According to Bernoulli's law of large numbers, in the coin tossing game the probability that the number of heads equal to the number of tails tends to $1$ as number of tosses increases.
In other words :
$\frac{H(n)}{T(n)}$ tends to $1$ as $n$ $\to$ $\infty$
$$\lim_{n\to\infty}P \bigg( \frac{H(n)}{T(n)}=1 \bigg)=1$$ where $H(n)$ and $T(n)$ denote the number of heads and tails, in $n$ tosses of coin.
On the other hand, the probability that the number of heads is exactly equal to the number of tails tends to zero.
when tossing $2n$ times, the the probability that it falls heads exactly $n$ times is $p(n)=2^{-2n}\binom{2n}{n}$
$$\lim_{n\to\infty}p(n)=0$$
The probability that the number of heads approximately equals the number of tails tends to one, whereas the probability that the number of heads is exactly equal to the number of tails tends to zero.
$$\lim_{n\to\infty}P \bigg( H(n)=T(n) \bigg)=0$$
Can anyone explain this contradiction?
The strong law of large numbers says $$ \Pr\left( \lim_{n\to\infty} \frac{H(n)}{T(n)} = 1 \right) = 1. $$ The law does not say: $$ \lim_{n\to\infty}\Pr \left( \frac{H(n)}{T(n)}=1 \right)=1, $$ which is different and in fact false.
(The weak law of large numbers says $\displaystyle\forall\varepsilon>0\ \lim_{n\to\infty}\Pr\left( 1-\varepsilon<\frac{H(n)}{T(n)}<1+\varepsilon \right)=1$.)