(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

Given a fair coin, how many times will you flip it (on average) before you get heads?

2. Relevant equations

E(x) = x_{1}p_{1}+ x_{2}p_{2}+ x_{3}p_{3}+ ... + x_{n}p_{n}

3. The attempt at a solution

I know this is easy, but I'm stumped with the probability part.

Probability of heads is 1/2 and probability of tails is 1/2. So, after 1 flip, the probability of having landed heads right away is 1/2. So x_{1}p_{1}= (1)(1/2). Otherwise, flip again: x_{2}= 2. My question: does p_{2}= (1/2)^{2}? I got this by arguing that the probability of the sequence TH is P(T) times P(H). Similarly, for p_{3}, the sequence is TTH, so does p_{3}= (1/2)^{3}?

I'm also guessing that arbitrarily long sequences TTT...TH (of length n) are possible. Assuming I'm right so far, would I take the limit of E(x) as n approaches infinity?

In symbols,

[tex] \lim_{n \to \infty} E(x) := E = \sum_{n=1}^\infty \frac{n}{2^n} = 2 [/tex]

So is the expected value 2 tosses? Will I usually have gotten heads on my second toss?

Thanks!

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# Homework Help: Expected number of coin tosses before landing heads

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