Question

In a game a coin is tossed `2n+m` times and a player wins if he does not get any two consecutive outcomes same for at least `2n` times in a row. The probability that player wins the game is `(m+2)/(2^(2n)+1)` b. `(2n+2)/(2^(2n))` c. `(2n+2)/(2^(2n+1))` d. `(m+2)/(2^(2n))`
A. `(m+2)/(2^(2n)+1)`
B. `(2n+2)/(2^(2n))`
C. `(2n+2)/(2^(2n+1))`
D. `(m+2)/(2^(2n))`

Answer 1

Correct Answer - D
Player should get `(HT, HT, HT,...)orTH, TH,...)` at least 2n times. If the sequence starts from first place, then the probability is `1//2^(2n)` and if starts from any other place, then the probability is `1//2^(2n+1).` Hence, required probability is
`2((1)/(2^(2n))+(m)/(2^(2n+1)))=(m+32)/(2^(2n))`