Question

A coin is flipped repeatedly until either two heads appear in a row or two tails appear in a row(and then stop). Find the exact answer for P(two heads in a row appears before two tails in a row) for a coin with probability p of getting heads.

Answer 1

Probability of getting a head = p ; Probability of getting a tail = 1-p

Lets say we get two heads when we flip twice , probability of two heads in 2 flips ,P(2) = p * p = p²

Similarly, P(3) = (1-p) * p *p = p² * (1-p) ( The first flip has to be tail)

Similarly, P(4) = p * (1-p) * p *p = p³ * (1-p). ( The first flip has to be head , 2nd has to be tail , 3rd & 4th has to be heads)

Similarly, P(5) = (1-p) * p * (1-p) * p *p = p³ * (1-p)². (The first flip has to be tail , 2nd has to be head , 3rd has to be tail, 4th & 5th has to be heads).

Similarly, P(6) = p * (1-p) * p * (1-p) * p *p = p⁴ * (1-p)². (The first flip has to be head , 2nd has to be tail , 3rd has to be head, 4th has to be tail ,5th & 6th has to be heads).

Similarly, P(7) = (1-p) * p * (1-p) * p * (1-p) * p *p = p⁴ * (1-p)³. (The first flip has to be tail , 2nd has to be head , 3rd has to be tail, 4th has to be head ,5th has to be tail, 6th & 7th has to be heads).

& so on.

Therefore the total probability , P = p² + p² * (1-p) + p³ * (1-p) + p³ * (1-p)²  + p⁴ * (1-p)²  + p⁴ * (1-p)³ + ....

P = ( p² + p³ * (1-p)+ p⁴ * (1-p)² + .... ) + (p² * (1-p) + p³ * (1-p)²+ p⁴ * (1-p)³ + .... )

P = p²(1+ p(1-p) + p²(1-p)² + ..) + p²(1-p) (1+ p(1-p) + p²(1-p)² + ..)

P = + = =

Therefore, total probability =