Pyroxene's Stupid Poker Challenge

The challenge:

Create a Texas Hold'em hand deal such that if all the players go to the river there is one hand that can neither win nor tie any boards.

I fooled around with it a few minutes and I have a 9 way deal where one hand has 0 wins and 0 ties. Doing it with fewer players would be a bonus.

EDIT: And it has to be only one hand that has no possible EV. Having two hands where neither has any EV does not satisfy the challenge.

Hand 1 - 2s 2h
Hand 2 - 4h 3s
Hand 3 - 6c 5d
Hand 4 - 8h 7s
Hand 5 - Tc 9d
Hand 6 - Qh Js
Hand 7 - Ac Kd
Hand 8 - 2c 3h
Hand 9 - 2d 3d

Doesn't matter what the last 2 hands are as long as they each have a 2.

Hand 1 can't win or tie.

that board can tie, any straight 10-a ties

Originally Posted by FishMagician

Hand 1 - 2s 2h
Hand 2 - 4h 3s
Hand 3 - 6c 5d
Hand 4 - 8h 7s
Hand 5 - Tc 9d
Hand 6 - Qh Js
Hand 7 - Ac Kd
Hand 8 - 2c 3h
Hand 9 - 2d 3d

Doesn't matter what the last 2 hands are as long as they each have a 2.

Hand 1 can't win or tie.

Ad 3s 4c 5h 10d -- Hands 1 8 and 9 tie.

hmmmm......pokerstove had 0% 0%

you guys are both right though. I will try again.

Ok pretty sure I really got it this time

Hand 1 - 2d 2c
Hand 2 - 3d 2h
Hand 3 - 4c 2s
Hand 4 - Ts Th
Hand 5 - Td Tc
Hand 6 - 7h 6s
Hand 7 - 9c 8d
Hand 8 - Ad Ac
Hand 9 - As Ah

Hand 1 no wins no ties.

your new one works fishmagician. can you find one with 8 hands? this is tough, i couldn't even come up with that one.

1 2d 2h
2 2c 3s
3 2s 4d
4 Ts Th
5 Td Tc
6 5h 5s
7 5c 5d
8 Ad Ac
9 Ah As

Originally Posted by FishMagician

Ok pretty sure I really got it this time

Hand 1 - 2d 2c
Hand 2 - 3d 2h
Hand 3 - 4c 2s
Hand 4 - Ts Th
Hand 5 - Td Tc
Hand 6 - 7h 6s
Hand 7 - 9c 8d
Hand 8 - Ad Ac
Hand 9 - As Ah

Hand 1 no wins no ties.

BOARD: 3h-4h-5d-6c-7c
Board ties. Does that count?

no, 89 wins

ahhh [dumb]didn't think about that. [/dumb]

FishMagician and Greddo017 met the challenge! My answer was similar to Greedo's; while FishMagician used a different method of accounting for board straights.

Bonus points to anyone that can do it in less than 9 hands.

Here's a similar dilemma: can anyone think of a group of starting hands at a full ring (10 handed) where a player with pocket aces couldn't possibly win? (He is however allowed to tie.) I cheated a little bit to make this happen, but I'm curious if anyone else can come up with a better solution.

I think I have it. *crosses fingers* 2d Kc never wins.

Hand 1 - 2d Kc
Hand 2 - 2h 6d
Hand 3 - 2c Ah
Hand 4 - 7s 8c
Hand 5 - 9d Tc
Hand 6 - Jh Qs
Hand 7 - Kd Ad
Hand 8 - As Ac

Originally Posted by mattperkins26

I think I have it. *crosses fingers* 2d Kc never wins.

Hand 1 - 2d Kc
Hand 2 - 2h 6d
Hand 3 - 2s 7c
Hand 4 - 2h Ah
Hand 5 - 9d Tc
Hand 6 - Jh Qs
Hand 7 - Kc Ad
Hand 8 - As Ac

You've got two repeat cards there (Kc and 2h)...

Edited above, but here it is again


Hand 1 - 2d Kc
Hand 2 - 2h 6d
Hand 3 - 2c Ah
Hand 4 - 7s 8c
Hand 5 - 9d Tc
Hand 6 - Jh Qs
Hand 7 - Kd Ad
Hand 8 - As Ac

Originally Posted by mattperkins26

I think I have it. *crosses fingers* 2d Kc never wins.

Hand 1 - 2d Kc
Hand 2 - 2h 6d
Hand 3 - 2c Ah
Hand 4 - 7s 8c
Hand 5 - 9d Tc
Hand 6 - Jh Qs
Hand 7 - Kd Ad
Hand 8 - As Ac

Matt,

The 2d Kc hand can win with a K2 on the board for two pair. Running it through one of the calculators on the net, Hand 1 can win 6949 ways and tie 1206 ways. I think most of the ties are places where the other two Kings fall and then the boar pairs up for a Kings over X full house.

Originally Posted by dalecooper

Here's a similar dilemma: can anyone think of a group of starting hands at a full ring (10 handed) where a player with pocket aces couldn't possibly win? (He is however allowed to tie.) I cheated a little bit to make this happen, but I'm curious if anyone else can come up with a better solution.

Ah Ad
Ac As
Kh Qh
Jh Th
9h 8h
7h 6h
Kd Qd
Jd Td
9d 8d
7d 6d

Ah Ad
Ac As
Kh Qh
Jh Th
9h 8h
7h 6h
Kd Qd
Jd Td
9d 8d
7d 6d[/quote]

Excellent. Mine was similar but involved a lot of pocket pairs using up the hearts and diamonds. Same effect I guess.

I'm probably missing something, but does this do it with 7 hands?

2c 2s (dead)
2d 2h
Ac As
Ad Ah
8d 8h
7c 8s
7s 8c

- Nate

hmm, i wonder if that one counts. Original rules said you can't have two of the same hand, but only one of the pairs of two's is dead, the other wins sometimes

Originally Posted by Greedo017

hmm, i wonder if that one counts. Original rules said you can't have two of the same hand, but only one of the pairs of two's is dead, the other wins sometimes

In that case, how about

2c 2s
Ac As
Ad Ah
8c 8s
8d 8h
2d 7c
2h 7s

More evidence that you should never go all-in against the almighty 27o!

- Nate

Or, if you insist on NO two hands being the same:

2s 2c
As Ac
8s 8c
Ad 8h
Ah 7s
2h 8d
2d 7c

- Nate