How to set probability of n-th trial so that first success index distribution will be normal? Suppose we toss an unfair coin until we get heads.
For every toss we can change the probability of success according to some function F(trial_idx, *function_parameters).
We repeat this process and each time note at which try we succeeded.
What should $F$ look like to obtain some normal distribution N(μ, σ^2) for success indices?
EDIT: whuber is right, this will never be exactly normal distribution. What I could call "good enough approximation" would work in "easy" cases - in which you can expect that the mean of the distribution is positive and at least 3 times greater than the standard deviation.
 A: You don't need parameters.
At trial $t = 1, 2, 3, \ldots,$ let $f(t)$ be the chance of success.  Write $X_f$ for the (random) time of first success.  Assuming the trials are independent, the chance that the first success will not occur by trial $t$ is the chance that the first $t$ trials are failures, which by independence is
$$\Pr(X_f \gt t) = \prod_{i=1}^t (1-f(i)) = S_{X_f}(t),$$
the survival function of $X_f.$  For future reference (see the code for f.make below), notice that $f$ can be recovered from any such function $S$ as
$$f(t) = 1 - \frac{S(t)}{S(t-1)}.$$
Because "first success at trial $t$" is equivalent to "no successes through trial $t-1$ and not (no successes through trial $t$)," subtracting gives the probability function for $X_f$ as
$$\Pr(X_f=t) = \Pr(X_f \gt t-1) - \Pr(X_f \gt t) = S_{X_f}(t-1) - S_{X_f}(t).$$
So, let $S$ be the survival function of any distribution whatsoever you wish to approximate and let $\phi$ measure the quality of approximation, with lower values of $\phi$ corresponding to better approximations.  All we need do is find $f$ to minimize $\phi(S_{X_f}, S).$
Some of the more natural choices of $\phi$ allow simple optimization algorithms.  For instance, suppose $\phi(S, S^\prime)$ is the largest absolute difference among the discrepancies $S(x) - S^\prime(x)$ ranging over all real numbers $x$ (the $L^1$ norm or KS statistic). Given a survival function $S$ for a random variable supported on the natural numbers, we can recursively construct $f$ to make $S_{X_f}$ agree exactly with the values of $S$ at all trials and obviously it's impossible to do better than that.
For the survival function $S$ of a continuous distribution, such as a Normal, we will do well by reproducing $S$ at the midway points between the trial indexes (a "continuity correction").
Here is an R algorithm.  It's really just one line, implementing the formula for $f$ in terms of $S,$ with a special case for the starting value to handle the probabilities assigned to values of $x$ preceding trial $1.$
f.make <- function(S, t.max=1e4, ...) {
  i <- seq_len(t.max)
  f <- 1 - S(i+1/2, ...) / S(i-1/2, ...)
  f[1] <- 1-S(3/2, ...)
  f
}

(The caller needs to specify the maximum number of trials, t.max, as appropriate for the support of the distribution.  S is the survival function of that distribution.)
I used this algorithm to compute $f$ in two cases, for Normal distributions with standard deviations of $3.$  One of them has a mean of $0$ and the other a mean of $10.$  The latter satisfies the criteria of the question insofar as most of its probability is positive.  Here is an example:
mu <- 10; sigma <- 3
f <- f.make(S, ceiling(qnorm(1-1e-6, mu, sigma)), mean=mu, sd=sigma)

To check the computations, I performed 100,000 iterations of the coin-flipping experiment in each case.
X <- replicate(n.sim, which.min(rbinom(length(f), 1, f) == 0))
k <- tabulate(X, length(f)) / n.sim # Array of frequencies by trial number 1, 2, 3, ...

Here are the results. They look very good: the Normal distributions, whose densities and survival functions are plotted in red, are reproduced as well as one could hope.

The distribution with mean $0$ has large probability, so the best we can do is to stop the coin flipping at the first trial with a probability equal to the chance assigned to all values less than $t=1.$  You can see that in the spikes at trial $1$ in the plots of $f$ and the "probability functions" ($\Pr(X_f=t)$).
