If the coin is thrown 2n times, then the number of distinct outcomes (each being a distinct sequence of 2n occurrences of heads or tails) is 2^2n (where ^ means power), because each occurrence has two possible states (heads or tails) and there are 2n such occurrences in total.
If an outcome (being a sequence of 2n occurrences of heads or tails) comprises h heads and t tails, with h+t=2n, then the number of such outcomes is (2n)!/(h!t!), because the number of ways of arranging 2n occurrences is 2n! but each such result comprises h occurrences of heads, which can be arranged in h! ways, and t occurrences of tails, which can be arranged in t! ways, so we have counted each distinct result h!t! times. If we are looking for cases where h=n and t=n, then the number of such outcomes is (2n)!/(n!n!), or (2n!)/(n!)^2.
So, the proportion of possible outcomes from 2n throws which comprise exactly n heads and n tails is (2n!)/((n!)^2)(2^2n), or (2n!)/(n!(2^n))^2 (because 2^2n is (2^n)^2).
This gives us the following results:
- For 2 throws, the probability of 1 heads and 1 tails is 0.5, or 1 in 2 (which is pretty obvious, because the outcomes are HH, HT, TH, TT, and 2 of those 4 outcomes comprise 1 head and 1 tail).
- For 4 throws, the probability of 2 heads and 2 tails is 0.375, or 1 in 2.666...
- For 6 throws, the probability of 3 heads and 3 tails is 0.3125, or 1 in 3.2
- For 8 throws, the probability of 4 heads and 4 tails is 0.2734375, or 1 in 3.657...
- For 10 throws, the probability of 5 heads and 5 tails is 0.24609375, or 1 in 4.063...
To answer our original question:
- For 100 throws, the probability of 50 heads and 50 tails is 0.079..., or 1 in 12.564...
Looking further ahead:
- For 1000 throws, the probability of 500 heads and 500 tails is 0.025..., or 1 in 39.643...
So there is a little more than a one in thirteen chance of getting exactly fifty heads and fifty tails in one hundred throws of the coin. Far from a dead cert, it is true, but more likely than I would have expected.
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