Two "Heads" are better than one

Relevant wiki: Expected Value - Problem Solving

Suppose that I toss a coin $n$ times in a row. For any $1 \le j \le n-1$ , let $X_j$ be the random variable which is equal to $1$ if Heads is tossed by the $j^\mathrm{th}$ and $(j+1)^\mathrm{st}$ coins, and $0$ otherwise, so that the total number of occurrences of consecutive Heads is $X \; = \; X_1 + X_2 + \cdots + X_{n-1} \;.$ It is clear that $E[X_j] \,=\, P[X_j = 1] \; = \; \tfrac14$ for each $1 \le j \le n-1$ , and hence that $N \; = \; E[X] \; = \; E[X_1] + E[X_2] + \cdots + E[X_{n-1}] \; = \; \tfrac14(n-1) \;.$

For $n=100$ , the value of $100N$ is $25 \times 99 = \boxed{2475}$ .

For those who are not comfortable with using random variables like this, here is another approach. Let $H_n$ be the expected number of consecutive Heads obtained in $n$ tosses, given that the first toss is Heads, and let $T_n$ be the expected number of consecutive Heads obtained in $n$ tosses, given that the first toss is Tails. If $N_n$ is the number of consecutive Heads obtained in $n$ tosses, then (conditioning on the outcome of the first toss) we have $N_n \,=\, \tfrac12(H_n + T_n)$ .

Conditioning on the outcome of the second toss tells us that $H_n \; = \; \tfrac12(1 + H_{n-1}) + \tfrac12T_{n-1} \; = \; \tfrac12 + N_{n-1} \qquad \qquad T_n \; =\; \tfrac12H_{n-1} + \tfrac12T_{n-1} \; = \; N_{n-1}$ Averaging these equations gives $N_n \,=\, \tfrac14 + N_{n-1}$ . Since $N_1 = 0$ we obtain $N_n = \tfrac14(n-1)$ .

In $100$ tosses, we have $99$ "double tosses". In each one of these $99$ "double tosses", two heads will come up with probability $1/4$ . Then the expected value of two heads in a row is $99*(1/4) = 24.75$

The total different result we can have in $100$ toss is $2^{100}$ .

If we write the result of $100$ toss in a row, we can see that there are $99$ positions for two heads.

Example:

First position : HH####...

Second position : #HH###...

Third position : ##HH##...

........

Each position has $2^{98}$ different path.

The the total number of two heads in all result is $2^{98}*99$

The expected number $(N)$ is $2^{98}*99/2^{100}$ = $24.75$

$100N$ = $24.75*100$ = $2475$