Can we prove that a fair coin tends to have 50% of relative frequency when number of trials tends to infinity ??

My first idea that was clear that wouldn't work is to try to prove that the limit of $${{n}\choose{n*p}}* p^{np}*(1-p)^{n - n*p}$$ tends to 1 when n goes to infinity. But it's obvious that it goes to 0 when we think in the plot of the binomial distribution PDF when the n(number of trials) goes to infinity. This was the python code I used to see it going to 0:

import math
from decimal import Decimal

def stirlingFactorial(n:Decimal):
    coeficient = Decimal(math.sqrt(2 * math.pi))*n

    e = Decimal(math.e)

    return coeficient * (n/e)**n

def stirlingCombination(n:Decimal, k:Decimal):
    return stirlingFactorial(n)/(stirlingFactorial(k)*stirlingFactorial(n-k))


def stirlingBinomialProbability(n:Decimal, k:Decimal, p:Decimal):
    return stirlingCombination(n, k)*p**k * (1-p)**(n-k)


n = Decimal(100000)
p = Decimal(0.5)

print(stirlingBinomialProbability(n, n*p, p)) #0.00001595769121605730877792391774

But this made me think: if this doesn't go to 1, why do we trust the assumption that a dice rolled n times, for big values of n, will probably have approximately n/2 successes??

Solution 1:

This is a very common misconception of the law of large numbers. Think of a coin flip. The geometric mean of $n$ coin flips ($head=1$ and $tail=0$) can be expressed as $\bar{x}_{n}=\frac{\frac{1}{2}n+z_{n}}{n}$ where $z_{n}$ captures the deviation from $50\%$ heads (i.e. the excess number of heads over 50 percent). Then note that $\lim_{n\rightarrow \infty }\bar{x}_{n}=% \frac{1}{2}$ does not require $z_{n}=0$. It merely requires that $% \lim_{n\rightarrow \infty }\frac{z_{n}}{n}=0$.