Poker puzzle challenge - Fundamental theorem

[reubenf]

This effect, on a much more subtle scale, is a reason collusion can be so effective. Colluding suboptimal strategies can be winners against a perfectly optimal non-colluding player. The optimal player just chooses which of the colluding players is the "loser" and which is the "winner".

Jerrod Ankenman on his blog got me thinking about this in the first place:
http://www.livejournal.com/users/hgfalling/14818.html

[jipster]

[ QUOTE ]
as far as i can see this is one:

you have Q[img]/images/graemlins/heart.gif[/img]Q[img]/images/graemlins/spade.gif[/img], and the board after the turn is:

Q[img]/images/graemlins/club.gif[/img]9[img]/images/graemlins/club.gif[/img]6[img]/images/graemlins/diamond.gif[/img]3[img]/images/graemlins/diamond.gif[/img]

you are up against 9 opponents who hold:

99
66
33
4[img]/images/graemlins/club.gif[/img]5[img]/images/graemlins/club.gif[/img]
7[img]/images/graemlins/diamond.gif[/img]8[img]/images/graemlins/diamond.gif[/img]
J[img]/images/graemlins/spade.gif[/img]T
A[img]/images/graemlins/heart.gif[/img]A[img]/images/graemlins/spade.gif[/img]
57
QJ[img]/images/graemlins/heart.gif[/img]

[/ QUOTE ]

there isnt two queens of hearts in a deck, hypothetically or otherwise.... i'm a newbie here, enjoy the post!

[PennyAnte]

From Poker: How to Win at the Great American Game (David A. Daniel)

You're dealt a pair of aces.
9 callers:
Mabel has 3-6
Fred has 3-7
Ralph has J-10
Blanche has 6-7
Irving has Q-J
Byron has Jc-4c
Ziggy has Kh-3h
Elmo has 9-8
Mel has 5-8

Flop 4h 5h 9c
Turn Kc

[FredBoots]

Interesting post. At first glance, I can't think of any situation where you hold the nuts on the flop but if everyone calls, you can't win. I think the problem is no collection of hands can guarantee to improve enough to beat the nuts. However, if we don’t restrict the table to 10 players, I guess this would work:

Qs Qh
Qd 7c 2c flop

Then one player could hold two clubs and 18 other players could hold the remaining spades, hearts, and diamonds. The only cards remaining for the flop and turn will give the guy with two clubs a flush.

Just for interest, this is what I figure the nuts are for different flops:

If the flop is X, the nuts are Y, so you have Z:
Unconnected rainbow flop, trips, pocket pair matching high card on flop.
Straight flop, straight, connectors to make high straight.
Flush flop, flush, AKs in suit of flop.
Paired flop, quads, pocket pair matching flop pair.
Trip flop, quads, Ax where x matches the trips.
Straight flush flop, straight flush, suited connectors to make high straight flush.

[Klyka]

I might be even more retarded, but I can't see any hand that would beat your set of queens if the fifth street comes a jack.

[rchandra]

[ QUOTE ]
I might be even more retarded, but I can't see any hand that would beat your set of queens if the fifth street comes a jack.

[/ QUOTE ]

I felt that way too, then noticed that both remaining jacks complete a flush.