$E(X)$ of number of coin tosses to get at least one head and one tail

Question

Letting $X$ be the number of flips of a fair coin it’ll take to get at least one head and at least one tail, find the value of $E(X)$.

Attempted Solution:

Event Space: ${HT}$, ${TH}$, ${HHT}$, ${TTH}$, ${HHHT}$, ${TTTH}$$,...,$

The value for $X$ for these would be $2,2,3,3,4,4,..,$

The probabilities would be $1\over{4}$,$1\over{4}$,$1\over{8}$,$1\over{8}$,$1\over{16}$,$1\over{16}$$,...,$

Then multiplying the $X$ values by their respective probabilities and summing, we get

$2\over{4}$+$2\over{4}$+$3\over{8}$+$3\over{8}$+$4\over{16}$+$4\over{16}$+...+

= $2$($2\over{4}$+$3\over{8}$+$4\over{16}$+...+)

= $2$($\sum_{i=2}^{\infty}$${i\over{2^i}})$

= $2(1.5)$

= $3$

Did I do this correctly?

Anyone know how to do the infinite sum without an infinite sum calculator? It's been a while since I've done those. — Remy, Oct 01 '17 at 18:52
@JohnH See, e.g., https://math.stackexchange.com/questions/30732/how-can-i-evaluate-sum-n-0-inftyn1xn — Travis Willse, Oct 01 '17 at 18:55

score 1 · Accepted Answer · answered Oct 01 '17 at 19:00

In response to your comment, the infinite sum is a little tricky, but you can break it down:

$\frac{2}{4}+\frac{3}{8}+\frac{4}{16}+...$

$=\frac{2}{4}+\frac{2}{8}+\frac{2}{16}+...+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+...+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+...$

With a series of geometric sums. We then have

$= 2(\frac{\frac{1}{4}}{1-\frac{1}{2}}) + \frac{\frac{1}{8}}{1-\frac{1}{2}}+ \frac{\frac{1}{16}}{1-\frac{1}{2}}+...$

$= \frac{1}{2} + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} +...$

$=\frac{1}{2} + \frac{\frac{1}{2}}{1-\frac{1}{2}}$

$=1.5$

Thanks, it makes a lot more sense after breaking down the fractions in that way. — Remy, Oct 01 '17 at 19:04

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