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  #1  
Old 11-20-2005, 05:13 AM
CrazyEvan CrazyEvan is offline
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Default side bet, may need help with the math

I noticed at Party Poker they have the side bet of bet on whether or not the flop will be all red or all black saying it pays "8 for 1".

if I pick red, then that's 26 red out of 52 for the first card, 25 out of 51 for the second card and 24 out of 50 for the last. Which equals 15600/132600 = .11764

now if it pays 8 to 1 doesn't that mean that you only have to be right once every 9 times ( or decimal .11111 repeating ). That is assuming that the time you win, you get to keep your original bet. If the side bet will work .1176 and it only needs to work .1111 isn't it positive EV? and if so Why is it offered.

I assume that A) I have misinterpreted there "Pays 8 for 1"
B) have done the math wrong.

Please tell me where I've gone wrong.

P.S. I don't even play on party, just curious.
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  #2  
Old 11-20-2005, 07:24 AM
ThinkQuick ThinkQuick is offline
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Location: Alberta, Canada
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Default Re: side bet, may need help with the math

[ QUOTE ]
I noticed at Party Poker they have the side bet of bet on whether or not the flop will be all red or all black saying it pays "8 for 1".

if I pick red, then that's 26 red out of 52 for the first card, 25 out of 51 for the second card and 24 out of 50 for the last. Which equals 15600/132600 = .11764

now if it pays 8 to 1 doesn't that mean that you only have to be right once every 9 times ( or decimal .11111 repeating ). That is assuming that the time you win, you get to keep your original bet. If the side bet will work .1176 and it only needs to work .1111 isn't it positive EV? and if so Why is it offered.

I assume that A) I have misinterpreted there "Pays 8 for 1"
B) have done the math wrong.

Please tell me where I've gone wrong.

P.S. I don't even play on party, just curious.

[/ QUOTE ]

A. You have misinterpreted "Pays 8 for 1"
Good job realizing that this is a possible error.

You assumed it meant "pays 8 to 1" which means you are returned $9 for a $1 bet.

This is incorrect. Since it pays 8 for 1, you receive 8 for your 1, therefore You are returned $8 for a $1 bet. This is equivalent to saying a 7 to 1 payout (or 7:1).

so..... 1/8 payout = 0.125

Therefore, in your terminology, side bet works .1176 but would need to work .125 for you to break even
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  #3  
Old 11-20-2005, 07:28 AM
ThinkQuick ThinkQuick is offline
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Location: Alberta, Canada
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Default Re: side bet, may need help with the math

From Wizard of Odds hold'em page

The probability of that the flop will all be the same of a particular color is combin(26,3)/combin(52,3) = 2600/22100 = 2/17 = 11.765%. The expected return on this bet is (2/17)*7 - (15/17) = -1/17 = -5.882%. Oct. 18, 2005
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  #4  
Old 11-20-2005, 07:41 AM
CrazyEvan CrazyEvan is offline
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Default Re: side bet, may need help with the math

Yeah , I thought as much.

(I'm also pleased I had the math right on the probability of the one colored flop. I have Getting the Best of It coming in the mail so I think that will help me out quite a bit on these types of things.)

Thanks for the reply.
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