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?

Basic algebra solves this one:

1st die:
5 red
1 blue
Requirements: matches = non-matches

r = red sides of second die
b = blue sides of second die
r + b = 6
(5 x r) red matches, (1 x b) blue matches
(1 x r), (5 x b) = non-matches
(note, fractions were removed as a function of the algebra for clarity)

(5 x r) + (1 x b) = (1 x r) + (5 x b)
5r + b = r + 5b
5(6-b) + b = 6-b + 5b
30 - 5b + b = 6 - b + 5b
24 - 4b = 4b
24 = 8b
b = 3

6 - 3 = 3 = r

second die = 3 blue; 3 red

Posted by Eric on 2003-03-30 11:57:32