[voldacar]

The key assumption is that T and H may not have the same probability, but each flip isn't correlated with past or future flips. Therefore, TH and HT have the same probability. So you can think of TH as "A" and HT as "B" then you repeatedly flip twice until you get one of those outcomes. So now your coin outputs A and B with equal probability.

[Jerrrrrrry]

I feel like I am missing something so obvious that I feel the need to correct wiki, but that likely means I am fundamentally missing the point.

"The Von Neumann extractor can be shown to produce a uniform output even if the distribution of input bits is not uniform so long as each bit has the same probability of being "one"->[first] and there is no correlation between successive bits.[7]"

As long as the person doesn't favor which of the two bits they chose is "first", then it should appear as random.

But that is self-defeating, as if the person had the capability to unbiased-ly choose between two binaries, they wouldn't need the coin.

But since the only way to determine the variation from expectation is repeatedly increasing sample size, I don't see how doing it twice, and just taking encoding of the bits, then...

Is the magic in the XOR step? To eliminate the most obvious bias (1v5 coin), until all that could had been left was incidental? Then, always taking the first bit, to avoid the prior/a priori requisite of not having a fair coin/choosing between two options?

and it clicked. Rubber duck debugging, chain of thought, etc.

I will actually feel better now.

[Jerrrrrrry]

>To eliminate the most obvious bias (1v5 coin), until all that could had been left was incidental?

There is only one coin, flipped _twice_; not a running occurrence, but in couples, perfectly simulating two coins functionally.

Once a literal couple of coins result in a XOR'd result eventually, no matter how biased - they differ - the exact ordinality of which will be random.

Two sides to a coin, no matter how random, still half the chance.

(for lurkers cringing at my subtle mis-understanding)

[ljsprague]

Maybe I don't understand why or what you don't understand but...

Say you have a biased coin. It lands heads 55% of the time (but you don't know that.) Then the probabilities are:

HH = (0.55 * 0.55) = 0.3025

TT = (0.45 * 0.45) = 0.2025

HT = (0.55 * 0.45) = 0.2475

TH = (0.45 * 0.55) = 0.2475

If you disregard the HH and TT results then the equal probabilities of HT and TH result in a perfect binary decider using a biased coin. You assign HT to one result and TH to the other.