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# A coin with a conscience

## Question

Suppose we have a coin that "tries" to be fair. To be more specific, if we flip the coin n times, and have X heads, then the probability of getting a heads in the (n+1)st toss is 1-(X/n).
[The first time we flip the coin, it truly is fair, with p=1/2.]
A short example is in order. Each row here represents the i-th coin toss, with associated probability of heads and its outcome:
p=1/2: H
p=0: T
p=1/2: H
p=1/3: T
p=1/2: T
p=3/5: T
It certainly would seem that the expected value of X should be n/2. Is this the case? If so (and even if not), then think of this coin as following a sort of altered binomial distribution.
How does its variance compare to that of a binomial [=np(1-p)]?
Does the coin that tries to be fair become unfair in the process? Or does it quicken the convergence to fairness?

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1. The expected value is n/2.
2. The variance is smaller than np(1-p) (the unbiased case)
3. The coin that "tries" to be fair does not become unfair.

Simulations were run over a wide range of coin tosses for both cases.
For an unbiased coin the variance was 0.25045n, as expected from np(1-p) = 0.25n.
For the "coin with a conscience," the variance was 0.0846n.

The ratio of the variances was 0.338, or very nearly 1/3.

My guess would be that the "Conscience coin" variance is exactly 1/3 of a fair coin's variance.

One can bias an "unconscionable coin," away from a central outcome by using the complementary probability:

Instead of making p(h) = t/n as in this case, we make p(h) = h/n.

This produces a uniform or flat distribution.

That is, it produces a bell-shaped curve with no fall-off away from the mean.

Simulation confirmed the variance just equaled the number coin flips.

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