What's the probability that n random numbers from [0,1] will sum less than 1?

(For purists: "uniformly distributed, independent" random numbers are assumed.)

The distribution density of the sum of the n independent identically-distributed variates, each uniform on [0,1], is the n-fold convolution of the uniform density with itself. By convolution of two functions f and g is meant the integral over all y of f(y)g(x-y)dy. The convolution result at the value x is obtained by taking the total area under f times a reversed and right-shifted copy of g, the amount of the shift being x. Depicting the successive  convolution processes graphically, it is easy to see that the result for values of x between 0 and 1 will be to integrate the previous result from 0 to x and thus raise x to the next higher power and divide it by n. The result of the nth convolution will then be (x^(n-1))/(n-1)!. Integrating this from 0 to 1 then gives the probability we seek, namely 1/n!.

Posted by Richard on 2004-08-02 22:05:38