A Fair Coin is Continually Flipped Until Heads Appears for the 10th Time

A fair coin is continually flipped until heads appears for the 10th time. Find the number of expected tails

Solution 1

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.

Solution 2

GrahamKemp, Could you expand on that a bit? I didn't understand the part about the variables having a geometric distribution. – shardulc

Sure. By definition: a (zero-based) geometric random variable is a count of failures before the first success in an indefinite sequence of iid Bernoulli trials. So if a fair coin is flipped indefinitely, the count of tails before the first head has a Geometric$_0(1/2)$ Distribution. And so too does the count of tails between the first and second head, and so on.

Let $T_1$ be the count of tails before the first head, and $T_k$ the count of tails between the $k-1$ and $k$-th heads.

$$T_k~\sim~\mathcal{Geo}_0(1/2)~\iff~ \mathsf P({T_k}=t)~=~\tfrac {1}{2}(1-\tfrac 12)^{t}~=~(\tfrac 1 2)^{t+1}$$

The expectation of such a random variable is $$\begin{align}\mathsf E(T_k)~=~&\tfrac{1-\tfrac 12}{\tfrac 12}\\[1ex]~=~&1\end{align}$$

Now if we count tails until the first head, then repeat times ten, we have counted of tails until the tenth head. Thus this is a sum of the ten iid geometric random variables. $~T~=~\sum_{k=1}^{10}T_k$

Hence by the Linearity of Expectation the expected count of tails before the tenth head is: $$\begin{align}\mathsf E(T)~=~&\sum_{n=1}^{10}\mathsf E(T_k)\\[1ex]~=~&10\end{align}$$

That is all.

$\Box$

Solution 3

Let $e$ = expected number of tails till first head.

Either you get heads on first toss with $Pr=1/2$ or need $e$ more tosses with $Pr= 1/2$

Thus $e = \frac12(1 +e) \to e = 1$

This is also the expected number of tails till the next head, thus ans $=10$

Comments

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!
Or use Linearity of Expectation. $~\mathsf E(T)~=~\mathsf E(\sum_{k=1}^{10}T_k)$ where $T_k$ is the number of tails between the $k-1$st and $k$th heads, each a Geometric$_0(\tfrac 12)$ distributed random variable -- hence having a mean of $1$. $~$ So $\mathsf E(T)~=~10$
and how the formula is?
@Mark Edited to make it clearer.
@GrahamKemp Could you expand on that a bit? I didn't understand the part about the variables having a geometric distribution.

Recents

masseyfrotteart.blogspot.com

Source: https://9to5science.com/a-fair-coin-is-continually-flipped-until-heads-appears-for-the-10th-time-find-the-number-of-expected-tails

A Fair Coin is Continually Flipped Until Heads Appears for the 10th Time