Question

Suppose that we flip a fair coin until either it comes up tails twice or we have flipped it six times. What is the expected number of times we flip the coin?

Answer 1

Let n be the number flips

then, we have to start from n =2 because we need to tails, so tails on first flip then tails on second flip

so, we will start from n = 2

Expected number of flips required to come up with two tailes before flipping it for 6 time =

[(n-1)/2^n)]

So, the expected number = n*probability

For n=2, expected number is n*[(n-1)/2^n)] = 2*[(2-1)/2^2] = 2/4

similarly for 3,4 and 5, we can write

For n=3, expected number is n*[(n-1)/2^n)] = 3*[(3-1)/2^3] = 3*(2/8)

For n=4, expected number is n*[(n-1)/2^n)] = 4*[(4-1)/2^4] = 4*(3/16)

For n=5, expected number is n*[(n-1)/2^n)] = 5*[(5-1)/2^5] = 5*(4/32)

Now, for the 6th flip, we have to stop

so, Probability of flips for 6th time = 1 - Probability of 2nd, 3rd, 4th and 5th flip = 1 - [1/4+2/8+3/16+4/32] = 1 - 13/16

so, probability of flip for 6th time = 3/16

expected number of flips for the 6th time to get two tails = 6*(3/16)

so, the total expected number of times we flip the coin = 2/4 + 3*(2/8)+ 4*(3/16)+5*(4/32)+6*(3/16) = 15/4

or we can write it as 3.75

so, the required answer is 3.75 or 15/4 flips