1 hand vs one player with 2 hands

[paulish]

Player A is dealt one hold'em hand, and Player B is dealt two separate hands (but he can only play one of them). Which hands are now +EV for Player A?

[Hellmouth]

I think that basically the same cards that you would normally play at a 10 person table would still be profitable in this situation. However, the player who sees 4 cards would have an advantage in some cases (Ie if he/she is delt 4 hearts then he can discount his chances of making a flush or if he has pocket 22 in one hand and A2 in the other he knows that his chances of hitting trips are deminished.)

Greg

Player A should go find another table!

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...find another table!


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Hell, I think he should find another casino!

[Xhad]

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I think that basically the same cards that you would normally play at a 10 person table would still be profitable in this situation.

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If this is a heads-up match, you're going to lose your shirt and it isn't even close.

[phish]

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I think that basically the same cards that you would normally play at a 10 person table would still be profitable in this situation.

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If this is a heads-up match, you're going to lose your shirt and it isn't even close.

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I remember a question that arose in another forum awhile back, where one guy would get two hands to pick from and the other guy would get the button every time (didn't specify whether button posts big or small blind) and they asked which would you choose.
And surprisingly, a lot of people chose the button. They think that position was enough to offset the hand disadvantage.

[paulish]

I don't think you guys understand me. I think there shouold be a way of calculating the exact hand group that is a favorite in this case.
I did some calculations, and I came to the conclusion that the top 50 "all-in-heads-up-race-hands" shod have an +EV, when the other player can chose to play the best out two seprate hands.

+EV:
[4[img]/images/graemlins/spade.gif[/img] 4[img]/images/graemlins/diamond.gif[/img]],
[A[img]/images/graemlins/heart.gif[/img] 2[img]/images/graemlins/heart.gif[/img]], [A[img]/images/graemlins/club.gif[/img] 4[img]/images/graemlins/spade.gif[/img]],
[K[img]/images/graemlins/heart.gif[/img] 6[img]/images/graemlins/heart.gif[/img]], [K[img]/images/graemlins/spade.gif[/img] 9[img]/images/graemlins/diamond.gif[/img]],
[Q[img]/images/graemlins/diamond.gif[/img] 9[img]/images/graemlins/diamond.gif[/img]], [Q[img]/images/graemlins/heart.gif[/img] T[img]/images/graemlins/club.gif[/img]],
[J[img]/images/graemlins/spade.gif[/img] T[img]/images/graemlins/spade.gif[/img]] or better.

But I'm not sure if my numbers are right

[Alex/Mugaaz]

Any hands that has an equity edge vs 2 random hands.

When does he choose which hand he will play?

[nate1729]

Interesting. This seems tough, so maybe we should start with an analogized situation.

I get a number in [0,1] and you get the max of two in [0,1]. When should I think mine has a good shot to be bigger? If I have x, then the probability I win is x squared. I'm better than even-money if x > (1/2)sqrt(2) or about .71.

Of course this isn't quite that easy; there's pot odds and position and whatnot. Say you have the button and are getting 3-1 immediately; then perhaps you might be interested in "hands" that have a .25 chance of being ahead. Sqrt(1/4) = 1/2, so you'd "play" x > .5, whatever exactly "playing" is. But then there are playing advantages and information gaps and all that fun stuff that makes poker poker.

Because I'd be at a playing/informational disadvantage I'd probably want to have what would correspond in the analogical case to a 2/7 "chance" of winning on the button, and 1/3 out of position, which would suggest I'd want to play something like my top 47% of hands in position and top 43% out of position.

Too tight? Too loose?