Timothy and Urban play a game with two dice. But they do not use the numbers. Some of the faces are painted red and the others blue. Each player throws the dice in turn. Timothy wins when the two top faces are the same color. Urban wins when the colors are different. Their chances are even.
The first die has 5 red faces and 1 blue face. How many red and how many blue are there on the second die?
(In reply to answer
by K Sengupta)
Since each die consist of 6 faces, it follows that there are precisely 36 equiprobable appearance of the respective face combinations whenever two dice are thrown simultaneously.
By the problem, the probability that any one of the two players will win are equal. Accrdingly, there are precisely 36/2 = 18 instances corresponding to appearance of the same color on the top face of each die.
Let b = # blue faces in the second die, so that:
# red faces in the second die = 6-b
Accordingly, we obtain:
1*b + 5(6-b) = 18, so that:
4b = 12, giving b=3, and hence:
Consequently, precisely three of the faces of the the second die are colored red while the remaiing 3 faces are colored blue.