Three boxes. Six coins. You reach into one at random and pull out gold. What is the chance the other coin in that box is gold as well? Almost everyone says one half. Almost everyone is wrong. The correct answer is two thirds, and the reason it surprises us says a lot about how human intuition handles evidence.
The puzzle, exactly as it is usually told
The French mathematician Joseph Bertrand published this puzzle in 1889, and more than a century later it still catches smart people out. I have used it in workshops for years, with analysts, teachers and engineers, and the pattern is remarkably stable: about four in five people answer 50% within seconds and defend it with real conviction.
Here is the setup. There are three identical boxes, each with two drawers. Inside them:
You choose a box at random, open one drawer at random, and see a gold coin. What is the probability that the other drawer in the same box also holds gold?
The answer in one line. It is 2/3, or about 66.7%. The tempting answer of 1/2 is wrong because the coin you saw is more likely to have come from the gold-gold box than from the mixed box.
Before I explain, I want you to feel why 50/50 is so seductive, because if you understand the temptation you will spot the same mistake in a medical report or a court case.
The argument that feels airtight
Here is the reasoning almost everybody uses. “I drew a gold coin, so this cannot be the silver-silver box. That leaves two boxes: gold-gold and gold-silver. They were equally likely at the start, so it is a coin toss whether I am holding the gold-gold box or the mixed one. Half the time the other coin is gold.”
Every sentence in that paragraph sounds fine. The first step is correct: silver-silver is out. The second step is where it quietly goes wrong. The two remaining boxes were equally likely before you saw a coin. But you did not just learn that the box is not silver-silver. You learned something more specific: a gold coin came out. And the two boxes are not equally good at producing gold coins.
- The gold-gold box can only ever give you gold.
- The gold-silver box gives you gold only half the time.
So the observation of gold is stronger evidence for the box that always produces it. That is the whole secret. Evidence does not just eliminate possibilities; it reweights the ones that survive.
The clean way to solve it: count coins, not boxes
When I teach this, I tell people to stop thinking about boxes altogether. Boxes are a distraction. The thing that is randomly chosen, in effect, is one of six coins, each equally likely to be the coin you touch. Label them and list their situations.
Now apply the evidence. You saw gold, so throw away every case where you drew silver. Three outlined cases remain:
- Gold coin A from the gold-gold box. The other coin is gold.
- Gold coin B from the gold-gold box. The other coin is gold.
- The gold coin from the mixed box. The other coin is silver.
Three equally likely possibilities, and in two of them the other coin is gold. That gives 2/3. No formulas, no jargon, just counting the things that really are equally likely.
The reason I love this puzzle as a teaching tool is that the correct method is a habit rather than a trick: list the equally likely basic outcomes, cross out the ones that contradict what you saw, and count what is left. Once you adopt that habit, entire families of confusing probability questions become easy.
Do not trust me: run the experiment
Whenever a probability answer feels wrong, do not argue about it, simulate it. I wrote a short program that repeats the experiment: pick a box at random, open a random drawer, keep the run only if the coin is gold, and record whether the other coin was also gold. Here are the results, at increasing numbers of gold-first draws.
| Gold-first draws | Other coin also gold | Share |
|---|---|---|
| 10 | 5 | 50.0% |
| 100 | 62 | 62.0% |
| 1,000 | 666 | 66.6% |
| 10,000 | 6,640 | 66.4% |
| 100,000 | 66,681 | 66.7% |
At 10 draws the number can bounce around, because small samples are noisy. At 1,000 draws it sits close to two thirds, and by 100,000 it is unmistakable. This is the law of large numbers doing what it does. If you want to convince a stubborn colleague, a simulation is more persuasive than any argument, because it removes the feeling that you are just trying to trick them with words.
You can even do a physical version at home. Put two gold and one silver stickers on a few index cards, or use red and white cards, and repeat it about fifty times. You will see about two thirds emerge, not exactly, but clearly.
The same idea, five different costumes
Bertrand’s puzzle is not really about coins. It is about how new information changes probabilities when the information arrives through a process that favours some cases over others. Once you have the pattern, you will start noticing it everywhere. Tap through the tabs below. Each one is a real, well-known version of the same reasoning, with the numbers worked out.
