Question

What is the expected number of heads(until any tail comes up) as a result of tossing a bias coin that gives heads 1/3 of the time?

What about the coin gives head 3/4 of the time?

Answer 1

Expected heads = weighted average of number of heads and probability of that happening

So, P(1 heads , 1 tail) = 1/3 * 2/3

So, P(2 heads , 1 tail) = (1/3)^2 *(2/3)

.

.

So, P(n heads, 1 tail) = (1/3)^n (2/3) ( where, n is infinite)

So, Expected number of heads is =1* (1/3)*(2/3) + 2* (1/3)^2*(2/3) +... + n*(1/3)^n (2/3)

= (2/3)*( 1/3 + 2*1/3^2 + 3*1/3^3 +... + n*1/3^n)

Let s = ( 1/3 + 2*1/3^2 + 3*1/3^3 +... + n*1/3^n)

So, s/3 = (1/3)*(( 1/3 + 2*1/3^2 + 3*1/3^3 +... + n*1/3^n))

s-s/3 = 2s/3 = (1/3)+(1/3)^2 +... (1/3)^n

s =(3/2)*(1/3 + ..1/3^n)

So, Expected number of heads is = (2/3)*s = (2/3)*(3/2)*(1/3+...+1/3^n)

= (1/3) / (1-1/3)

= (1/3)/(2/3)

= 1/2 [Answer]

--------------------------

Similarly, if for 3/4 probability of heads of biased coin , we will have :

Expected number of heads =

1*(3/4)*(1/4) + 2*(3/4)^2*(1/4) +... n*(3/4)^n*(1/4)

= (1/4)*( 3/4 + 2*(3/4)^2 + .. n*(3/4)^n)).......1

Let s = ( 3/4 + 2*3/4^2 + 3*3/4^3 +... + n*3/4^n)

So, 3s/4 = (3/4)*(( 3/4 + 2*3/4^2 + 3*3/4^3 +... + n*3/4^n))

s - 3s/4 = s/4 = 3/4 + (3/4)^2 + ... + (3/4)^n

s/4 = (3/4)/(1-(3/4)) = (3/4)/(1/4) = 3
=> s = 12

=> Expected number of heads = 1/4*(12) = 3 [Answer