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Old 08-20-2017, 07:41 AM   #1
MonkeyVase
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Poker side bet

I'm playing in a live cash game the other day when this old grinder offers me a side bet - if an J 10 or 2 comes on the flop I pay him £10, if not he pays me £10. Seems like my side is good so I take the bet. After playing for 7 hours I'm down a lot, just variance or am I getting hustled?
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Old 08-20-2017, 07:59 AM   #2
NewOldGuy
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Re: Poker side bet

Neither. He knows the math and you don't. He has a huge edge on you in this bet. The chance any one of three selected ranks shows up on the flop is 55%. The chance none do is 45%.

You can pick any three ranks for the bet.
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Old 08-20-2017, 09:52 AM   #3
MonkeyVase
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Re: Poker side bet

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Originally Posted by NewOldGuy View Post
Neither. He knows the math and you don't. He has a huge edge on you in this bet. The chance any one of three selected ranks shows up on the flop is 55%. The chance none do is 45%.

You can pick any three ranks for the bet.
Thanks, I'll add it to my side game hustle!
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Old 08-20-2017, 01:15 PM   #4
whosnext
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Re: Poker side bet

In case anyone wants to know how these probs can be derived, in this case it is easiest to think of deriving the prob of NO Jack, Ten, or Deuce on the flop.

Since there are 3 ranks in question, and 4 cards per rank, there are obviously 12 total cards in question here (before any of the hole cards are known). So there are 52-12=40 "other" cards. For there to appear NO J,T,2, all of the three flop cards must be "other".

So the prob of No J,T,2 = C(40,3) / C(52,3) = 9880 / 22100 = 44.71%

where the C(X,Y) is the number of ways to draw Y items out of X total items, when the order of the drawn items is irrelevant. C(X,Y) is called the Choose function and gives the number of Combinations. The formula for C(X,Y) is straightforward and can be found in many places and is available on many calculators.

Of course, the prob of one or more J,T,2 appearing on the flop is 1 minus that amount, or 55.29%.

P.S. The above ignores the slight "skew" in the distribution of flop-cards due to players generally playing (holding) high cards, rather than low cards. I don't think this card-removal effect has an appreciable effect on the probabilities derived above.
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Old 08-21-2017, 05:33 AM   #5
MonkeyVase
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Re: Poker side bet

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Originally Posted by whosnext View Post
In case anyone wants to know how these probs can be derived, in this case it is easiest to think of deriving the prob of NO Jack, Ten, or Deuce on the flop.

Since there are 3 ranks in question, and 4 cards per rank, there are obviously 12 total cards in question here (before any of the hole cards are known). So there are 52-12=40 "other" cards. For there to appear NO J,T,2, all of the three flop cards must be "other".

So the prob of No J,T,2 = C(40,3) / C(52,3) = 9880 / 22100 = 44.71%

where the C(X,Y) is the number of ways to draw Y items out of X total items, when the order of the drawn items is irrelevant. C(X,Y) is called the Choose function and gives the number of Combinations. The formula for C(X,Y) is straightforward and can be found in many places and is available on many calculators.

Of course, the prob of one or more J,T,2 appearing on the flop is 1 minus that amount, or 55.29%.

P.S. The above ignores the slight "skew" in the distribution of flop-cards due to players generally playing (holding) high cards, rather than low cards. I don't think this card-removal effect has an appreciable effect on the probabilities derived above.
Thanks for that
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