Three cards
A hat holds three cards: one red on both sides, one white on both sides, one red on one side and white on the other. You draw one, look at one face at random, and it is red. What is the chance the other face is red?
Working it out. Count the red faces, not the cards. There are three red faces in the hat. Two of them belong to the double-red card and one belongs to the mixed card. Given that you are looking at a red face, there is a 2 in 3 chance that the back is red as well. Same structure as the boxes, just with paper and ink.
Monty Hall
You pick one of three doors. The host, who knows where the car is, opens a different door showing a goat and offers a switch. Should you switch?
Working it out. Your first pick was right one time in three, and that does not change because the host opened a door. Since the host never opens the car door, the remaining 2/3 of probability piles onto the other closed door. It is Bertrand’s logic in a game-show costume: the host’s reveal is new information, but it is not information that treats all cases equally.
Two children
A family has two children. You learn that at least one is a boy. What is the chance both are boys?
Working it out. List the four equally likely families: BB, BG, GB, GG. “At least one boy” removes GG and leaves three, of which only BB has two boys. So the answer is 1/3, not 1/2. But if you are told the older child is a boy, only BB and BG remain and the answer really is 1/2. Small changes in wording change which cases survive, which is the point of the whole paradox.
Medical test
A screening test is 90% sensitive with a 9% false-positive rate for a condition that 1% of people have. You test positive. What is the chance you are actually sick?
Working it out. Of 100 sick people, 90 test positive. Of 9,900 healthy people, 891 test positive by error. So of the 981 positives, only 90 are genuinely sick, about 9.2%. Most people, including many doctors in classic studies, guess something near 90%. It is the same trap: reading the reliability of the test as the probability of the condition.
Spam filter
An inbox gets 1,000 emails. 20 are phishing. A filter flags 95% of phishing and wrongly flags 5% of the rest. One email is flagged. How likely is it phishing?
Working it out. 19 phishing emails are flagged, and 5% of the 980 legitimate ones, which is 49, are flagged by mistake. That makes 68 flagged emails in total, and only 19 are phishing, roughly 28%. The filter is very good and still wrong about most of its flags because real phishing is rare. Whenever the thing you are hunting is rare, expect this.
Notice how each tab has the same three moves: list the equally likely cases, remove the ones contradicted by the evidence, and count what remains. The medical and spam examples add a fourth idea, that rare things stay rare even after a positive signal, which brings us to Bayes.
The Bayes view: updating your beliefs
Statisticians phrase all of this in terms of Bayes’ theorem. Do not let the name scare you. It is a rule for updating a belief when you get new evidence. Start with what you believed before (the prior), ask how likely the evidence is under each possibility (the likelihood), and rescale.
For the boxes:
- Prior: each box is chosen with probability 1/3.
- Likelihood of drawing gold: gold-gold gives 1, gold-silver gives 1/2, silver-silver gives 0.
- Multiply: 1/3 × 1 = 1/3 for gold-gold, 1/3 × 1/2 = 1/6 for gold-silver, 0 for silver-silver.
- Rescale so the total is 1: gold-gold gets (1/3) / (1/2) = 2/3, gold-silver gets 1/3.
The two thirds falls right out. The gold-gold box started at one third and rose to two thirds because it was better at explaining the evidence. Same answer, different language. Some people find the coin-counting version more natural, others the Bayes version. I suggest learning both, because when the numbers get bigger, Bayes becomes the safer bookkeeping.
Where this goes wrong in real life
Medical screening
Consider a test for a condition that affects 1% of people. The test catches 90% of true cases and wrongly flags 9% of healthy people. You test positive. Intuition shouts that you are 90% likely to be sick. Let us count in a population of 10,000.
Of the 100 sick people, 90 test positive. Of the 9,900 healthy people, 891 also test positive. So among 981 positives, only 90 are truly sick: about 9.2%. The test is not bad. The condition is just rare, so false alarms outnumber true detections. Doctors in famous studies have made this exact error, which is why good clinics follow a positive screening with a confirmatory test rather than reacting to the first result.
Fraud alerts and spam filters
Banks send you a “suspicious transaction” text. Most of the time it is a false alarm, because fraud is rare among millions of transactions. This does not mean the fraud system is broken. It means its precision is limited by the base rate, and it is a deliberate trade: a few annoying alerts for a lot of caught fraud.
