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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?

Juniven Acapulco
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communistpancake
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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 05 '14 at 01:44
  • And why is that? – communistpancake Mar 05 '14 at 01:46
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    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 05 '14 at 01:49
  • 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 05 '14 at 01:50
  • Thanks! Could you make that an answer so I can close this question? – communistpancake Mar 05 '14 at 01:50
  • I was going to suggest an understanding of Pascal's Triangle would really answer this question. – JTP - Apologise to Monica Mar 05 '14 at 01:53

1 Answers1

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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.

Nick Peterson
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André Nicolas
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