[jvanderbot]

It may be more likely that H or T happens (an unfair coin), but in a pair of H and T, both HT and TH are equally likely. Therefore which is "first" is equally likely H or T.

Only holds if no spooky effects change results based on last result. (like a magic die that counts upwards or a magic coin that flips T after H no matter what)

P(TH) = p(T)*p(H) = P(HT)

[vikingerik]

Your second paragraph is correct and may be where the previous poster's intuition was disagreeing, that the method doesn't necessarily hold for repeated iterations in a physical system where one trial starts from where the last one ended.

It's not even really "spooky" - all you need is a flipping apparatus that's biased towards an odd number of rotations, and so then THTH is more common than THHT and you get a bias towards repeating your last result.

[eddd-ddde]

Exactly right, I was thinking an unfair coin could have "memory" but then the method doesn't hold.

[guenthert]

What about a 'dirty' coin or dice, where the dirt falls off during the run?

[AJTSheppard]

That's a clever point. But I think a corner case.

I suspect that when the user is loading coins or dice in the machine, they would notice any dirt that was significant enough to look as though it might be a problem.

And oil deposits from your fingerprints I would imagine are so minuscule as to be insignificant in creating varying bias.

Even then, in both cases, you could wipe the objects with an alcohol swab before putting them into the shaker cups.

It could be argued, I suppose, that every micro-collision of the coin or die with the cup removes a few atoms, but I would suggest that its effect on the bias of the coin or die over time is again minuscule. Indeed, unmeasurable over a full sequence of cycles (128 for example) of the machine when generating a Bitcoin key.

But an interesting point. Keep 'em coming!

[jvanderbot]

That'd do it.

P(H|N) != P(T|N)

And

P(H|N) != P(H|N-1) (and visa versa)

Means that

P(HT) = P(H|N-1, T|N) != P(TH)