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## prop bet math question

1. I was playing poker at my local 1/2 game and was asked to do a prop bet with a fellow poker buddy.

he gets to choose any 3 cards and if any of the 3 cards hit the flop he wins the bet. if none hit, then i win. i'm quite sure i'm behind but would like to know the math behind it.
2. Do you get to see your hole cards when making this bet?
3. its the same 3 cards every time. The cards do not change.
4. OK, unless my math is wrong, you are getting a bit screwed.

Odds of you losing this bet on the first card of the flop are 12/52 or 23.1%
Odds of you losing this bet on the 2nd card of the flop are 12/51 or 23.5%
Odds of you losing this bet on the 3rd card of the flop are 12/50 or 24%

So looks like you would lose this bet 70.6% and should be getting approx 7 to 3 odds. My math might be off though, so would like someone to correct me if I am off-base.
5. I'm not sure how to write this equation, but 12/52 is ~23%, so he has a 23% chance 3 times. From there I'm not sure how it goes as I'm pretty sure you don't just multiply it by 3.

6. Oh right, I forgot that each card gone is one subtracted from the deck. I don't think you can add these 3 %'s up because what if instead of 12 it's 20? It comes out to over 100%.

7. True it might be something more like this..

Odds of you losing this bet on the first card of the flop are 12/52 or 23.1%
76.9% of the time, need to see card 2 - then Odds of you losing this bet on the 2nd card of the flop are 12/51 or 23.5% (18% when factoring in the 76.9%) or 40/52*12/51
58.9% of the time, need to see card 3 - Odds of you losing this bet on the 3rd card of the flop are 12/50 or 24% (14.1% when factoring in 58.9%) or 40/52*39/51*12/50

So looks like if this math is correct you would lose 55.2% of the time which isn't as bad as the 7 to 3 I was originally coming up with. I could still be wrong, its 1am here, and might not be calculating to the best of my ability at the moment.
8. You're not from Michigan are you? I know of this guy tthat offers that prop bet all the time, and then later he'll tell the rest of the table about how rigged that prop bet is and how it's so much in his favor.
9.
##### Originally Posted by wackyJaxon

True it might be something more like this..

Odds of you losing this bet on the first card of the flop are 12/52 or 23.1%
76.9% of the time, need to see card 2 - then Odds of you losing this bet on the 2nd card of the flop are 12/51 or 23.5% (18% when factoring in the 76.9%) or 40/52*12/51
58.9% of the time, need to see card 3 - Odds of you losing this bet on the 3rd card of the flop are 12/50 or 24% (14.1% when factoring in 58.9%) or 40/52*39/51*12/50

So looks like if this math is correct you would lose 55.2% of the time which isn't as bad as the 7 to 3 I was originally coming up with. I could still be wrong, its 1am here, and might not be calculating to the best of my ability at the moment.

Ya that sounds a lot closer than the first number, but I'm not familiar with this type of calculating.

10.
##### Originally Posted by wackyJaxon

True it might be something more like this..

Odds of you losing this bet on the first card of the flop are 12/52 or 23.1%
76.9% of the time, need to see card 2 - then Odds of you losing this bet on the 2nd card of the flop are 12/51 or 23.5% (18% when factoring in the 76.9%) or 40/52*12/51
58.9% of the time, need to see card 3 - Odds of you losing this bet on the 3rd card of the flop are 12/50 or 24% (14.1% when factoring in 58.9%) or 40/52*39/51*12/50

So looks like if this math is correct you would lose 55.2% of the time which isn't as bad as the 7 to 3 I was originally coming up with. I could still be wrong, its 1am here, and might not be calculating to the best of my ability at the moment.

I think you should just add up the probabilities like the first example. Since we are trying to predict it coming out based on a three card flop, you treat it as one event, not three separate. we willl always see the second and third, so there is no sense in calculating the cumulative probability of the event based on the priors.
Edited By: rayspizza Jul 7th, 2012 at 04:15 AM
11.
##### Originally Posted by rayspizza

I think you should just add up the probabilities like the first example. Since we are trying to predict it coming out based on a three card flop, you treat it as one event, not three separate. we willl always see the second and third, so there is no sense in calculating the cumulative probability of the event based on the priors.

so if we had 5 cards instead of 3, one of our cards would come out on the flop 115% of the time?

12. ha did not think it would be that complicated. I Think its around 60/40ish from what i could tell doing it for an hour or so. def not 7/3. i remember its a 40% chance for any two cards to pop up on the flop so in between 55-60% would make sense. Just funny no ones got an equation for it.

thanks for the input
13.
##### Originally Posted by wackyJaxon

True it might be something more like this..

Odds of you losing this bet on the first card of the flop are 12/52 or 23.1%
76.9% of the time, need to see card 2 - then Odds of you losing this bet on the 2nd card of the flop are 12/51 or 23.5% (18% when factoring in the 76.9%) or 40/52*12/51
58.9% of the time, need to see card 3 - Odds of you losing this bet on the 3rd card of the flop are 12/50 or 24% (14.1% when factoring in 58.9%) or 40/52*39/51*12/50

So looks like if this math is correct you would lose 55.2% of the time which isn't as bad as the 7 to 3 I was originally coming up with. I could still be wrong, its 1am here, and might not be calculating to the best of my ability at the moment.

missed this post. that seems correct. Thanks Wacky
14. Ignoring what your own cards are: this is a quicker method to work it out = 1 - probability he misses, ie. 3 blanks come = probability he wins

1 - (40/52 x 39/51 x 38/50) = 55.294..%, 55% he wins, 45% you lose. You'd need odds of say 3:2 for the bet to be in your favour.
15. if you are accepting 1-1 odds on his cards missing the flop, you are betting into a 10.59% hold. That means he is getting the best of it, and he can expect to make \$10.59 for every \$100 in bets.

