[bonanova]

Two friends A and B play poker in a way that eliminates

the element of chance. They spread the 52 cards face up

on the table so all cards can be see by both players.

Play proceeds as follows:

1. A selects any five cards for his hand
   
2. B does the same
   
3. A exchanges 0 1 2 3 4 or 5 of his cards for new ones, discarding the old ones, which are then no longer in play
   
4. B does the same
   

Highest hand wins. Suits have equal value.

With best play, can A always win?

Since the cards are visible, you can assume both hands are exposed at all times.

Edited March 2, 2009 by bonanova
Clarifying step 3 [red]

[Izzy]

If A gets four 10's and A ace, B can pick up KKKKA. A will need something to beat this, and since he doesn't have all the A's, a straight flush or royal flush is the only way to do it. A can't get a K, so no flush to ace for him, so I'd go for 8910JQ, same suit. After this, B can get whatever cards he likes, among with a royal flush. I don't see how that works?

[Guest]

If A gets four 10's and A ace, B can pick up KKKKA. A will need something to beat this, and since he doesn't have all the A's, a straight flush or royal flush is the only way to do it. A can't get a K, so no flush to ace for him, so I'd go for 8910JQ, same suit. After this, B can get whatever cards he likes, among with a royal flush. I don't see how that works?

the four tens that were originally taken by A are no longer in play once discarded, which is how I've always played poker.

[bonanova]

Was there another poker puzzle where the answer hinged on picking the ten(s)?

This one

?

[Guest]

If players cannot use discarded cards, Player A would draw all 4 Aces and some other random card. The highest hand that player B could draw that would beat player A would be a King high straight flush. Player A would then discard his 3 aces and the random card (keeping the Ace that is not the suit of player Bs straight flush). Player A would then draw the cards to complete his Royal Flush winning the game.

If A chooses 4 Aces (+random), B can easily counteract with 4 Kings (+random). Now the best A can get is a Queen high straight flush and B can get a King high straight flush.

Edited March 3, 2009 by BrainFreeze

[Guest]

The answer IS:

A draws 4 tens thus ensure B can not have a straight flush higher than 9-high. It also ensures that he will at least

have a 10-high flush on any given hand.

Two friends A and B play poker in a way that eliminates

the element of chance. They spread the 52 cards face up

on the table so all cards can be see by both players.

Play proceeds as follows:

1. A selects any five cards for his hand
   
2. B does the same
   
3. A exchanges 0 1 2 3 4 or 5 of his cards for new ones, discarding the old ones, which are then no longer in play
   
4. B does the same
   

Highest hand wins. Suits have equal value.

With best play, can A always win?

Since the cards are visible, you can assume both hands are exposed at all times.