doesn't matter), the last flip is a suspenseful 50-50 chance to win.

b) heads, heads, "oh no!"

Edit: we should do a bar bets thread...
"In my own case, when it seemed to me after a long illness that death was close at hand, I found no little solace in playing constantly at dice." -- Girolamo Cardano, 1563

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