Given a fair coin, on average what is the number of tosses needed to get 3 heads in a row?

Question

My attempt:

Let $H$ be the outcome heads, and $T$ be tails. Let $X$ be the random variable "number of tosses required to get 3 consecutive heads". The minimum number of trials (tosses) is 3. Thus we have infinite sample space:

$$\{HHH, THHH, TTHHH, HTHHH, TTTHHH, THTHHH, HTTHHH, HHTHHH,....\}$$

For the first 2 elements of sample space, we have the probabilities $(1/2)^{3}$ and $(1/2)^{4}$ respectively. Then we have 2 ways to get 3 consecutive heads out of 5 tosses, hence the probability is $2(1/2)^{5}$. Similar for 4th, 5th and 6th tosses. For $n = 7, 8, 9,... \infty$ tosses, we have

$$2^{n-4} - \frac{(n-5)(n-6)}{2}$$ ways of getting 3 consecutive heads.

Thus the expectation is

$$E(X) = 3(1/2)^{3} + 4(1/2)^{4} + 5 \cdot 2 (1/2)^{5} + 6 \cdot 4(1/2)^{6} + \sum_{n=7}^{\infty} \frac{n}{2^{n}}\left(2^{n-4} - \frac{(n-5)(n-6)}{2}\right) $$

However the infinite sum above doesn't converge, so obviously the approach is wrong.

I have looked at related posts (i.e. Expected Number of Coin Tosses to Get Five Consecutive Heads) but I don't see how my approach fails.

Edit: computation of $$2^{n-4} - \frac{(n-5)(n-6)}{2}, \qquad n \geq 7.$$

I need to compute the number of ways that X -values can be next to each other for a given number of tosses $n \geq 7$. For this $n$, I have a sample space element of the form $...TXXX$, consisting of $n$ symbols $T$ and $H$. Thus, I have a choice of $2^{n-4}$ arrangements for the first $n-4$ symbols. Out of those, I need to exclude the sequences of symbols of the form $XXX...X$, for 3 or more values of $X$. There are $(n-6)$ such arrangements for 3 consecutive $X$ values, $(n-7)$ arrangements for 4 consecutive $X$s, $(n-8)$ arrangements for 5 $X$s, and so on. Hence for the total of such arrangements, we have the arithmetic progression

$$n-6, n-7, n-8,...0.$$ For a given $n$, the sum turns out to be $$\frac{(n-5)(n-6)}{2}$$ Hence, we have the result $2^{n-4} - \frac{(n-5)(n-6)}{2}$ as stated above...

Answer 1

Here is a standard trick for this kind of situation. It hinges on the fact that expectation satisfies a law similar to the law of total probability. You'll see what I mean below.

Let $A$ be the event "the first toss is a heads". Then we have $$ E(X)=E(X\mid A)P(A)+E(X\mid \bar A)P(\bar A)\\ =\frac12E(X\mid A)+\frac12(1+E(X)) $$ since the expectation of $X$ given that the first toss is a tails is clearly one more than the expectation of $X$; you just wasted a throw.

Now, for $E(X\mid A)$, we can do a similar trick: let $B$ be the event "the second toss is a heads", and we get $$ E(X\mid A)=E(X\mid A,B)P(B)+E(X\mid A,\bar B)P(\bar B)\\ =\frac12E(X\mid A,B)+\frac12(2+E(X)) $$ Finally, we must calculate $E(X\mid A,B)$. You can probably guess how: let $C$ be the event "the third toss is heads", and we get $$ E(X\mid A,B)=E(X\mid A,B,C)P(C)+E(X\mid A,B,\bar C)P(\bar C)\\ =\frac12\cdot 3+\frac12(3+E(X)) $$ Since $A,B,C$ together means you succeeded to get three heads in a row on the first three throws, we must have $E(X\mid A,B,C)=3$.

Now you can just insert everything into the original equation, and solve for $E(X)$.

Answer 2

Let's define the states as the expected stopping time from given n consecutive Heads, i.e. $S_n = E\{\tau \mid \text{n consecutive H} \}$

Starting with zero Heads, we can write transition recurrence equations.

\begin{align} S_0 &= \frac12 S_0 + \frac12 S_1 + 1 \\ S_1 &= \frac12 S_0 + \frac12 S_2 + 1 \\ S_2 &= \frac12 S_0 + \frac12 S_3 + 1 \end{align}

Note that $S_3=0$ by definition (we're done after $HHH$). From where you can solve $S_0$

Answer 3

Let $e_k$ $(0\leq k\leq2)$ be the expected number of additional tosses when we have $k$ heads "on the stock". We then have the following system of equations: $$ e_0=1+{1\over2}e_0+{1\over2}e_1,\qquad e_1=1+{1\over2}e_0+{1\over2}e_2,\qquad e_2=1+{1\over2}e_0\ .$$ Solving gives $e_0=14$.

Answer 4

You are not subtracting off conditions like XXXTXXXTXXX, since you are only counting the variations where there is only 1 XXX. As n gets large, you incur an exponential penalty from all these cases you are ignoring.

Answer 5

The more general case of asking for the average number $E$ of rolls of a fair $X$-sided die ending with a run of $N$ consecutive heads is answered in this MSE post as \begin{align*} E=\frac{X(X^N-1)}{X-1}\tag{1} \end{align*}

Here we apply (1) by considering a coin as two-sided die ($X=2$) and tosses ending in heads with run-length $N=3$. We conclude the average number of trials is \begin{align*} \color{blue}{E(X=2,N=3)=\frac{2\left(2^3-1\right)}{2-1}=14} \end{align*}