Question

What is the probability in n flips of a fair coin that there will be two heads in a row?

Answer 1

To do this we need to find the probability that there are NO two consecutive heads. Let’s build a recursion relation:

In order to get such a string of length n, then either the last throw was T and there were no consecutive heads in the first n-1 throws, or the two last throws were TH and the first n-2 throws had no two consecutive throws. The probability than obeys the recursion relation

P(n)=P(n−1)2+P(n−2)4P(n)=P(n−1)2+P(n−2)4

With the initial condition

P(2)=34,P(1)=1.P(2)=34,P(1)=1.

To solve the recursion equation we assume P(n)=xnP(n)=xn and get

x2=x2+14x2=x2+14.

The solutions to the equation are

x±=1±5√4x±=1±54 .

Therefore the solution is

P(n)=a+xn++a−xn−.P(n)=a+x+n+a−x−n.

Setting this in the initial conditions yields

1=a+x++a−x−1=a+x++a−x−

34=a+x2++a−x2−34=a+x+2+a−x−2

and we get

a±=5±35√10.a±=5±3510.

Therefore the probability to have two consecutive heads in a string of n throws is

1−P(n)=1−5+35√10(1+5√4)n−5−35√10(1−5√4)n