Puzzle:
I was playing a game of five card draw poker with a bunch of logicians. By the time we had finished bidding and were just about to reveal our cards, I was pretty confident I would win of the four of us remaining. I had three nines, some face card (I can't remember what suit or even whether it was a jack, queen, or king) and a four. (Or was it a five? I can't remember.)
I was even more sure when two of my opponents laid down their cards. One had a pair of fours and a pair of sevens, the other had a pair of twos and a pair of eights. My third opponent, however, laid down his five cards face down in a row. He said, "I have a straight, and the cards are, from lowest to highest: a ten, a jack, a queen, a king, and an ace. I have at least one card of each of the four suits: clubs, spades, hearts, and diamonds. I am fairly certain that this is the winning hand, but I'm feeling generous today, and I will give a third of the pot to whoever can determine which suit I have two cards of.
Now I know you can't figure it out without some clues. Here they are:
1. The king is next to at least one diamond.
2. The queen is next to exactly one heart.
3. The jack is next to at least one spade, but is not next to any hearts.
4. The ten is next to at least one club.
5. The ace does not border any black cards, nor does it border any diamonds.
6. My two cards of the same suit are not next to each other.
7. Of the ten possible pairings of cards, only one pair, when removed, leaves three cards in ascending order from left to right.
8. My ace is not the card on the far left."
There was a minute's silence. One of the other logicians said, "I give up! There's no way to figure that out!"
The other agreed. But I didn't. I had just figured out which suit he had two of.
Which suit is it?
Solution:
The two cards on the ends and the card in the middle border a combined number of two suits. So at least one pair of them should be listed as bordering a card of the same suit in clues 1-5. Of the five, only the ace and the queen meet this criterion, so they must be two cards apart. One of them must be in the center, but it can't be the ace (the two cards on the ends would only border hearts - and one of them has to be bordering something else from clues 1-4), so it must be the queen. The ace is not on the left end from clue 8, so it must be on the right end, and the card next to it must be a heart. So far we have this(X represents unknown):
X X Q X A
X X X H X
At least one of the two cards to the left of the queen is lower than the queen (two of the remaining cards are lower, and only one can be right of the queen), so the triplet that is in ascending order is the queen, the ace, and the single card to the left of the queen that is lower than the queen. That means the king is left of the queen, and is also left of both the jack and the ten (otherwise, jack king ace or ten king ace would be a second triplet), so it must be on the far left. The jack must be left of the ten (or else ten jack ace would be a second triplet), so it must be second from the left, and the ten must be second from the right. The only card next to the king is the jack of diamonds (it's a diamond from clue 1). So far we have(T is ten):
K J Q T A
X D X H X
Now we seem to be stuck. We don't have any more clues that can be used. So how did I figure it out?
I held a face card (I said so myself in the intro). I must have known it cannot be that card and deduced the correct answer from there.
But which card did I have? We already know the suit of the jack, so my holding a jack would not help. Could I have held a queen? No, because I know the queen to either be a spade or club (clue 6), so one of the cards next to the queen would have its clue satisfied (see clues 3 and 4), and I could not determine the suit of the card on the other side of it.
Therefore, I must have held a king. But which king? I know the king is not a diamond (clue 6) or a heart (clue 3). If I held the king of clubs, then the king next to the jack would have satisfied the jack's clue, and I could not have determine the suit of either card next to the ten.
Therefore, I held the king of spades. The king of clubs must have been on the far left (only suit available), so the only card next to the jack that could be a spade is the queen, so the only card next to the ten that could be a club is the ace. In summary:
K J Q T A
C D S H C
Therefore there are two CLUBS.
Tuesday, April 27, 2010
Subscribe to:
Post Comments (Atom)
I came up we a set of 4 possible solutions:
ReplyDeleteK-spade J-diamond Q-spade 10-hearts A-clubs
K-spade J-diamond Q-clubs 10-hearts A-diamonds
K-clubs J-diamond Q-spade 10-hearts A-clubs
K-clubs J-diamond Q-spade 10-hearts A-diamonds
which gives us 3 possible answers: spade, diamonds or clubs.
So even if players played of one deck of cards, and a player had K-clubs then there are still 2 possible answers.
I don't think the info given in this puzzle is sufficient to solve it