A fair coin is tossed until both heads and tails have been obtained. Let x = the number of tosses needed (for example, if the sequence of tosses is HHT, then x = 3; If the sequence is TTTTTH, then x = 6).

Find the expected value of x aka E(x).

I think E(x) = 1/P, where P is the probability of getting a heads or tails.


closed as off-topic by Nate Eldredge, Thomas, Davide Giraudo, azimut, Vedran Šego Oct 2 '13 at 15:44

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    $\begingroup$ What is your attempt? $\endgroup$ – tylerc0816 Oct 2 '13 at 14:55
  • $\begingroup$ Look at the definition of the expectation. $E(x) = (0 \times \text{probability of stopping after 0 tosses}) + (1 \times \text{probability of stopping after 1 tosses}) + \ldots$. $\endgroup$ – utdiscant Oct 2 '13 at 14:58

Whatever the result of the first toss is, say tail, we have used up one toss, and are waiting for head to show up.

The waiting time after the first toss has geometric distribution, with mean $\frac{1}{1/2}=2$.

So the expected waiting time until each of head and tail has shown up is $3$.


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