[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]

Thinking as I go along. A picks up all the aces, and a king of his choice, leaving B with no chance of getting a Royal flush, meaning he'll, at best, Pick up KQJ109 of the same suit, but different from the one A picked, or all Q's and K kicker. ..And from here on I can't think of a way that putting down cards would make A win. Obviously, it was to be a forced win, making B draw some incredibly crappy cards, but I don't think it can be done? Because all cards A doesn't have at one time will be on the table, which are enough to get a Royal Flush, or, unless A goes for an off-suit flush to A, B can tie it.

So.. No?

[Guest]

A royal straight flush is the highest hand in poker. If player A picks a royal straight flush of any suit and trades no cards, then player B can do the same, ending the hand in a tie.

If anything player B has an advantage because he can see the hand he has to meet or beat, and he can always do so since the suits are of equal value.

[Guest]

Both players will royal flush each hand and push. No winners.

[Guest]

I'm not sure the puzzle is answerable in this form, maybe it is and I'm just missing something; not thinking it through. Your puzzle doesn’t specify whether the discarded selections of player A are available to player B and vice versa. If the cards are available perpetually, and four sets of five high cards can have no higher authority than each other, then no economic scarcity is ever created and thus no way of flanking the opponent. However, if those cards cannot be reused, player A could permanently eliminate one key cards from each suit that would allow player B to never have the highest hand of that suit, and thus the first player would always win through eliminating the possibility of competition by the second. If the cards can be reused and suits have no rank, between only two players I do not see how any player could ever win.

Edited March 2, 2009 by A Smith

[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.

[Guest]

I'm not sure the puzzle is answerable in this form, maybe it is and I'm just missing something; not thinking it through. Your puzzle doesnt specify whether the discarded selections of player A are available to player B and vice versa. If the cards are available perpetually, and four sets of five high cards can have no higher authority than each other, then no economic scarcity is ever created and thus no way of flanking the opponent. However, if those cards cannot be reused, player A could permanently eliminate one key cards from each suit that would allow player B to never have the highest hand of that suit, and thus the first player would always win through eliminating the possibility of competition by the second. If the cards can be reused and suits have no rank, between only two players I do not see how any player could ever win.

This was the dilemma I was facing. If by discard, we mean removed from game, then it is a different scenario altogether.

discards are put back in to "play".

A and B will tie each game.

discards are taken out of play.

A draws 4 aces and king.

B draws king high straight flush.

A discards to Ace, King, and draws to Royal Flush winning.

Edited March 2, 2009 by theharangue

[Guest]

A chooses the four 10's and an arbitrary card.

No matter what B chooses at this point, A can win. A can always win with either a straight flush A-10 or 10-6 depending on what cards B takes. However, at this point B's best possible hand is a straight flush 9-5.

Suck it Trebek!

[Guest]

Suck it Trebek!

A chooses the four 10's and an arbitrary card.

No matter what B chooses at this point, A can win. A can always win with either a straight flush A-10 or 10-6 depending on what cards B takes. However, at this point B's best possible hand is a straight flush 9-5.

Agreed.

[Guest]

Then A cannot ensure any win

1. A picks up all aces and one other card

2. B picks up all kings and one other card

Now A can either sit on hist aces and see B get a straight flush, or get himself a straight flush, but then B will have access to the discarded cards and get Royal

A can only ensure a tie by pick himself a Royal from the beginning

Edited March 2, 2009 by TrueMaja

[Guest]

Suck it Trebek!

A chooses the four 10's and an arbitrary card.

No matter what B chooses at this point, A can win. A can always win with either a straight flush A-10 or 10-6 depending on what cards B takes. However, at this point B's best possible hand is a straight flush 9-5.

I agree with this solution. I don't see another way A can guarantee a win. If A chooses four Aces and a King, B can follow up by taking the remaining Kings as well as the Queen of the same suit as A's King along with one more Queen. Unless A wants to lose, A is forced to take J-7 of the suit that B now has the K-Q of. Both players end up with J-7. (at this point if A decides to go with a Q-8, B can get a K-9)

[Guest]

A exchanges 0 1 2 3 4 or 5 of his cards for new ones, discarding the old onesHighest hand wins. Suits have equal value.

Clarification please? I take "exchange" to mean that A puts his old cards back into the mix, but "discarding" could also mean that those cards are subsequently taken out of play.

because, lets face it, it's not much of a puzzle the other way.

(A chooses Royal flush and ties every time)

Choosing four 10's and any other card should ensure a win every time if the traded cards are discarded. The highest hand then available would be straight flush to the 9. As long as A's discards aren't reintroduced, he can safely make his hand a royal flush without worry and win. If the cards go back in the pile, then B readjusts to a Royal Flush and pushes, as stated before.

Of course, if B chooses four jacks and an Ace after A chooses his 10's, then the best hand A can make is a 10-high straight flush, followed by.... 9-high straight flush for B. Yup, I think that gets it.

[Guest]

you pick up all kings and the ace of spades, leaving the other player with no possible higher hand, at best they can get 3 aces and a queen or a strait flush ending at queen which will always be beaten by four kings ace high...

i dunno if you meant it to have some sly answer with tradin cards and stuff, but i don't see hw you could loose with this meathod, considering that A always chooses first

[Guest]

you pick up all kings and the ace of spades, leaving the other player with no possible higher hand, at best they can get 3 aces and a queen or a strait flush ending at queen which will always be beaten by four kings ace high...

i dunno if you meant it to have some sly answer with tradin cards and stuff, but i don't see hw you could loose with this meathod, considering that A always chooses first

Any strait flush is better than any four of a kind, regardless of the kicker.

