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I am trying to work through this problem from Sheldon's 10th Ed., 7.61. It states

A coin that comes up heads with probability p is continually flipped. Let N be the number of flips until there have been both at least n heads and at least m tails. Derive an expression for E[N] by conditioning on the number of heads in the first n + m flips.

Following their suggestion, if we let H be the number of heads obtained in the first $n+m$ flips, then $$E[N]=\sum_{j=0}^{n+m} E[N|H=j]P\{H=j\} = \sum_{j=0}^{n+m} E[N|H=j]\binom{n+m}{j}p^j(1-p)^{n+m-j}$$ Clearly, $n+m$ is the minimum number of flips required to obtain n heads and m tails, and if we assume that we have already obtained h heads, then we need $n-h$ more, while we must already have $n+m-h$ tails, so we need $m-(n+m-h)=h-n$ more. However, at that point I am stuck; it seems to me that I have not reduced the complexity of the problem since in obtaining $n-h$ more heads, we may or may not obtain $h-n$ more tails, and vice versa. I have access to the answer, and it states that $$\sum_{j=0}^{n+m} E[N|H=j]\binom{n+m}{j}p^j(1-p)^{n+m-j}=\sum_{j=0}^n \left(n+m+\frac{n-j}{p}\right)\binom{n+m}{j}p^j(1-p)^{n+m-j} + \sum_{j=n+1}^{n+m} \left(n+m+\frac{j-n}{1-p}\right)\binom{n+m}{j}p^j(1-p)^{n+m-j}$$ However, I cannot make the mental leap and see how $E[N|H=j]=\left(n+m+\frac{n-j}{p}\right)$ when $0≤j≤n$ and $E[N|H=j]=\left(n+m+\frac{j-n}{1-p}\right)$ when $n+1≤j≤n+m$.

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    $\begingroup$ Note that, with $n+m$ trials you must have completed one of your tasks and either you completed the other one on the last toss or you still have more to go on that one. $\endgroup$
    – lulu
    Commented Mar 3, 2023 at 0:25
  • $\begingroup$ should say: it's hard to follow what you wrote since you have $h$ as the summation variable but then your expressions depend on the undefined variable $j$. Perhaps you meant $h=j$? $\endgroup$
    – lulu
    Commented Mar 3, 2023 at 0:27
  • $\begingroup$ My apologies, when I worked it I used h as the number of heads, but Ross used j. I will fix it. Also, why would we necessarily be close to finishing a task? Suppose we had n+m-1 heads and only one tail? We would still have at least m-1 tails to go? $\endgroup$
    – Ted Lorance
    Commented Mar 3, 2023 at 0:49
  • $\begingroup$ If the number of Heads is $<n$ and the number of Tails is $<m$ then you can't have tossed the coin $n+m$ times. $\endgroup$
    – lulu
    Commented Mar 3, 2023 at 0:50
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    $\begingroup$ Take $h<n$. Then you have finished Tails but need $n-h$ more Heads. We know how long we expect it to take to do that... $\endgroup$
    – lulu
    Commented Mar 3, 2023 at 0:56

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I finally understood what lulu was trying to tell me (see the comments on the original post). If $j<n$, then we can be sure that $n+m-j>m$, so we have enough tails and we need only consider how many trials to get $n-h$ more heads. This is a negative binomial random variable and therefore has an expected value of $\frac{n-j}{p}$, so the expected number of total flips – which is $E[N|H=j]$ – is $n+m+\frac{n-j}{p}$.

Likewise, if $j≥n$, then we can be sure that we have enough tails and we need only consider how many trials to get $h-n$ more tails. The expected value of the additional number of flips is $\frac{j-n}{1-p}$ for a total number of flips $n+m+\frac{j-n}{1-p}$.

Thank you for your assistance.

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