Evidence in court
Lawyers call the mistaken version the prosecutor’s fallacy: taking “the chance of this evidence if the person were innocent is one in a million” and hearing it as “the chance the person is innocent is one in a million.” Those are two different conditional probabilities. Confusing them has contributed to real miscarriages of justice, and it is the same logical slip as reading “gold came out of the gold-gold box” as if it were “the box is gold-gold.”
Hiring and screening
A company designs a screening test that 95% of great candidates pass. Then it assumes anyone who passes is 95% likely to be great. If only a small percentage of applicants are great, most passers are not. Again the rate at which the thing occurs in the population is the piece people forget.
Why our brains keep choosing 50/50
Psychologists have a few explanations, and I find them all helpful when I am teaching this.
- We count the visible options, not the weights. After the silver box is excluded, two boxes are visible, so we say half and half. The weights are invisible unless you deliberately look for them.
- We confuse the box with the coin. The question is about boxes in our heads, but the random selection was really about coins.
- We love symmetry. Two options often feel symmetrical even when they are not. The mind reaches for 50/50 when it is unsure.
- We ignore how evidence was generated. A gold coin is more likely to appear from a box with more gold in it. Whenever a signal is easier to get from one hypothesis than another, seeing the signal shifts the odds.
Once you know these four traps, you can build a checklist. It is what I use before I trust any probability I have just worked out in my head.
The four-step checklist.
1. Write down the equally likely basic outcomes (not the tidy groups).
2. Remove the ones that contradict the evidence.
3. Count or weight what is left.
4. Ask whether the rate of the thing in the general population changes the answer.
Variations that test your understanding
A good way to make sure you really understand a puzzle is to change it slightly and predict what happens. Try these.
What if you are told only that the box is not silver-silver?
Then the two remaining boxes really are equally likely, and the chance that the box is gold-gold is exactly 1/2. The difference between this and the original is that you did not see a coin. The coin observation is what carries the extra weight. This variation is a fantastic reminder that how you learned something can matter as much as what you learned.
What if a coin is chosen at random from the gold coins?
Suppose someone gathers all three gold coins and hands you one at random, then asks whether its box-mate is gold. You would get the same 2/3, because you are effectively picking among the same three cases.
What if there are four boxes?
Add a second gold-silver box. Now there are four gold coins in total, two in the gold-gold box and two in the mixed boxes. Draw gold, and the chance the other coin is gold is 2 out of 4, or 1/2. The count changes, so the answer changes. This shows the method is flexible: nothing magical about two thirds, it is simply the result of the count.
What if the coin-drawing is not random?
If somebody peeked and deliberately showed you a gold coin whenever they could, the mechanism changes and the probabilities can change again. This is the same subtlety that makes the Monty Hall problem sensitive to the host’s rules. Always ask how did this evidence reach me?
A short history and why the name “paradox” is fair
Strictly speaking it is not a contradiction; the mathematics is completely consistent. It is usually called a veridical paradox: the answer is true, but it clashes with a strong intuition. The name “paradox” was attached to the puzzle later, and it is not something I would claim Bertrand himself chose. What he did give us is a lovely argument for why 1/2 cannot be right, and it needs no counting at all.
The symmetry argument against 1/2
Choose a box. Before you look at anything, the chance that it holds two coins of the same kind is 2/3, because two of the three boxes are matched. Now point to a drawer. You will see gold or silver. If seeing gold moved the chance of a matching box down to 1/2, then by symmetry seeing silver would move it to 1/2 as well. But if it moves to 1/2 whichever coin you see, you did not need to look: the answer was already 1/2 before opening the drawer, which contradicts the 2/3 we started with. The probability cannot change by looking when every possible look changes it the same way. So it stays at 2/3, and the matching box is just as likely to contain gold as silver. That is the cleanest way I know to see why the tempting answer fails.
The same family includes the Monty Hall problem and the two-child problem. If you enjoy Bertrand’s boxes, those are the natural next puzzles. They share a structure: some information is revealed, and the trap is to treat the remaining cases as equally likely when they are not. One caution on the two-child family: small changes in wording can really change the answer. “At least one is a boy” gives 1/3 for two boys, while “at least one is a boy born on a Tuesday” gives 13/27. The day is not irrelevant, because it changes which families qualify, and the symmetry argument above does not carry over: a family can have boys born on several different days, so the possible Tuesday-style announcements overlap instead of splitting the families into clean groups. The answer also depends on how you learned the fact. Meeting a random child who turns out to be a boy born on a Tuesday leads back to 1/2 for the other child.
