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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?

 See The Solution Submitted by Brian Smith No Rating

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 analytical solution | Comment 3 of 4 |
die 1: r1 red, b1 blue
die 2: r2 red, b2 blue

P(Tim win)=(r1*r2+b1*b2)/[(r1+b1)*(r2+b2)]=1/2
2(r1r2+b1b2)=(r1+b1)(r2+b2)
2r1r2+2b1b2=r1r2+r1b2+b1r2+b1b2
r1r2+b1b2=r1b2+b1r2
r1r2-r1b2=b1r2-b1b2
r1(r2-b2)=b1(r2-b2)
r1(r2-b2)-b1(r2-b2)=0
(r1-b1)(r2-b2)=0
so either r1=b1 or r2=b2
thus at least one of the dice must have
an equal number of red and blue faces.

 Posted by Daniel on 2017-06-20 18:48:19

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