Give him a counter offer, tell him you will pay him 2-1 odds on him winning twice in a row. Then you'll have the edge with an 8.29% expected edge.

If you want a fair gamble, demand to get 6/5 on your money if your going to be taking the short side. He'll still have a 1.65% edge, but you at least will be given a chance.

The best way to play this game is to accept the game but with an added rule. Any time the flop is 3 cards of the same suit, he loses and pays double. This is regardless of whether he hits a card or not. This added variant makes this game almost statistically 50/50. If you can get him to accept these terms, you will have the very slightest of an edge. You will have a 0.58% edge in the game. But essentially, this is fair for both players. There are 22,100 flop possibilities. 12,220 of those, he will hit one of his 3 cards, 9880 of those he wont. 1140 out of the 22,100 will be the same suit. 664 of his 12,220 wins gets trumped and he goes -2 units on those instead of +1 unit. 476 of his losing 9880 flops he goes from -1 unit to -2 units. If he's leary of this new offer, tell him you'll take the other side, because it's that even. If he only wants to play this game by gouging the other player with a 10.59% hold on 1-1 payouts, tell him to #%*& off.
16.
##### Originally Posted by shakhtar

if you are accepting 1-1 odds on his cards missing the flop, you are betting into a 10.59% hold. That means he is getting the best of it, and he can expect to make \$10.59 for every \$100 in bets.

Give him a counter offer, tell him you will pay him 2-1 odds on him winning twice in a row. Then you'll have the edge with an 8.29% expected edge.

If you want a fair gamble, demand to get 6/5 on your money if your going to be taking the short side. He'll still have a 1.65% edge, but you at least will be given a chance.

The best way to play this game is to accept the game but with an added rule. Any time the flop is 3 cards of the same suit, he loses and pays double. This is regardless of whether he hits a card or not. This added variant makes this game almost statistically 50/50. If you can get him to accept these terms, you will have the very slightest of an edge. You will have a 0.58% edge in the game. But essentially, this is fair for both players. There are 22,100 flop possibilities. 12,220 of those, he will hit one of his 3 cards, 9880 of those he wont. 1140 out of the 22,100 will be the same suit. 664 of his 12,220 wins gets trumped and he goes -2 units on those instead of +1 unit. 476 of his losing 9880 flops he goes from -1 unit to -2 units. If he's leary of this new offer, tell him you'll take the other side, because it's that even. If he only wants to play this game by gouging the other player with a 10.59% hold on 1-1 payouts, tell him to #%*& off.

17.
##### Originally Posted by bazingaking

Ignoring what your own cards are: this is a quicker method to work it out = 1 - probability he misses, ie. 3 blanks come = probability he wins

1 - (40/52 x 39/51 x 38/50) = 55.294..%, 55% he wins, 45% you lose. You'd need odds of say 3:2 for the bet to be in your favour.

Uhhhh, absolutely not. Why are you multiplying the probabilities? It is an OR case

First card: 12/52
Second Card: 12/51
Third Card: 12/50

You add the probabilities. So it's about a 72% chance he hits the cards. This is a known suckers bet. Similar to the "67" hits the flop more than it misses
18.
##### Originally Posted by edwardt1988

Uhhhh, absolutely not. Why are you multiplying the probabilities? It is an OR case

First card: 12/52
Second Card: 12/51
Third Card: 12/50

You add the probabilities. So it's about a 72% chance he hits the cards. This is a known suckers bet. Similar to the "67" hits the flop more than it misses

no. (not that you are the only wrong one..)

i'm assuming OP meant three ranks of cards.

40/52 x 39/51 x 38/50 = .4471

44.7% all three cards will blank for the proposer of the bet, 55.3% he hits.
Edited By: EyeKnows Jul 9th, 2012 at 11:21 PM
19.
##### Originally Posted by edwardt1988

Uhhhh, absolutely not. Why are you multiplying the probabilities? It is an OR case

First card: 12/52
Second Card: 12/51
Third Card: 12/50

You add the probabilities. So it's about a 72% chance he hits the cards. This is a known suckers bet. Similar to the "67" hits the flop more than it misses

I'm assuming you flunked math a few times.

Under your formula, he'd be over 100% to catch one of his 12 cards by the 5th draw out of a 52 card deck. I'm curious, do you even attempt to think or reason things out before you embarrass yourself with moronic posts like this? Or are you just so ignorant that you just assume you know something for sure, when you clearly have no grasp of it.

The correct answer has already been established. There are 22,100 flop possibilities, 12,220 contain one of his 3 ranked cards. This is simple math. Please, go away, and go back to your leggos or picking radishes, or whatever it is you normally do throughout the day.
Edited By: shakhtar Jul 10th, 2012 at 01:40 AM
20.
##### Originally Posted by shakhtar

I'm assuming you flunked math a few times.

Under your formula, he'd be over 100% to catch one of his 12 cards by the 5th draw out of a 52 card deck. I'm curious, do you even attempt to think or reason things out before you embarrass yourself with moronic posts like this? Or are you just so ignorant that you just assume you know something for sure, when you clearly have no grasp of it.

The correct answer has already been established. There are 22,100 flop possibilities, 12,220 contain one of his 3 ranked cards. This is simple math. Please, go away, and go back to your leggos or picking radishes, or whatever it is you normally do throughout the day.

Looooooool. Ya I did mess that up. Ooops. Not the first time I fuck up simple probability questions, so my apologies to the guy I originally quoted. Don't worry, I never flunked math. I was one course away from a math major before giving up on it, but I always avoided probability classes because of stuff like this.