TITLE: Suppose that we flip a coin until either it comes up tails twice or we have flipped six times. What is the expected number of times we flip the coin?
QUESTION [7 upvotes]: So this is the answer that I got (where $N$ represents the number of coin flips after which you can stop) but can someone tell me if this looks right?
$P(N=2) = P(TT) = \dfrac{1}{4} $
$P(N=3) = P(HTT , THT) = \dfrac{2}{8} $
$P(N=4) = P(HHTT;HTHT;THHT) = \dfrac{3}{16} $
$P(N=5) = \dfrac{4}{32} $
$P(N=6) = 1 - \dfrac{1}{4} - \dfrac{2}{8} - \dfrac{3}{16} - \dfrac{4}{32} = \dfrac{3}{16}$
$E(N) = 2\left(\dfrac{1}{4}\right) + 3\left(\dfrac{2}{8}\right) + 4\left(\dfrac{3}{16}\right) + 5\left(\dfrac{4}{32}\right) + 6\left(\dfrac{3}{16}\right) = \dfrac{15}{4} = 3.75$

REPLY [2 votes]: What you are looking at is a right-censored negative binomial distribution.  That is, let $Y = X \wedge 6 = \min\{X, 6\}$ where $X \sim {\rm NegBinomial}(r = 2, p = 1/2)$, where the particular choice of parametrization for $X$ is $$\Pr[X = k] = \binom{k-1}{k-r} (1-p)^r p^{k-r}, \quad k = r, r+1, r+2, \ldots.$$  That is, $X$ counts the total number of trials needed to observe $r = 2$ tails, where each trial is an independent Bernoulli trial with probability of observing tails equal to $p = 1/2$.  Then $Y$ is $X$ right-censored at $6$.  So we can calculate $$\begin{align*} {\rm E}[Y] &= {\rm E}[X \wedge 6] \\ &= \sum_{k=2}^5 k \Pr[X = k] + \sum_{k=6}^\infty 6 \Pr[X = k] \\ &= 6 - \sum_{k=2}^5 (6-k)\Pr[X = k] \\ &= 6 - \frac{9}{4} = \frac{15}{4}.\end{align*}$$