• one token is red on both sides,
• one token is white on both sides,
• one token is red on one side and white on the other.
A token is drawn completely at random and laid flat, with each of its two faces equally likely to land on top. The face showing is red.
What is the probability that the hidden face is also red?
💡 Hint
Count faces, not tokens. List the six painted faces, keep only the ones that could be showing, and see what lies behind each.
Reveal answer
The probability is 2/3, not 1/2.
Step-by-step solution
- Label the faces. The red-red token has faces R1 and R2. The white-white token has W1 and W2. The mixed token has R3 and W3. Each of the six faces is equally likely to end up on top.
- You see red, so the top face is R1, R2 or R3. These three cases are equally likely.
- If the top is R1, the bottom is R2 (red). If the top is R2, the bottom is R1 (red). If the top is R3, the bottom is W3 (white).
- Two of the three equally likely cases have a red bottom, so the probability is 2/3.
- Another way to see it: the red-red token produces a red top every time it is drawn, the mixed token only half the time. Seeing red makes the red-red token twice as likely as the mixed one.
Why it fools people
The puzzle invites you to count objects when you should count outcomes. “Two tokens left, so fifty-fifty” feels like common sense, but probabilities attach to equally likely events, and the tokens are not equally likely once you know what you saw. The evidence you observed is more typical of one token than the other, and that shifts the odds. This is the same idea behind Bayes’ rule.
Did you get it?
About This Puzzle
At a county fair in late summer, a booth run by a retired math teacher offers no prizes, only a question. On the counter sits a cloth bag holding three wooden tokens, each painted on both faces. She shakes the bag, pulls a token without looking, and sets it flat on the counter so only the top face shows. Then she asks the crowd to bet on the color underneath. People are sure they know the odds, and they are usually wrong. The puzzle is fair, simple to state and completely precise, and it teaches one of the most useful habits in probability: count the right things.
How to Think About It
The tempting move is to count tokens. A sharper move is to count faces. There are six painted faces in the bag, and the drawing process makes each of them equally likely to be the one facing up. Now use the information you were given: the face showing is red. Throw away every face that is not red, and look at what is left. For each remaining red face, ask what is painted on the other side of that same token. The answer is a simple ratio of equally likely cases.
Common Traps
Nearly everyone says one half. The reasoning sounds airtight: the white-white token is ruled out, so the token is either the red-red one or the red-white one, and those seem equally likely. The flaw is the word “seem.” Two tokens remain, but they did not have the same chance of showing red. One of them shows red every single time it is drawn, while the other shows red only half the time. Treating the two survivors as a coin flip throws that difference away.
Where This Puzzle Comes From
This is a retelling of a famous puzzle usually called Bertrand’s box paradox, after the French mathematician Joseph Bertrand, who described a version with boxes of gold and silver coins in the late 1800s. It is a close relative of the Monty Hall problem, and teachers use it to show why counting outcomes, not objects, is the safe way to handle conditional probability.
Try This Variation
Add a fourth token to the bag, another one that is red on both sides. Now the bag holds two red-red tokens, one white-white token and one red-white token. A token is drawn at random and shows red. What is the probability the hidden face is red?
Show the answer to the variation
It is 4/5. There are now five red faces in the bag: four on the two red-red tokens and one on the mixed token. Each is equally likely to be on top. Four of those five red faces have red on the back, so the probability is 4 out of 5.