Whats the best possible odds to give?

ninjia3x · #1 08-09-2005, 04:23 PM

Heads Up NL holdem.

Each player has 40 dollars to start, blinds are 1-2.

Player A will always get to see Player B's hands b4 preflop betting occurs.

Whats the best possible odds player A can give to B and yet still be marginal +EV?

Math orientiated proof is prefered.

thanks

schwza · #2 08-09-2005, 04:44 PM

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Math orientiated proof is prefered.

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hahaha

AaronBrown · #3 08-09-2005, 06:13 PM

Let's start by assuming they play straightforwardly. If B's hand is weak, A will bet regardless of his hand, and B will fold. If B's hand is strong and A's is weaker, A will fold. If B's hand is strong and A's is stronger, A will bet. B will have to call these bets, because otherwise he's sure to lose. If he folds, he loses 3/4 of the blinds and has no chance of winning.

A wants to keep the bets as small as possible, just big enough to ensure B will fold with a weak hand. A has the advantage, the more bets, the more chance he has of winning. It's like the casino would rather you bet $1 on 100 spins of the roulette wheel rather than $100 on one spin.

B is in the opposite position, that of the casino patron. He wants one big bet to settle it. When A bets and B has a moderately strong hand, B will go all in. Also, B cannot afford to let A make decisions after seeing board cards, there's too much advantage when he knows both sets of pocket cards. So if B is going to be in the hand at all, he wants to be all-in preflop.

Thinking this through a little more, B might reason that it doesn't pay to wait for a moderately strong hand. He's losing almost 4% of his chips per hand, maybe he should start going all-in from hand one. That means he wins the blinds until A finds a hand he likes to bet on.

My guess is both players are better off with moderate strategies. B has to be willing to go all-in at any decent opportunity. A has to be willing to call often enough that he doesn't end up getting behind on blinds. So my guess is this comes down to an all in with A holding a 55% or 60% advantage.

ninjia3x · #4 08-09-2005, 08:54 PM

Thanks for the reply,

My friend asked me to approximate the odds, and i gave around 2:1 to 3:1.

I started by assuming the odds, then finding if it will yield a positive result for player A.

what i did was Player won't play a hand unless he is a 2:1 favourite. (if you choose to go all in) That leaves pocket pairs of your highest card and up, plus hands that has one your cards with a higher kicker.

with the intial pot at 3. my EV was 3*(probability he doesn't have the right odds) - (probablity he does have the right odds)*(6/7)*(1/3)

when he calls i'm assuming his odds of winning are 6:1 on avg. the 1/3 is the odds he gave me.

Does anyone think this makes sense?

meow_meow · #5 08-09-2005, 09:58 PM

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when he calls i'm assuming his odds of winning are 6:1 on avg. the 1/3 is the odds he gave me.

Does anyone think this makes sense?

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I don't understand what you mean here. His odds of winning are 6:1?? I don't understand.

The only combination that A has a 3:1 advantage over B is a pocket pair higher than both of Bs hole cards. Even something like KT vs T7 doesn't give A a 3:1 advantage.
Given that A will have a pocket pair only 5.8% of the time (and only a fraction of those times will it be higher than both of Bs hole cards), and that the odds of having one matched card is something like 25%, with A on the better side half the time, A will only have a "very dominating" hand something like 1 time in 7.

The best strategy for B appears to be to push every hand. I think with these blinds I would take the B at 2:1.

I could be wrong.

AaronBrown · #6 08-09-2005, 10:02 PM

I think it makes sense in general, but I would do the figures a bit differently.

Say B goes all in on every hand (if he goes, he should go all in, and do it preflop), and A only calls when he has a 2:1 edge. My guess is that will happen about 1 time in 6. So A has 1/9 chance of winning on the first hand, B has 1/18. If that doesn't happen, A will go all in with a 2:1 edge with fewer chips than B, so B gets another 1/3 chance of winning. If A wins that all-in bet, he's probably got something like 60 chips, at which point B's chances drop to a small value. So B has 1/18 + 1/3 plus some small value, call it 1/18; that adds up to 4/9. That says the odds are 5:4.

My guess is that however you set A's strategy, if B goes all-in on any decent hand, A has trouble getting even a 2:1 advantage. 3:2 to 5:4 seem more reasonable to me.