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.)

Let X1, ..., Xn be iid with U[0,1]

P(X1 + ... + Xn<1)
=¡Ò...¡ÒP(X1<1 - y2 - ... - yn|X2¡Öy2, ..., Xn¡Öyn)P(X2¡Öy2, ..., Xn¡Öyn), integrals all from 0 to 1
where X¡Öx means X in [x, x+dx]

P(X2¡Öy2, ..., Xn¡Öyn)=P(X2¡Öy2) x ... x P(Xn¡Öyn)
and P(Xi¡Öyi) = f(yi) = 1 (pdf for all Xi =1)

P(X1 + ... + Xn<1)
=¡Ò...¡Ò1 d(yn) ... d(y1)
where the ith integrand range from 0 to 1 - y1 - ... - y(i-1)

Now with the following change of varibles:
ri = 1 - y1 - ... - yi, thus d(ri) = -d(yi)
When yi = 0, ri = 1 - y1 - ... - y(i-1) = r(i-1)
When yi = 1 - y1 - ... - y(i-1), ri = 0

Therefore, the above becomes
=¡Ò...¡Òd(rn) ... d(r1), with ith integrand range from 0 to r(i-1)
=¡Ò...¡Òr(n-1) d(r(n-1)) ... d(r1)
=¡Ò...¡Ò[r(n-2)]^2/2! d(r(n-2)) ... d(r1)
...
=¡Òr1^(n-1)/(n-1)! d(r1)
= r1^n/n!, r1 from 0 to 1
= 1/n!

which is exactly the solution given by Richard right below.

Posted by Bon on 2004-08-05 19:35:54