It's getting a bit crowded in here

I'll look first at the general case of $n$ numbers chosen uniformly and at random from the interval $[0,1]$ .

Let the largest of these $n$ numbers be $M$ and the smallest $m$ . Now the probability that $m \gt x$ for $0 \lt x \lt 1$ is $p(m \gt x) = (1 - x)^{n}$ . By symmetry, the distribution of $(1 - M)$ will be the same as that of $m$ . We can then say that

$E[W] = E[M - m] = E[1 - 2m] = 1 - 2*\displaystyle\int_{0}^{1} p(m \gt x) dx =$

$1 - 2*\displaystyle\int_{0}^{1} (1 - x)^{n} dx = 1 - \dfrac{2}{n + 1} = \dfrac{n - 1}{n + 1}.$ .

For $n = 9$ we thus have $E[W] = \dfrac{8}{10} = \dfrac{4}{5}$ , and so $a*b = 4*5 = \boxed{20}$ .

(Note that as $n \rightarrow \infty$ we have $E[W] \rightarrow 1$ , as would be expected.)

Consider the measure of the sample space when W is a fixed. The smallest variable is uniformly distributed on [0, 1-W], and the largest variable is fixed by the smallest. This gives a measure of 1-W. The remaining 7 numbers are uniformly distributed on an interval of length W. Thus the measure for the remaining 7 numbers is W^7 K, where K is the measure for 7 numbers uniformly distributed on [0,1]. The total measure for any specific W is thus (1-W) W^7 K. Thus we get the expected value of

$\dfrac{\displaystyle\int_0^1 w (1-w)w^7 k dw}{\displaystyle\int_0^1 (1-w)w^7 k dw}$

$=\dfrac{\displaystyle\int_0^1 (1-w)w^8 dw}{\displaystyle\int_0^1 (1-w)w^7 dw}$

$=\dfrac{1/9-1/10}{1/8-1/9}=\dfrac{1/90}{1/72}=4/5$