mathschallenge.net logo

Coloured Dice

Problem

Two dice have each face coloured either red or blue. The first die has five red faces and one blue face.

The two dice are rolled.

How many faces on the second die need be red so that there is an equal chance of getting two faces of the same colour?


Solution

If P(Red on 2nd die)=p, then P(Blue on 2nd die)=1minusp.

P(same colour)=P(RR)+P(BB)=(5/6)p+(1/6)(1minusp)=(4p+1)/6.

But as we are trying to get, P(same colour)=1/2=3/6, it follows that 4p+1=3, 4p=2, and p=1/2; that is, three red and three blue faces.

This can be solved in a surprisingly simple way. Regardless of obtaining red or blue on the first die, the only way we would have equal chance of getting two faces of the same colour is if the second die has an equal number of red and blue faces.

Problem ID: 124 (Oct 2003)     Difficulty: 2 Star

Only Show Problem