MM029 - Tenth-Round Ace

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Tenth-Round Ace
MM029puzzle1.jpg
MM029puzzle2.jpg
Puzzle Number 029
Puzzle Name Tenth-Round Ace
Picarats Given 30 Picarats
Type Multiple Choice
Location Merchant District
Previous Puzzle MM028 - Bunny-Hop Swap 2
Next Puzzle MM030 - Big Cake, Little Cake

This is the twenty-ninth puzzle you'll encounter in Professor Layton and the Miracle Mask. To access this puzzle, you must talk to Frankie. In order to solve this puzzle, you must figure out who is the winner after 10 rounds of a card game.

Contents

[edit] Hints

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

Hint Two
    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 of card that A drew with total of the other two types that B drew, that's half the battle.

Hint Three
    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.

Super Hint
    The five time 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?


[edit] Messages

[edit] When Failed

Too bad.

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

[edit] When Completed

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.

[edit] Solution

A is the winner.

[edit] Progress

865 Picarats and 111 Hint Coins.



Last edited by Squiggle on 20 August 2015 at 03:22
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