Expected number of coin tosses
What is the average number of coin tosses needed to throw a head?
The trial is a series of n tosses the first n-1 yielding tails and the nth yielding a head. The random variable is n with p(n)=(1/2)^n.
So the expectation is by definition sum_(n=1)^inf(n*(1/2)^n).
How do we compute this? Feed it into Alpha 
http://www.wolframalpha.com/input/?i=sum_%28n%3D0%29%5Einf%28n*%281%2F2%29%5En%29
http://www.wolframalpha.com/input/?i=sum_%28n%3D0%29%5Einf%28n*%281%2F2%29%5En%29The answer is 2.
But really, how do we sum the series? The only method I know is to take the geometric series
sum_(n=1)^inf(x^n) = x/(1-x)
and differentiate it.
I don’t have a fully rigorous solution for the expected number of tosses to get m heads in a row.
[...] leave a comment » I wrote before how to compute the expected number of coin tosses needed to throw a head: http://bbzippo.wordpress.com/2009/11/05/expected-number-of-coin-tosses/ [...]
Expected number of tosses to get N heads in a row « Drawing Blanks
11/25/2009 at 7:20 pm