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What is the expected number of tosses of a biased coin to get the $n^{th}$ head?

So my reasoning is as follow: Let $X_n$ be the number of tosses to get the $n^{th}$ head and let $p$ be the probability of getting head in a single toss. If the first toss is head, then $E[X_n]=p(1+E[X_{n-1}])+(1-p)(1+E[X_n])\Rightarrow E[X_n]=\frac{1}{p}+E[X_{n-1}]$. As $E[X_1]=\frac{1}{p}$, the recurrence gives $E[X_n]=\frac{n}{p}$.

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  • $\begingroup$ You sort of gloss over why $E[X_1]=1/p,$ but you have $E[X_1]=p (1+E[X_0])+(1-p)(1+E[X_1]).$ But $E[X_0]=0,$ so this is $E[X_1]=p+(1-p)(1+E[X_1])$ or $pE[X_1]=1$ or $E[X_1]=1/p.$ $\endgroup$ – Thomas Andrews Jul 3 '19 at 14:52
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    $\begingroup$ This is one of the four ways of looking at the negative binomial distribution: Wikipedia has another, with mean $\frac{pr}{1-p}$ where $r=n$ and where Wikipedia's $1-p$ is your $p$ and where you add an additional $n$ by counting the heads as well as tails, but apart from that is the same $\endgroup$ – Henry Jul 3 '19 at 14:58
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Your answer is correct. You might want to argue why $E[X_1]=\frac{1}{p}.$

The result is the expected time $E[X_1]$ times $n,$ because you just sum the expected time between the $i$th heads and the $i+1$th head. The value will be $nE[X_1].$

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