Party Poker Sidebets

[rwesty]

What is the EV of these bets?

[Vincent Lepore]

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Vince

[tonypaladino]

-15%

[TomCollins]

EV = $8 *(1/8.5) - 1 *(7.5/8.5) = .94 - .88 = -6.25%

[yellowjack]

I hope this clears things up. Here is the player's EV for $1 wagered on red (black is the same):

P(all red) = (26/52)*(25/51)*(24/50) = 2/17 = 1/8.5
Payoff: +$7
P(not all red) = 1 - P(all red) = 7.5/8.5
Payoff: -$1

EV: (1/8.5)*(+$7) + (7.5/8.5)*(-$1) = -$0.05(8823)

[PokerGoblin]

This is the way I figured it:

If you place the bet after seeing your hole cards it is very slightly +EV.

If you have two black hole cards and bet red your probabilities are (26/50)*(25/49)*(24/48) = 0.1326.


the payout, 1/8, is .125, so you would have .7% +EV.

Still it doesn't seem right. It's as if I am forgetting something.

PG

[yellowjack]

You decide on a bet before getting your hole cards.

My math:

26/52 * 25/51 * 24/50 = 0.11764706 or 1 in just a hair more than 8.5

This gives odds of 7.5 to 1 against your making your flop.

By my line of thinking, for every 17 bets you place, you will win twice. This means you will lose 15 bets. On the last two, you will win and be paid 8 to 1, making 16 bets, for a total profit of 1 bet.

However, I'm quite convinced I'm missing something, as Party Poker would be very foolish to offer this sort of bet.

[Gabe]

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Vince

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Care to share your work?

I don't see why they would offer a bet and not have the best of it.

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Vince

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Care to share your work?

I don't see why they would offer a bet and not have the best of it.

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Likewise.