# Another Probability Question(Not so simple)

1. Nov 26, 2007

### Beowulf2007

1. The problem statement, all variables and given/known data

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 $$n = 2,3,...$$

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

Last edited: Nov 26, 2007
2. Nov 27, 2007

### 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).