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Comment by voldacar | Hacker News Reader

This result that the number of random reals on the interval (0,1) you need to form a sum > 1 is exactly equal to e is fascinating and beautiful.

Does anyone know of an accessible proof of this statement?

edit: the tricky part is recognizing that P(sum < 1) is equal to 1/n!, after that, it's the world's most recognizable taylor series

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This was framed as a problem on 538's The Riddler column a few years ago: https://fivethirtyeight.com/features/should-you-pay-250-to-p...

One solver in particular submitted several elegant solutions: http://math.uchicago.edu/~timblack/blog/addtoone.html

The compound interest explanation was particularly intriguing. In part because I haven't fully grokked the discrete-vs-continuous interest difference yet (also haven't tried very hard -- just remember being a bit surprised by some things mentioned in Paul Wilmott Introduces Quantitative Finance). I liked the intuition of simply "owning very many such bills" reducing to the expectation.

Nice fact! How about this for a proof: Let X_j be a sequence for unif(0,1) random variables, S_n the sequence of partial sums (so S_4=X_1+X_2+X_3+X_4).

If N is the number of uniforms needed to have a sum bigger than 1, you can see that P(N>n)=P(S_n<1).

P(S_n<1) is not too bad to compute, or you can wikipedia it (Irwin-Hall distribution) to find the expression P(S_n<1)=1/n!

Combining the things, E[N]=sum(P(N>n),n=0..infinity)=sum(1/n!,n=0..infinity)=exp(1).

Thank you, I will look into this. I've taken discrete math and some graph theory, but never stats, so all those families of distributions are alien to me :)

You can use the relation f(x) = 1 + integral_0^1 f(t) dt, then just check that e^x is a solution or use taylor expansion.

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