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This sounds like a simple question, but here's the gist:

Given a coin flip (or some other random process that can result in one of two outcomes) that has a perfect 50-50 probability of landing on heads or tails (the probability of heads is 50%, the probability of tails is 50%), if I were to flip the coin 10 times, the results would be close to 5-5. If I flip it 100 times, the results would be close to 50-50. The larger my sample size, the closer the results reflect the probability.

But if I flip this coin once, there's a 50-50 chance of landing on either heads or tails. The next time I flip the coin, the probability is the same. This means that each result of, say, 20 flips would be equally likely (8 heads and 12 tails and 10 heads and 10 tails would be equally likely).

If this is true, why do the results of flipping a coin many times trend towards an equal split? If this isn't true, why not?

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Eight heads and $12$ tails are less likely than $10$ heads and $10$ tails. And as we get further from an "even split" the probabilities get smaller. For example $20$ heads and $0$ tails has probability $\frac{1}{2^{20}}$, while $10$ and $10$ has probability $\binom{20}{10}\frac{1}{2^{20}}$, hugely bigger. –  André Nicolas Mar 5 at 1:44
    
And why is that? –  communistpancake Mar 5 at 1:46
    
Because there are many patterns in which we can get $10$ and $10$, for there are many places we can put the $10$ heads. Each pattern has probability $\frac{1}{2^{20}}$. By way of contrast, there is only one pattern that yields $0$ heads, and only a small number ($20$) that yield $1$ head. –  André Nicolas Mar 5 at 1:49
    
Using formulas, and in other ways, we can show that the binomial coefficient $\binom{2n}{k}$ increases until $k=n$, and then decreases. –  André Nicolas Mar 5 at 1:50
    
Thanks! Could you make that an answer so I can close this question? –  communistpancake Mar 5 at 1:50

2 Answers 2

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The events "the number of heads is $0$," "the number of heads is $1$," "the number of heads is $2$." and so on, are very far from being equally likely.

For a fair coin, the probability of $k$ heads in $2n$ tosses is $\binom{2n}{k}\cdot \frac{1}{2^{2n}}$. This is because any particular sequence of heads and/or tails has probability $\frac{1}{2^n}$, and there are $\binom{2n}{k}$ sequences of length $2n$ that have $k$ heads and the rest tails.

As an extreme example, the probability of $0$ heads in $20$ tosses is $\frac{1}{2^{20}}$, because there is only $1$ pattern of tosses that yields $0$ heads. The probability of $10$ heads is $\binom{20}{10}\frac{1}{2^{20}}$, very much larger, since $\binom{20}{10}$ is large.

Computation will show that $8$ heads and $12$ tails is a fair bit less likely than $10$ and $10$.

The binomial coefficients $\binom{2n}{k}$ climb from $k=0$ to $k=n$, and then decrease symmetrically.

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If you flip a coin 10 times, then you have $2^{10} = 1024$ different outcomes that are equally likely. If $H$ is head and $T$ is tail, then the string of $HTTTHHTTHT$ is just as likely as for example $HTHHTTTHHT$ or $TTTTTTTTTT$.

Now, if you make a list of all the $2^{10}$ different strings and if you count the number of heads and tails in each string, then you will see why the most likely outcome is $5$ $H$ and $5$ $T$:

$$ \begin{array}{ccc} \text{Outcome} & H & T \\ \hline HHHHHHHHHH & 10 & 0\\ HHHHHHHHHT & 9 & 1 \\ HHHHHHHHTH & 9 & 1 \\ HHHHHHHHTT & 8 & 2 \\ \vdots \end{array} $$ So, for example, only in $1$ out of the $1024$ will you have $10$ heads. If you are considering the probability that you have $9$ $H$ and $1$ $T$, then you have $10/1024$ and so on.

It is not hard to convince yourself that the most occurring outcome is $5$ $H$ and $5$ $T$.

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