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#1
10-10-2005, 01:18 PM
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Party sidebets
I was wondering if someone can check my math about betting on the party sidebets. If you don't know what that is you can bet $5 or $10 that the flop will be all black or all red, and it pays "8 for 1" which I think is 7-1. If you weren't playing in the game and didn't know any cards then the odds of winning would be (26/52)*(25/51)*(24/50) which works out to 7.5-1. Now if you were playing in the game and had two red cards in your hand then the odds of all black on the flop would be (26/50)*(25/49)*(24/48) which works out to around 6.54-1. So it looks like if you knew only two of the cards were either red or black then you'd have a small edge. It doesn't seem like party poker would offer something that gives the player an edge without even having to collude with other players, so I must be missing something.
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#2
10-10-2005, 01:22 PM
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Re: Party sidebets
The side bet is for the next hand.
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#4
10-10-2005, 01:24 PM
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Re: Party sidebets
[ QUOTE ]
The side bet is for the next hand. [/ QUOTE ] That'd do it. Oh well, that was 5 minutes of my life I can never get back.
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#5
10-10-2005, 02:21 PM
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Re: Party sidebets
[ QUOTE ]
The side bet is for the next hand. [/ QUOTE ] That was the first thing I checked. I'm not sure if it would be +EV betting the other color with your hole cards one color (I couldn't be bothered to do the math once I realized it was a moot point), but if you knew the color of more than one player's hole cards ... Unfortunately, Party isn't quite that stupid. While I would never collude against other poker players, I would have no reservations about doing it against a house banked sucker bet like this.
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#8
10-10-2005, 02:51 PM
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Re: Party sidebets
You need to get the latest version of the Party Pattern Mapper, then these bets most definitley become +EV.
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