[susam]

And for people who like equations, here is my attempt at explaining it.

Assume each flip is independent and the bias remains same in each flip.

Let

  P(H) = p,
  P(T) = 1 - p.

Then

  P(HH) = p^2,
  P(HT) = p(1 - p),
  P(TH) = (1 - p)p,
  P(TT) = (1 - p)^2.

Therefore

  P(HT or TH) = 2p(1 - p).

Now calculate

  P(HT | HT or TH) = p(1 - p) / (2p(1 - p)) = 1/2,
  P(TH | HT or TH) = (1 - p)p / (2p(1 - p)) = 1/2.

[taegee]

You don't need conditional probability here, as the flips are independent.

It's just p(H)p(T).

And p(H)p(T) = p(T)p(H), thus 2*p(H)p(T) = 2p(1-p).

[susam]

Independence tells us how to compute the probability of a sequence like HT or TH:

  P(HT) = P(H)P(T) = p(1 - p)

But the question I am addressing is not just "what is the probability of HT?" It is "given that the two flips are different, what is the probability that the order was HT rather than TH?"

That is a conditional probability:

  P(HT | HT or TH)

[clickety_clack]

That wasn’t what he was trying to prove, but the proof could be done without conditionals like this:

If: p(H)p(T) = p(T)p(H)

And: p(H)p(T) + p(T)p(H) = 1

Then: p(H)p(T) = p(T)p(H) = 0.5

[cdaringe]

Thats how i noodled thru it internally