The Sleeping Beauty problem is a different kind of puzzle, and I would not file it here. It asks about a self-locating belief after memory erasure, and thoughtful people still disagree. “Thirders” say 1/3 and “halfers” say 1/2, depending on how they model what waking up tells her. There is no settled consensus, so treat anyone who calls it closed with some suspicion.
Five mistakes people make with conditional probability
1. Treating the remaining cases as equally likely
After eliminating cases, do not assume what is left has equal weights. Check how likely each surviving case was to produce the evidence.
2. Mixing up P(A given B) with P(B given A)
The chance of a positive test given illness is not the chance of illness given a positive test. The two can differ enormously, especially when the condition is rare.
3. Ignoring the base rate
If the thing you are detecting is uncommon, even an accurate test will produce more false alarms than true finds. Always ask how common the condition is.
4. Forgetting how the evidence was produced
The same fact can mean different things depending on whether it was revealed at random or by someone who knew the answer. Monty Hall lives entirely in that difference.
5. Trusting intuition over a quick count
When two methods disagree with your gut, do the count or a simulation. It takes five minutes and it prevents embarrassing conclusions in a report.
Quick quiz: test yourself
Tap a question to reveal the answer and its reasoning.
In Bertrand’s box problem, you draw a gold coin. What is the chance the other coin in the same box is gold?
- 1/3
- 1/2
- 2/3
- 3/4
Three gold coins could be the one you drew. Two sit in the all-gold box. So 2 out of 3.
Why is the answer 1/2 tempting?
- Because the math is unfair
- Because it seems only two boxes remain and they look equally likely
- Because coins are random
- Because silver is impossible
After seeing gold, you can rule out the silver-silver box, which leaves two boxes. But the boxes are not equally likely to have produced a gold coin: the all-gold box has twice the chances.
Which change would make the answer exactly 1/2?
- Learning the box you picked is not all-silver, without seeing a coin
- Drawing two coins
- Adding a fourth box of silver
- Painting the coins
If all you know is that the box is not silver-silver, then the two remaining boxes are equally likely. It is the coin evidence that tilts the odds.
A test is 99% accurate but the disease affects 1 in 1,000 people. A positive result means the chance of disease is closest to:
- 99%
- About 9%
- 50%
- 1%
Per 100,000 people, 100 are sick and 99 test positive. Of the 99,900 healthy, about 999 test positive by mistake. So 99 of 1,098 positives are sick, about 9%. Base rates matter.
What is the general principle behind Bertrand’s paradox?
- Probabilities never change
- Always choose 50%
- Small samples are useless
- Condition on the evidence in terms of equally likely basic outcomes
List outcomes that are truly equally likely (here, the six coins), remove those that contradict the evidence, and count what remains.
Frequently asked questions
What is Bertrand’s box paradox?
A probability puzzle by Joseph Bertrand from 1889. Three boxes contain two gold, two silver and one of each. You pick a box at random, draw one coin, and it is gold. The chance that the other coin is also gold is 2/3, not the tempting 1/2.
Why is the answer 2/3 and not 1/2?
Because three gold coins could have been drawn, and two of them belong to the gold-gold box. Each coin was equally likely to be pulled, so the coin, not the box, is the right thing to count.
Is Bertrand’s box the same as the Monty Hall problem?
They share the same logic. Both involve information that arrives through a process that is not neutral, and both reward counting equally likely basic outcomes. Monty Hall adds a host who knows where the prize is.
Who was Joseph Bertrand?
A French mathematician (1822–1900) who published the puzzle in his book on probability in 1889. He was also known for the Bertrand paradox about random chords in a circle, which is a different puzzle.
How does this connect to Bayes’ theorem?
Bayes’ theorem updates a probability after new evidence. Here the prior chance of each box is 1/3, but a gold coin is twice as likely to come from the gold-gold box, so the posterior chance of that box becomes 2/3.
Where does this matter in real life?
Medical screening, spam filters, fraud alerts, court evidence and any situation where a positive signal is interpreted without considering how common the underlying thing is.
