Suppose you're watching a series of coin tosses and you see 8 heads in a row come up. Should you be surprised? The answer depends on the probability of heads and the number of times the coin has been tossed. If \(p\) is the probability of heads, how many times do you have to toss the coin, on average, to get a run of \(r\) heads? Let \(\mu\) and \(\sigma\) be the mean and standard deviation of the number of tosses required, then:
\[\mu = \frac{1-p^r}{qp^r}\]
\[\sigma = \sqrt{\frac{1}{(qp^r)^2} - \frac{2r+1}{qp^r} - \frac{p}{q^2}}\]
where \(q = (1-p)\) is the probability of tails.
The table below shows the results for \(p\) = 0.3, 0.4, \(\ldots\), 0.7 with 8 heads in a row.
| p | \(\mu\) | \(\sigma\) |
|---|---|---|
| 0.3 | 21,772 | 21,765 |
| 0.4 | 2,541 | 2,535 |
| 0.5 | 510 | 503 |
| 0.6 | 146 | 140 |
| 0.7 | 54 | 48 |
From the table we see that for a fair coin, the mean number of tosses for getting 8 heads in a row is 510. Note that the standard deviation is comparable to the mean, being less than the mean by only 6 or 7 for each probability in the table. This implies that the first 8 coin tosses could be 8 heads in a row, but it's unlikely unless the coin is heavily biased toward heads.
The QuantWolf Coin Toss Runs Calculator will calculate the mean given probability of heads and run length.
The above formulas and more are found in our book The Coin Toss: The Hydrogen Atom of Probability.
© 2010-2012 Stefan Hollos and Richard Hollos
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