Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Suppose flipping a coin with probability $p$ to get a head. Suppose we flip it until a head appear. What is the mean number of flip required getting a head? (Better to use conditional expectation to show the mean)

share|cite|improve this question
What do you mean by the last sentence? Is it a question? – Stefan Hansen Sep 29 '12 at 10:07
Ya, but i am asking if someone can use conditional expectation to solve the problem – Mathematics Sep 29 '12 at 10:20

If $E$ is the expected number of flips, then we have following relation $$ E = p\cdot 1+ (1-p)\cdot(1+E)$$ because with probability $p$ we succeed at first try and with probability $1-p$ we have "wasted" one try and start again. Once we agree that $E$ is finite, this produces $$ E = \frac1p.$$

share|cite|improve this answer

It may be helpful to use a tree diagram and consider the various conditional probabilities. If you do this then you sum the following $E(X) = p + 2qp + 3q^2p + 4q^3p + \dots$

This can then easily be written as $E(X) = p\times(1 + 2q + 3q^2 + 4q^3 + \dots) = \frac{p}{(1-q)^2} = \frac{p}{p^2} = \frac{1}{p}$.

This result can be interpreted as follows:

$E(X) = 1\times p + (2qp + 3q^2p + 4q^3p + \dots)$

and the expression in brackets can be written as $(1 + \frac{1}{p})\times q$ so that we have

$E(X) = 1\times p + (1 + \frac{1}{p})\times q$

The first term is just $E(X|H) P(H)$ while the second term is $E(X|T) P(T)$.

share|cite|improve this answer
It might be worth stating explicitly that $q:=1-p$, for the sake of a more immediate understanding. – G. Sassatelli Nov 15 '15 at 17:15

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.