Dealing with normal (Gaussian) distribution - random

Question

I basically stucked on rather simple problem:
Toss N coins and discover, how many of them land heads
Solution performance must not depend on N, so we can't just call Math.random() < 0.5 N times. Obviously, there is Gaussian distribution to the rescue.
I used Box-Muller method for it:
function gaussian_random(mean, variance) {
var s;
var x;
var y;
do {
x = Math.random() * 2.0 - 1.0;
y = Math.random() * 2.0 - 1.0;
s = Math.pow(x, 2) + Math.pow(y, 2);
} while ( (s > 1) || (s == 0) );
var gaussian = x * Math.sqrt(-2*Math.log(s)/s);
return mean + gaussian * Math.sqrt(variance);
}
Math says, that mean of N coin tosses is N/2 and variance is N/4
Then, I made test, which tosses N coins M times, giving Minimum, Maximum, and Average number of heads.
I compared results of naive approach (Math.random() many times) and Gaussian Box-Muller approach.
There is typical output of tests:
Toss 1000 coins, 10000 times
Straight method:
Elapsed time: 127.330 ms
Minimum: 434
Maximum: 558
Average: 500.0306
Box-Muller method:
Elapsed time: 2.575 ms
Minimum: 438.0112461962819
Maximum: 562.9739632480057
Average: 499.96195358695064
Toss 10 coins, 10000 times
Straight method:
Elapsed time: 2.100 ms
Minimum: 0
Maximum: 10
Average: 5.024
Box-Muller method:
Elapsed time: 2.270 ms
Minimum: -1.1728354576573263
Maximum: 11.169478925333504
Average: 5.010078819562535
As we can see on N = 1000 it fits almost perfectly.
BUT, on N = 10 Box-Muller goes crazy: it allows such tests outcomes, where i can get (quite rarely, but it is possible) 11.17 heads from 10 coin tosses! :)
No doubt, i am doing something wrong. But what exactly?
There is source of test, and link to launch it
Updated x2: it seems, previously I didn't describe problem well. There is general version of it:
How to get sample mean of N uniform random values (either discrete or continuous) in amortized constant time. Gaussian distribution is efficient for large N, but is there a way to make it work good on small N? Or on which exactly N, solution should switch from Gaussian method to some other (for exampled straightforward).

Answer 1

Math says, that mean of N coin tosses is N/2 and variance is N/4.
Math only says that if it's a fair coin. And there's no way the solution doesn't depend on N.
Assuming all tosses are independent of each other, for exact behaviors use a binomial distribution rather than a normal distribution. Binomial has two parameters: N is the number of coin tosses, and p is the probability of getting heads (or tails if you prefer). In pseudocode...
function binomial(n, p) {
counter = 0
successes = 0
while counter < n {
if Math.random() <= p
successes += 1
counter += 1
}
return successes
}
There are faster algorithms for large N, but this is straightforward and mathematically correct.

Answer 2

Based on what discussed in approved answer i came up to this particular solution.
There is rule of thumb n*p >= 10 and n*(1-p) >= 10, but lets define stricter one.
First of all, Box-Muller gonna be hard capped at [-6,6], to ensure proper outcome (640 kB ought..., I mean, 6 sigmas ought to be enough for everybody).
Then, using 6 constant, we state, that in order for Box-Muller to produce valid results, Math.sqrt(variance) * 6 must not exceed mean. After using N/2 and N/4 as mean and variance respectively, and reductions, we end up with this:
Math.sqrt(N) * 6 <= N
N >= 36
While this condition is true, we can safely use capped Box-Muller Gaussian.
On any lower sample size, stick to the Binomial solution.
Just checked this rule statistically - on relatively large amount (10 million) of tests, minimum value stops falling out of boundaries from sample size 32 and above.