If everyone mixes their hats and receives one at random, will anyone receive their own?
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Derangements - Numberphile
Watch a shuffled card game turn into the same hat-check puzzle, then follow matches, fixed points, derangements, and the near two-in-three result.
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What else makes you wonder?
What changes in groups of two, three, or four?
List every one-to-one handout for a tiny group and mark each own-hat return.
What if nobody may receive a neighbor's hat?
Removing allowed routes changes which complete handouts can exist.
What if a hat choice could be copied and chosen again?
The choices would become independent, but the one-hat-per-person promise would disappear.
After you watchIf everyone mixes their hats and receives one at random, will anyone receive their own?
The short answer
Yes—surprisingly often. In a random one-to-one handout, the chance that at least one person receives their own hat is close to 63%, or roughly two times in three, once the group has more than a few people.
Try this next
- What if the group has only two or three people? Draw every possible handout with arrows, then mark each fixed point. Compare your exact tiny-group list with the large-group title pattern.
- What if everyone must avoid the person beside them? Cross out the forbidden recipient routes before counting. Predict whether that restriction makes an own-hat return more or less common.
- What if people choose independently instead of using each hat once? Model each person choosing from all hat types with replacement. Identify which original one-to-one assumption has changed before comparing chances.
Now you — bend it
- What if What if two people may receive the same copied hat design?Decide whether you still have a one-to-one handout before reusing the old counting method.
- What if What if every person is forbidden from receiving one particular other person's hat?Draw the allowed owner-to-recipient grid and remove one route in each row before counting.
- What if What if one labeled hat is secretly placed instead of mixed?A forced assignment changes both that child's chance and the possibilities left for everyone else.
Can you prove it?For one named child in an n-person one-to-one handout, exactly one of the n possible hats is their own. — Use the group-size toy or draw n distinct hat cards. Hold the child fixed, count all n hats that could arrive, and identify the single matching card.
Design your own test:Before counting, predict how forbidding a route will change the chance of at least one match, then test a small case by listing every allowed handout.
Explain it to a 6-year-old: Even when everyone's own hat is hard to pick, a whole group often has at least one lucky return.
The whole story
How it works
A handout uses every hat exactly once. One person's own-hat chance gets smaller as the group grows, but a larger group also supplies more people who could match. To find the group chance exactly, mathematicians count the handouts in which nobody matches, called derangements, and subtract that share from all possible handouts. The result quickly settles near 1 minus 1/e, about 0.632.
What people get wrong
It is easy to mix up one person's chance with the whole group's chance. Your personal chance is only 1 out of the group size, but the question asks whether anyone in the group matches. Many small individual chances can still make at least one group match fairly common.
The catch
The simple result assumes a perfectly random one-to-one handout: each person gets one hat, each hat is used once, and every possible assignment is equally likely. Real coatrooms may include labels, different sizes, familiar colors, or deliberate choices, so real handouts need not follow the mathematical model.
Questions kids ask
What counts as a match?
A match happens when one particular hat returns to the person who originally owned it. Mathematicians call that unchanged owner-to-item position a fixed point.
Why does the group chance stay near two in three?
Each person's own-hat chance shrinks as the group grows, while the number of possible people who could match grows. Exact counting of the no-match handouts shows that the two changes balance near 1 minus 1/e.
Is one child's own-hat chance also about 63%?
No. For one named child in a group of n, the chance is exactly 1/n. The near-63% result asks whether at least one person anywhere in the group matches.
Why must every hat be used exactly once?
That one-to-one rule makes the experiment a permutation. If hats could be copied, skipped, or chosen more than once, it would be a different probability model with a different answer.
Talk about it
- Before a handout, ask the child to explain how one person's chance differs from the chance that anyone in the group matches.
- At the three-bucket prediction, invite a reason for each option before choosing; a one-match, two-match, or larger-match story can all sound plausible.
- After the reveal, ask which evidence came from one sample and which evidence came from checking every possible successful handout.
For grown-ups
A random one-to-one handout is a permutation, and a person receiving their own hat is a fixed point. A permutation with no fixed points is a derangement. Inclusion–exclusion gives D(n)=n! times the alternating sum of 1/k!, so D(n)/n! approaches 1/e and the probability of at least one fixed point approaches 1−1/e. The story preserves that public title answer while reserving a different conditional fixed-point-distribution question for its prediction gate.