[Guest]

because, lets face it, it's not much of a puzzle the other way.

(A chooses Royal flush and ties every time)

Choosing four 10's and any other card should ensure a win every time if the traded cards are discarded. The highest hand then available would be straight flush to the 9. As long as A's discards aren't reintroduced, he can safely make his hand a royal flush without worry and win. If the cards go back in the pile, then B readjusts to a Royal Flush and pushes, as stated before.

Of course, if B chooses four jacks and an Ace after A chooses his 10's, then the best hand A can make is a 10-high straight flush, followed by.... 9-high straight flush for B. Yup, I think that gets it.

If they chose four tens, and any other card, player B could just choose four of the highest possible avalable card and any other card, and would win every time, blocking a royal flush. Lets say that A picks four tens and an ace, B picks 4 kings and an ace, PLayer A cannot four of a kind aces any longer because player B blocked that move, leaving no royal flushes and the highest available hand four of a kind in queens.... The tens meathod does not work, the only way i can see this puzzle working is picking four kings and an ace, or four aces and a king, blocking the royal flush, and blocking four of a kind of anything higher, then on the trade in, player a just doesn't trade in anything.

[Guest]

the discarded cards are out of play.

A will always win, if played right.

A: Takes four Kings and a 5th card, let's say a Queen.

B: Takes the next available four of a kind, in this case, Aces.

A thenw discards and draws to a King high strait flush.

Best B can get is a Queen high strait flush.

Edited March 2, 2009 by Jeff Dierking

[Guest]

the discarded cards are out of play.

B will always win, if played right.

A: Takes four Aces and a 5th card, let's say a King.

B: Takes the next available four of a kind, in this case, Queens.

Best hand that A can improve to is a Jack high strait flush, however, B will have ther ability to get a King high strait flush.

Unless A uses that suit to get the Jack high straight flush.

[Guest]

this is a confusing puzzle, i'm gonna go play this game with my friends and see what i come up with

[Guest]

First Post.

I think its a great riddle.

Player A Picks up the K, Q, J, 10 and 9 all different suits except the K and one other card.

Player B is now incapable if selecting a royal flush.

Player B has two choices. Stop player A from getting a royal flush or select the best hand possible.

By selecting the best hand possible (a low strait flush), Player B has left a royal flush available to player A and therefore player A will win.

By trying to stop player B from getting a royal flush, player B leaves open the K, Q, J, 10, 9 straight flush which is not available to player B so long as player A's discards are not available. (which they shouldn't be)

[Guest]

the discarded cards are out of play.

A will always win, if played right.

A: Takes four Kings and a 5th card, let's say a Queen.

B: Takes the next available four of a kind, in this case, Aces.

A thenw discards and draws to a King high strait flush.

Best B can get is a Queen high strait flush.

and A takes four kings and a queen, then I take four jacks. A can't make a royal without a jack, so I'm guaranteed the highest straight flush, unless A takes all of the 10s, in which case I keep my four jacks.

[Guest]

First Post.

I think its a great riddle.

Player A Picks up the K, Q, J, 10 and 9 all different suits except the K and one other card.

Player B is now incapable if selecting a royal flush.

Player B has two choices. Stop player A from getting a royal flush or select the best hand possible.

By selecting the best hand possible (a low strait flush), Player B has left a royal flush available to player A and therefore player A will win.

By trying to stop player B from getting a royal flush, player B leaves open the K, Q, J, 10, 9 straight flush which is not available to player B so long as player A's discards are not available. (which they shouldn't be)

Macdoc,

I take the other 3 10s, the jack of the suit of your 10 (which you do not have since only two of your cards share a suit, one of which is a king), and an arbitrary card. Now I can't lose to you.

[Guest]

Macdoc,

I take the other 3 10s, the jack of the suit of your 10 (which you do not have since only two of your cards share a suit, one of which is a king), and an arbitrary card. Now I can't lose to you.

make that arbitrary card the nine of the suit of your 10, and I'm guaranteed to win.

[Guest]

Unless A uses that suit to get the Jack high straight flush.

Actually, I edited and reversed my decision.

[Guest]

the discarded cards are out of play.

A will always win, if played right.

A: Takes four Kings and a 5th card, let's say a Queen.

B: Takes the next available four of a kind, in this case, Aces.

A thenw discards and draws to a King high strait flush.

Best B can get is a Queen high strait flush.

A: Takes four Kings and a 5th card, let's say a Queen.

B: Takes three Queens and two Jacks, making sure one of the Jacks is the same suit as A's Queen.

Best A can get is Jack high straight flush, but then B gets Queen high straight flush.

I don't see a way for A to prevent B from winning at this point.

[Guest]

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
   
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.

Yes,

A selects 4 Aces each round and waits for B. The only thing that can beat 4 aces is a Straight Flush. Nevermind, that

won't work.

[Guest]

I agree with several people - if A picks four 10s, then the best hand left on the table for B is 5,6,7,8,9. There is then no way that B can prevent A getting either 6,7,8,9,10 or 10,J,Q,K,A

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