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Lower and upper bounds for a gambler's ruin problem
#!/usr/bin/env python
from __future__ import division
from numpy import *
def iter(n,boundary):
p = ones(n)
b = boundary.copy()
j = 3*arange(1,n)
for i in xrange(1000):
b[:n] = p
p[1:] = .5*(p[:-1]+b[j])
return p
print repr(iter(1000,zeros(3000))[1])
print repr(iter(1000,((1-sqrt(5))/2)**arange(3000))[1])
@girving
Owner

Output:

cone:~% ./gambler 
0.57477536382958938
0.57477536382958938
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