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Make It Even Again (Posted on 2017-06-20) Difficulty: 3 of 5
Timothy and Urban are playing a dice game like they did before. As before the faces of the dice are colored red or blue but the dice could have any number of sides. Each die has at least 2 sides but the two dice do not necessarily the same number of faces. Both dice are fair.

The rules are the same: Timothy wins when the two top faces are the same color. Urban wins when the colors are different. Their chances are even with these dice.

Is it always the case that one of the dice has an equal number of red and blue faces?

See The Solution Submitted by Brian Smith    
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Some Thoughts back to the source | Comment 2 of 4 |


The original question appears in   " 100 Numerical Games,"         by Pierre Berloquin,1994.)

It addresses 2 six - sided dice,  one with  five red faces and one blue, the other  x red and 6-x blue.
The chances  are 50:50 to get equal or   different colors on both.

the original solution :

Throwing two six-sided dice produces 36 possible outcomes.
If Timothy and Urban have equal chances of winning, then there must be 18 outcomes in which both dice display the same color.
If 
x is the number of red faces on the second die, then:

18 = 5x + 1(6 - x)

x = 3

The second die must have 3 red faces and 3 blue faces.

Rem: "playing " with other distributions - gets always 1/2 ratio on the 2nd dice.

General solution looks complicated.

Edited on June 20, 2017, 4:37 pm
  Posted by Ady TZIDON on 2017-06-20 16:33:22

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