Coin equivalence: THT and TTH?

Question

This is a follow-up to the question:

http://answers.yahoo.com/question/index;_ylt=Ah6X0FC4qjh0w16E4QppFxPsy6IX;_ylv=3?qid=20080825143625AAdmSPo

which is, in turn, a follow-up to

http://answers.yahoo.com/question/index;_ylt=Anr2nrTJ85YJWm4CKDAmXIg8.Bd.;_ylv=3?qid=20080825135226AAIsU0t

Suppose we flip a coin until either the sequence THT or TTH shows up. If we want to have equal probabilities for either sequence to end our coin flipping, how should we weight the coin? That is, if the probability in any toss of heads is p, what value of p creates equal probabilities (of 1/2) of ending with THT or with TTH?

[I have two more follow-ups in mind; let me know if you like this enough to see some more...]

Answer 1

Have you actually worked this out? Because I'm getting p(H) = 1.

As with the previous question, we only start when we get our first T.

If we get TT, we are guaranteed to end with TTH rather than THT.

If we get TH, then there is q prob of ending with THT, and p prob of getting another H and starting all over again.

So far we have

P(TTH) = q

P(THT) = pq

P(starting over again) = p^2

Once we start over, these same probabilities keep repeating themselves.

So we require that q = pq

or q - pq = 0

q^2 = 0

p = 1

Of course physically with p = 1, we have zero chance of ending with either THT or TTH. That would be equal. I've even done  a monte carlo on this. The ratio THT / TTH is only approaching 1 as p approaches 1.