[DB152] The Card Tournament

152 - The Card Tournament

The Card Tournament is a puzzle in Professor Layton and the Diabolical Box.

Puzzle

US Version

Some people meet for a round-robin card tournament, where every person plays every other person once. Wallace has to leave after only a few hands, sitting out the remainder of the tournament. A total of 59 hands are played at the tournament. How many hands did Wallace play before leaving?

The card game in question is a two-player game, and no person played with the same opponent more than once. No one sat out any hands besides Wallace.

UK Version

Some people meet for a round-robin card tournament, where every player plays one hand against every other player. Monty has to leave after only a few hands, missing the remainder of the tournament. A total of 59 hands are played at the tournament. How many hands did Monty play before leaving?

The card game in question is a two-player game, and no person played with the same opponent more than once. No one missed any hands besides Monty.

Hints

Click a Tab to reveal the Hint.

US Version

Mock up an equation with the number of people and the number of matches played as variables. That would be a good place to start.

UK Version

Create an equation with the number of people and the number of hands played as variables. That would be a good place to start.

US Version

In order to create the equation for the number of hands, you would need to multiply the number of players by the number of players minus one, then divide this product by two. If there were three people, there would have been three hands. If there were four people, there would have been six hands. You can assume from the conditions given that if no one left, the minimum number of hands would still have to be at least 60.

UK Version

In order to create the equation for the number of hands, you would need to multiply the number of players minus one, then divide this product by two. If there were three people, there would have been three hands. If there were four people, there would have been six hands. You can assume from the conditions given that if no one left, the minimum number of hands would still have to be at least 60.

There were 12 people at the tournament. Now think it through!

Solution

Incorrect

Too bad!

This is a calculation problem, but it does take some ingenuity to solve.

Correct

Good job!

US Version

Wallace played four hands. First, you need to find the total number of hands if everyone stayed. You can find it with an equation like the one here. If there were 11 people, there would've been 55 matches, and if there were 12 people, there would've been 66. Since we know 59 hands were played, there must have been 12 people at the start. When we subtract the 59 hands played from the 66 ideal, we learn that Wallace missed seven hands. Since everyone would've played 11 hands ideally, that means Wallace only played four hands.

UK Version

Monty played four hands. First, you need to find the total number of hands if everyone stayed. You can find it with an equation like the one here. If there were 11 people, there would've been 55 hands, and if there were 12 people, there would've been 66. Since we know 59 hands were played, there must have been 12 people at the start. When we subtract the 59 hands played from the 66 ideal, we learn that Monty missed seven hands. Since everyone would've played 11 hands ideally, that means Monty only played four hands.

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