The Gambler's Fallacy: Why a Coin Has No Memory

The coin after ten heads

A fair coin has landed heads ten times running. What is the probability that the next flip is heads?

One half. It was one half before the run started and the coin has learned nothing since. The gambler's fallacy is the answer that tails is now more likely because it is due; the hot-hand version is the answer that heads is now more likely because the coin is on a streak. Both are wrong, and both are wrong for the same reason.

The number that feels as though it ought to matter is real, but it answers a different question. Asked before any flipping, the probability of eleven heads in a row is one in 2,048, which is genuinely tiny. But ten of those flips have already happened. The improbable part is behind you, and the only uncertainty left in front of you is a single flip.

The arithmetic that settles it

Conditional probability makes the point exactly rather than by assertion. Divide the probability of eleven heads by the probability of the ten that have already happened, and what is left is the probability of the next one.

The last line in the list is the one that does the most work. HHHHHHHHHHT and HHHHHHHHHHH are equally unlikely, both at one in 2,048, and so is every other specific sequence of eleven flips. Runs look special because we notice them, not because they are rarer than the shapeless sequences nobody remarks on.

• P(eleven heads in a row) = (1/2)¹¹ = 1/2048
• P(ten heads in a row) = (1/2)¹⁰ = 1/1024
• P(head on flip 11, given ten heads already) = (1/2048) ÷ (1/1024) = 1/2
• P(ten heads then a tail) = 1/2048 as well - exactly as unlikely as eleven heads

The long run corrects by dilution, not by repayment

There is a real theorem hiding behind the fallacy, which is why the intuition is so stubborn. Over a large number of flips, the proportion of heads does settle towards a half. Students hear that and conclude a surplus of heads must be paid back by a surplus of tails.

It is never paid back. It is swamped. Suppose after 20 flips you have 15 heads and 5 tails - ten ahead, a proportion of 0.75. Flip 980 more times and get roughly 490 of each. You now have 505 heads out of 1,000. You are still exactly ten ahead in absolute terms, and nothing has compensated for anything, but the proportion has fallen to 0.505.

That is the whole content of the law of large numbers as it applies here, and putting it this way removes the fallacy at the root. The average moves because more data arrives and drowns the early excess, not because the past gets corrected. Nothing is keeping score.

When the past genuinely does tell you something

The fallacy is not the claim that past results never matter. It is the claim that they matter for independent events. There are three situations on the syllabus where they matter a great deal, and telling them apart from the coin is the skill the exam is testing.

The question that separates them is short: could this trial have changed the situation for the next one? Removing a counter from a bag changes it. Flipping a coin does not. Where the answer is yes, the probabilities on the next branch of the tree are different, and where it is no, they are identical - which is exactly what a probability tree makes visible.

• Without replacement. Draw a red counter from a bag of 3 red and 5 blue and the bag now holds 2 red and 5 blue, so the probability of red has fallen from 3/8 to 2/7. Different probabilities on the second set of branches are the visible signature of dependence.
• Unknown fairness. A coin landing heads 90 times in 100 is not due for tails - it is probably not a fair coin. Relative frequency, successes divided by trials, is how a probability is estimated when it cannot be calculated theoretically, and more trials make the estimate better.
• Physical change. Dice chip, spinners bend, machines wear. Where the mechanism itself can alter, past results are evidence about the mechanism even though they say nothing about chance.