A fair coin is continually flipped until heads appears for the 10th time. Find the number of expected tails.
Im very lost in this problem, can someone help? I think I have to use neg binomial, but not sure, any help will be appreciated!
[Math] A fair coin is continually flipped until heads appears for the 10th time. Find the number of expected tails
probability
Best Answer
Informally, it will take 20 flips on average to get 10 heads since there is half a chance that it will be heads on every flip. You should, in that time, see $20-10 = 10$ tails. More rigorously:
Out of a total of $k+10$ flips, we want $k$ to be tails. The last flip must be a heads, so we have to choose $k$ places for the tails from $k+10-1 = 9 +k$ in $\binom{9+k}{k}$ ways. Heads and tails are equiprobable, so we have the appropriate exponent of $0.5$ in each case. Thus the probability of exactly $k \geq 0$ tails being seen before the 10th heads is $\binom{9+k}{k}\cdot 0.5^{10}\cdot 0.5^k$. The expected number of tails is $$ \begin{align} \sum_{k=0}^{\infty} \binom{9+k}{k}\cdot 0.5^{10}\cdot 0.5^k \cdot k &= 0.5^{10}\cdot \sum_{k=0}^{\infty} \binom{9+k}{k}\cdot 0.5^k \cdot k \\ &= 0.5^{10}\cdot \sum_{k=0}^{\infty} \frac{(9+k)!}{k!\,9!}\cdot 0.5^k \cdot k \\ &= \frac{0.5^{10}}{9!}\cdot \sum_{k=0}^{\infty} (k+9)\dotsm (k+1)k\cdot 0.5^k \end{align} $$
I do not know how to simplify this sum by hand, but WolframAlpha tells me that it evaluates to $9.99\ldots$ or almost 10. So you would expect to see 10 tails before the 10th head.
Edit: I do know how to figure this sum out using hypergeometric series. It turns out that $\sum_{k=0}^{\infty} (k+9)\dotsm (k+1)k\cdot 0.5^k$ can be rewritten as $\sum_{k=0}^{\infty} \frac{2}{10!}(k+1)_{10}0.5^{k+1}$. The hypergeometric form makes the original sum $S = \frac{10!}{2}\,{}_1\!F_0[11;;0.5\,] = \frac{10!}{2}(0.5)^{-11}$. So the expected number is $$\frac{0.5^{10}}{9!}\cdot \frac{10!}{2}(0.5)^{-11} = 10$$ which works out perfectly.