Isn't the pdf of a uniform distribution just f(x) = 1. What he gave was an exponential distribution with rate parameter lambda. The rest of the math seems correct. But the dependence on e is not surprising as it is relevant to the initial pdf. Am I missing something?
import random
totalsteps = 0
for i in range(100000000):
sum = 0.0
steps = 0
while sum < 1.0:
sum += random.random()
steps += 1
totalsteps += steps
print('AVG STEPS: ' + str(totalsteps / i))
It's funny. I read the problem, estimated that it was "probably 2 or 3" and then wasn't all that surprised to find that the solution was e. The derivation of it is pretty cool, too.
You're right. I think the article's reasoning is fallacious. At least, I don't see what the Poission distribution has to do with this at all.
One way to arrive at the answer correctly is to write the number of samples needed before they first sum to 1 as N and observe that the probability that P[N > k] = 1/k!. You can get this from a k-dimensional integral. A general formula for E[N] is Sum_k=0^infty P[N > k] thus E[N] = e.
