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 Make It Even Again (Posted on 2017-06-20)
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?

 Submitted by Brian Smith No Rating Solution: (Hide) Let p and q be the probabilities for each die to come up red. Then Timothy's chance of winning is p*q + (1-p)*(1-q). This is equal to 1/2 since the game is fair. Then apply some algebra: p*q + (1-p)*(1-q) = 1/2 p*q + 1 - p - q + p*q = 1/2 2*p*q - p - q + 1/2 = 0 4*p*q - 2*p - 2*q + 1 = 0 (2*p-1) * (2*q-1) = 0 From this last equation, p=1/2 or q=1/2. This implies one of the dice must have an equal number of red and blue faces.

 Subject Author Date re: analytical solution Ady TZIDON 2017-06-20 23:14:09 analytical solution Daniel 2017-06-20 18:48:19 back to the source Ady TZIDON 2017-06-20 16:33:22 computer findings Charlie 2017-06-20 15:18:19

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