This is pleasingly insane, congratulations! Is there a program to test the fairness of a given dice or coin? Is that a program that's even feasible to write?
You've always got the standard way to get fair random numbers from a fairness-unknown coin. Flip it twice. Restart if you get both heads or both tails. If you get H then T or T then H, those are equally probable, so take the first one of those as the final outcome.
This generalizes to a die of N sides. Roll it N times. If you don't get all N distinct results, restart. If you do, then take the first result as your final outcome.
(That may take a lot of trials for large N. It can be broken down by prime factorization, like roll 2-sided and 3-sided objects separately, and combine them for a d6 result.)
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.
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.
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.
Coins and dice and datums (solid objects with detectable outcomes) may, or may not have bias, it depends on how they were made and on manufacturing defects that resulted. But, at a minimum, such bias can oftentimes be side-stepped or bypassed.
Consider this argument from Johnny Von Neuman.
Suppose you have a single biased coin with these outcome probabilities:
A) Heads (1) 60% (Call this probability p.)
B) Tails (0) 40% (The probability of this outcome is q=(1-p), by definition.)
Now let us apply this algorithm to sequential tosses for this coin:
1) Toss the coin twice.
2) If you get heads followed by tails, return 1. (Say this outcome occurs with probability p’.)
3) If you get tails followed by heads, return 0. (The probability of this outcome is q’=(1-p’), by definition.)
4) Otherwise, ignore the outcome and go to step 1.
The bit stream that results is devoid of bias. Here’s why. The probabilities of obtaining (0 and 1) or (1 and 0) after two tosses of the coin are the same, namely p(1-p). On the other hand, if (1 and 1) or (0 and 0) are thrown, those outcomes are ignored and the algorithm loops around with probability 1 – 2p(1-p). So, the probability (p’) of getting a 1 using this algorithm after any sequential two tosses of the coin is p’ = p(1-p) + p’(1-2p(1-p)). The solution of which is p’=1/2, and since q’=(1-p’), then q’=1/2. A fair unbiased toss!
In fact, the example bias numbers given above don’t matter for the argument to hold (note that after solving for p’ it is independent of p). The outcome of the algorithm is a fair toss (in terms of the (0 and 1)-bit stream that results), regardless of the actual bias in the coin for a single toss. All the bias does is have an effect on the efficiency with which the bit stream is created, because each time we toss heads-heads or tails-tails we loop around and those two tosses are thrown away (lost). For an unbiased coin the algorithm is 50% efficient, but now has the guarantee of being unbiased. For a biased coin (or simply unknown bias) the algorithm is less than 50% efficient, but now has the guarantee of being unbiased.
This algorithm is trivial to implement for the Satoshi9000.
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)
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.
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.
I love the slow pace of the video, including a few minutes presentation of all available programs. And indeed, there are programs to test dice and coin bias:
....you can do that using our universe's physical constants too.
Care to elaborate? Or link?
I mean, everything that is, is just displaced temporarily homogeneous complexity, allowable between the fluctuations of gradients of temperature, allowing the illusion of work to appear more than just energy dissipating into the either of expanding space-time, dragged by the irreconcilability idea of "gravity".
But that doesn't help bake an Apple pie from scratch, as Carl Sagan would put it.
