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Tenth-Round Ace is a puzzle in Professor Layton and the Miracle Mask.

Puzzle

A and B are confronting each other in a 10-round card game. The rules of this game are simple: fire beats wood, wood beats water, and water beats fire.

Play A drew fire three times, wood five times, and water twice, while player B drew fire twice, wood five times, and water three times. There were no ties.

Can you work out who won over 10 rounds?

Hints




Click a Tab to reveal the Hint.

The fact that there were no draws is important. It means that for every card that A drew, B drew a different card.

For instance, if A drew fire, B must have drawn wood or water. If you know how many of each type of card each player got, you should be able to calculate the win/loss ratio.

Basically, if you match up the number of one type card that A drew with the total of the other two types that B drew, that's half the battle.

A drew wood five times. In those five rounds, B drew fire twice and water three times. From this, we can gather that A must have beaten B three times and lost twice.

You can use similar logic to determine the results of the five bouts when B drew wood. Then it's just a question of adding up all the results and working out who had the most wins.

The five times that A drew wood, B drew fire twice and water three times so A won 3-2.

The five times that B drew wood, A drew fire three times and water twice, meaning A won 3-2 again.

Now add them up. Who wins?


Solution

Incorrect

Too bad.

You had a 50/50 chance of guessing correctly, but you should really think about the answer!

Correct

Correct!

If A won 3-2 when A drew wood, and A also won 3-2 when B drew wood, player A must be the overall winner.

A big thanks to http://www.youtube.com/user/LaytonKyouju

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