Question

Let X equal the larger outcome when a pair of 6-sided dice are rolled.

(a) Assuming the two dice are independent, show that the probability function of \(X\) is \(f(x)=\frac{2 x-1}{36} \quad x=1, \ldots, 6\)

(b) Confirm that \(f(x)\) is a probability function.

(c) Find the mean of \(X\).

(d) Can you generalise \(E(X)\) to a pair of fair \(m\) -sided dice?

\(\left[\right.\) Hint: recall that \(\sum_{i=1}^{n} i=n(n+1) / 2\) and \(\left.\sum_{i=1}^{n} i^{2}=n(n+1)(2 n+1) / 6\right]\)

Answer 1

We see a pattem in frequencies which are 1,3,5,7,9,11 . which are odd number and hence the

\(f(x)=\frac{2 x-1}{36} \quad x=1,2,3,4,5,6 \ldots\)

b.

For \(\mathrm{f}(\mathrm{x})\) to be pdf

\(\sum \mathrm{f}(\mathrm{x})=1\)

\(\sum_{x=1}^{6} \frac{2 x-1}{36}=\frac{2 \sum_{x=1}^{6} x-\sum_{x-1}^{6} 1}{36}=\frac{2 \frac{6 * 7}{2}-6}{36}=\frac{36}{36}=1\)

c.

Mean

\(\sum x f(x)=\frac{2 \sum_{x=1}^{6} x^{2}-\sum_{x=1}^{6} x}{36}=\frac{2 \frac{6 * 7 * 13}{6}-\frac{6 * 7}{2}}{36}=4.472\)

The mean when number of sides are \(\mathrm{m}\) is given by

\(\sum \mathrm{xf}(\mathrm{x})=\frac{2 \sum_{x=1}^{m} x^{2}-\sum_{x=1}^{m} x}{36}=\frac{2 \frac{m(m+1)(2 m+1)}{6}-\frac{(m)(m+1)}{2}}{36}\)

\(\frac{1}{36}\left\{\frac{m(m+1)(2 m+1)}{3}-\frac{(m)(m+1)}{2}\right\}=\frac{1}{36}\left\{\frac{(m)(m+1)\{4 m+2-3\}}{6}\right\}\)

\(=\left\{\frac{(m)(m+1)\{4 m-1\}}{216}\right\}\)