Another Probability Question(Not so simple)

[Beowulf2007]

Homework Statement

An honest coin is throughn again and again until two consigutive heads or tails appears.
Assume that the troughs are independent. Let T be the number of throughs which can take on the values 2,3,...

Calculate $$P(T = n)$$ for $$n = 2,3,...$$

The Attempt at a Solution

P(HH,TT) = P(HH) + P(TT) = $$2^{-n} + 2^{-n} = \frac{1}{2^{(-n-1)}}$$

for n = 2,3,...

Does this look okay? Or do I need to add something more text?? If yes, could somebody please give me a hint or surgestion?

Sincerely Yours
Beowulf

 

[Shooting Star]

(you have summed the final answer wrongly. Also, please use a spell-checker.)

If I understand correctly, (n-1) throws have gone before the nth throw, with alternating H and T. The (n-1)th and the nth throw are HH or TT.

If it had started with H, then it has gone like HTHT... for n-1 throws. The P of that is (1/2)^(n-1). The P of the nth throw matching that of n-1 is 1/2. So, total P for this case is (1/2)^(n-1)*(1/2).

Similarly, if it had started with T, then the total P of nth being same as n-1 is again (1/2)^(n-1)*(1/2).

So, the reqd P is the sum of the two = (1/2)^(n-1).

This tallies with your answer.

(As a check, let n=2, which means that the first two throws will be the same. We know that P of HH or TT is 1/2. Similarly, test for n=3. HTT or THH has prob 1/4.)