- #1
Unit
- 182
- 0
Homework Statement
Given a fair coin, how many times will you flip it (on average) before you get heads?
Homework Equations
E(x) = x1p1 + x2p2 + x3p3 + ... + xnpn
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 x1p1 = (1)(1/2). Otherwise, flip again: x2 = 2. My question: does p2 = (1/2)2? I got this by arguing that the probability of the sequence TH is P(T) times P(H). Similarly, for p3, the sequence is TTH, so does p3 = (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!