Drawing Blanks

Premature Optimization is a Prerequisite for Success

Expected number of coin tosses

with 2 comments

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 Embarrassed http://www.wolframalpha.com/input/?i=sum_%28n%3D0%29%5Einf%28n*%281%2F2%29%5En%29
The 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.
Expected number of tosses to get N heads in a row:
An elementary way to compute the expectation:
“Expectaion” is an anagram of “Inexact Poet

Written by bbzippo

11/05/2009 at 3:45 pm

Posted in math

Tagged with

2 Responses

Subscribe to comments with RSS.

  1. […] leave a comment » I wrote before how to compute the expected number of coin tosses needed to throw a head: https://bbzippo.wordpress.com/2009/11/05/expected-number-of-coin-tosses/ […]

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

%d bloggers like this: