Question

You toss a fair coin till you get n consecutive heads. What is the expected number of coin tosses en to get n consecutive heads.

Answer 1

Hi there!
Here's the answer:

¶¶¶ Expected no. of tosses to get, 'n' consecutive heads:
=> n>1.
°Say Expected No. of tosses = f

¶¶ First Let's assume n=2 and analyse the possible cases:
n=2=> 2 consecutive heads
Expected no. of tosses to get 2 consecutive heads:

Case-1:
If the first toss is Tails.
{T...}
°Probability of this Event = 1/2
One toss is wasted. So f increases by 1.
°Tosses Required = (t+1)

Case-2:
If the first toss is head and second toss is Tails.
{H,T…}
°Probability of this Event = (1/2²) = 1/4
Here 2 tosses are wasted. So expected no. of tosses increases by 2.
°Tosses Required = (t+2)

Case-3
If the first toss is head and second toss is also head, then we are done.
{H,H,…}
°Probability = (1/2²) = 1/4
°Tosses required = 2

Expected No. of tosses is the sum of expected value of these 3 cases.

According to Rule of Linearity of Expectation, we have

t= [(1/2)(t+1)] + [(1/4)(t+2)] + [(1/4)(2)]
=>t = [(t+1)/2] + [(t+2)/4] + (1/2)
=>t = [2(t+1) + (t+2) + 2]/4
=> 4t = 3t +6
=> t = 6

•°• Expected No. of tosses for 2 consecutive heads= 6

¶¶ Now consider expected no. of tosses for 'n' consecutive heads:

Same cases as previous consideration

case-a)
If we get 1st, 2nd, 3rd,...,nth tail as the first tail in the experiment, then we have to start all over again.

case-b)
Else we are done.

°For the 1st flip as tail, the part of the equation is (1/2)(x+1)
°For the 2nd flip as tail, the part of the equation is (1/4)(x+2)
...
…
…
°For the kth toss as tail, the part of the equation is (1/(2 k ))(x+k)
...
…
°For the N'th flip as tail, the part of the equation is (1/(2 N))(x+N)

°The part of equation corresponding to case (b) is (1/(2^N ))(N)

Adding we get,

x = (1/2)(x+1) + (1/4)(x+2) + ... + (1/(2^k))(x+k) + .. + (1/(2^N))(x+N) + (1/(2^N))(N)

The entire equation can be very easily reduced to the following form:
x = 2^(N+1) -1.

Thus, the expected number of coin tosses for getting N consecutive heads is 2[(2^N) - 1].

:)
Hope it helps
Comment if you need to clear anything.