# Probability of rolling three different dice [duplicate]

Three fair dice, colored red, blue, and yellow, are rolled once. We denote it by R, B, Y the numbers appearing on the upper side of the three dice, respectively.

(a) $$P(R=B)=\frac{1}{6}$$

(b) $$P(R\lt B)=\frac{1}{2}(1-P(R=B))=\frac{5}{12}$$

(c) $$P(R=B=Y)=\frac{1}{36}$$

(d) $$P(R\lt B\lt Y)$$

I know how to do (a), (b), (c). I think (d) is $$\frac{\left(^{6}_{3}\right)}{216}$$

but this problem need us to solve it with (c) and symmetry. How to do it?

• All of your answers are correct. – N. F. Taussig Mar 10 at 16:33
• @N.F.Taussig No. It doesn't solve it with symmetry. Also thank you – Maggie Mar 10 at 17:01

You have to find $$P(R\lt B\lt Y)$$

All possible rolls can be divided into cases

$$P(R \lt B \lt Y)$$ has $$3!=6$$ permutations.

So you can subtract rolls in which any $$2$$ or all $$3$$ dice show the same number from the total arrangements and then divide by $$6$$

• thank you! $\frac{1}{6}(1-P(R=B=Y)-P(R=B\neq Y)\times 3)=\frac{1}{6}(1-\frac{1}{36}-\frac{5}{6\times 6}\times 3)=\frac{5}{54}$. That's correct. – Maggie Mar 10 at 16:58

The way you solve d) is correct.

It can also be solved with inclusion/exclusion and symmetry as:$$\frac1{3!}\left(1-P(R=B\text{ or } R=C\text{ or } C= B)\right)=$$$$\frac16(1-3P(R=B)+3P(R=B=C)-P(R=B=C))=$$$$\frac16(1-3P(R=B)+2P(R=B=C))$$