Unfairly fair coin?

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... show more 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 ith coin toss, with associated probability 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?

(I haven't worked this one out, it just occurred to me. If the algebra becomes intractable, computer solutions are allowable.)
Update: In light of John C's excel results, and some thinking of my own, I have the results for up to n=47 (and it wouldn't be too hard to expand those results). The mean is indeed the "correct" mean n/2 (as ksoileau has proved). The variance is exactly n/12, or 1/3 of an ordinary coin's variance,... show more In light of John C's excel results, and some thinking of my own, I have the results for up to n=47 (and it wouldn't be too hard to expand those results). The mean is indeed the "correct" mean n/2 (as ksoileau has proved). The variance is exactly n/12, or 1/3 of an ordinary coin's variance, except for n=2, where the variance is 0.

Astounding. Perhaps someone will get to an algebraic proof?
Update 2: To shed some light on my work: I figured that the probability P(X; n)...that is, the probability of having X heads, given the parameter of n tosses...should follow a recurrence relation. Specifically, P(X; n) = (1-X/n)*P(X-1; n-1) + X/n*P(X; n-1). Using this relation it was relatively easy to create an... show more To shed some light on my work:

I figured that the probability P(X; n)...that is, the probability of having X heads, given the parameter of n tosses...should follow a recurrence relation. Specifically,
P(X; n) = (1-X/n)*P(X-1; n-1) + X/n*P(X; n-1).

Using this relation it was relatively easy to create an excel spreadsheet with the probabilities, then to calculate E(X) and E(X^2). The really interesting thing is when John noticed the Eulerian triangular numbers, which follow a rather similar recurrence. Oddly though, I haven't been able to make the connection concrete.

Anyway, if we conclude that P(X; n) can be written as <n-1, x-1> [note that I'm using the angular brackets similar to the choose function, not A(n,k)], then we should be able to calculate E(X^2) and from there get the variance. I have unfortunately not been able to complete this work. It may include the use of the formula about the sum of triangular numbers being n!.
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