Got this riddle from Wrexen’s blog:
Two men play a game of draw poker in the following curious manner. The spread a deck of 52 cards face up on the table so that they can see all the cards. The first player draws a hand by picking any five cards he chooses. The second player does the same. The first player may now keep his original hand or draw up to five cards. His discards are put aside out of the game. The second player may now draw likewise. The person with the higher hand then wins. Suits have equal value, so that two flushes tie unless one is made of higher cards. After a while the players discover that the first player can always win if he draws his first hand correctly. What hand must this be?
I think I may have the answer to the riddle (will be posted in the comments).
Here’s the ranking of the hands if you don’t know about poker (suits don’t matter):
Royal Flush
Straight Flush
4 of a Kind
Full House
Flush
Straight
3 of a Kind
2 Pair
1 Pair
High Card
An interesting question someone raised was how will it turn out if there were 3 players.
Player 1: Choose all 4 T (10’s) and a random card. The reason for this is to leave only 4 consecutive cards above him so player 2 won’t be able to make a Royal Flush.
Player 2: He has really lost at this point. In order to win, he’d have to block both top and bottom. Block the top to stop player 1 from making any royal flush and blocking the bottom to stop player 1 from making a 10 high straight flush, because the highest hand player 2 can ever make at this point is a 9 high straight flush. However, to seal off both top and bottom, player 2 would need 8 cards (2 for each suit). Being only able to choose 5 cards, he has lost the game.
S H D CA A A A
K K K K
Q Q Q Q
J J J J <- player 2 will have to block
T T T T <- player 1 takes this row
9 9 9 9 <- player 2 will have to block
8 8 8 8
7 7 7 7
6 6 6 6
5 5 5 5
4 4 4 4
3 3 3 3
2 2 2 2
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As I mentioned, someone brought up the question of what would happen in the case of 3 players.
With 3, the game is completely changed, player 2 and player 3 now have a combined total of 10 cards they can choose (which is greater than the 8 needed). If player 1 grabs all the 10s again, he’s going to lose if player 2 and player 3 colludes. However, despite the fact player 1 is going to lose, he gets to choose the winner between player 2 or 3 or make them for a tie. So now player 1 can collude with either player 2 or 3 or neither.
However, according to game theory, you never suppose to take a path that you know there’s a 100% chance of losing, so player 1 should never choose all the 10s initially. I haven’t given much thought about 3 players yet, but given that I don’t see an immediate winning strategy for any player, they can all get their royal flushes on their first turn and all tie.
Player 1 takes 10 10 5 5. (every straight, and thus straight flush, must contain 10 or 5)
The best move player 2 could make would be to take the other two 10s and 5s, to prevent Player 1 from getting a straight flush.
Player 1 could then take the aces and whatever (a king for arguments sake).
If player 2 took anything other than the 10s and 5s, Player 1 would take the highest remaining straight flush possible.
If player 1 takes all 4 10’s, player 2 can still choose the cards to make a straight flush and win. ( ex 2-6 clubs )
He can, but player 1 can just get a royal flush, and a royal flush beats a 6-high straight flush.