The takeaway
Bertrand’s box paradox is small enough to fit in your pocket and big enough to change how you read a headline. Evidence does not merely eliminate options; it reshapes how much each remaining option deserves to be believed. Count the equally likely basic outcomes, remove what the evidence rules out, and only then divide.
Next time someone says “it has to be fifty-fifty, there are only two possibilities,” you can smile and ask the veteran’s question: are the two possibilities really equally likely?
Joseph Bertrand did not attach the name “paradox” to the problem. It’s why 1/2 can’t be right.
Choose a box. The probability that is has two of the same kind of coin is 2/3. Point to a drawer. If that probability changes to 1/2 were you to open the box and find a gold coin, it would also change to 1/2 were you to open it and see a silver coin. But if it changes regardless of which coin you see, you don’t need to open it. It is already 1/2. When did it change?
Answer: it can’t. That’s the paradox.
The Monty Hall problem, and the two-child problem, are versions. Some add irrelevant information to the two-child problem, like born on a Tuesday,” and so change their answer. You’d think that they would find that a similar paradox, so so suspect their solution.
But the Sleeping Beauty Problem is not the same. It is an example of not constructing a proper sample space, or underrstanding what “new information” means. There are four combinations of the random elements: {H&Mon, T&Mon, H&Tue, T&Tue}. Beauty is sampling one combination at random, regardless of whether she is awake. Here inability to observe H&Tue does not change the fact that all four have a prior probability of 1/4. Her “new information” – which means member(s) of the sample space are eliminated, not that what happened was not expected to happen – is that H&Tue is eliminated., allowing an updated probability. for the other three.
Thanks for this, really good comment. Your symmetry argument is the cleanest way to see why 1/2 can’t be right: if the answer became 1/2 whichever coin you saw, you wouldn’t need to look, and that contradicts the 2/3 you start with. I’ve added it to the article and fixed the line about Bertrand himself using the word “paradox”, since that label came later.
One small pushback on the Tuesday version. The symmetry trick works for the coin because gold and silver split the outcomes into two clean, non-overlapping groups. A family can have boys born on several different days, so “at least one boy born on a Tuesday” and “at least one boy born on a Wednesday” overlap, and the argument doesn’t carry over. The day isn’t irrelevant: it changes which families qualify. So 13/27 holds when the family is selected for having a Tuesday boy, and it goes back to 1/2 if you meet one random child who happens to be a Tuesday boy. It depends on how you learned the fact.
On Sleeping Beauty, the four-case model you describe is the thirder position, and it’s a good one. But halfers argue the real sample space is just {heads, tails} (Monday always happens, and tails gives two awakenings from one flip), so waking up adds no new information. Serious people still disagree, so I wouldn’t call it settled. I’ve taken it out of the “same family” list in the article for that reason.
The same trick works with the Tuesday version, but you have to use 14 combinations instead of two. The the probability of a boy and a girl is 14/27 for every statement of the form “I have two children, and at least one is a who was born on a ,” the it is 14/27 if you simply say “I know a gender and a day of the week”. But if I simply say that without revealing what I know, the probability is 1/2.
But this just explains why the change is wrong. The correct solution, is that the probability of receiving information in whatever format you choose is 1 if both children fit the description, but 1/2 if only one does. The answer to the Tuesday Boy Problem is not (1+12)/(1+`12+14), it is (1+12/2)/(1+12/2+14/2).
I’ll get back ti you about Sleeping Beauty
You are applying the symmetry to the number of children, when it should apply to the number of distinct descriptions. ” born on a ” creates a 14-fold symmetry. The same “trick” works because each of the 14 possibilities has to have the same answer when used the same way in the question, or when it is stated when that the information is known but it is not revealed. But this only proves that the probability can’t change when such information is provided. (This is actually an reason often used in the Monty Hall Problem, but never justified as I did.)
The correct solution to any version of the Two Child Problem needs to count more than the combinations that can exist, it needs to use the probability that each combination will produce the information provided. This is literally an example of Bayes Law. In the Tuesday version:
1) One combination has two Tuesday Boys. Probability of the information is 100%.
2) 12 combinations have a Tuesday Boy and another boy. 26 combinations have a Tuesday Boy and another child. Probability of the information is 50%.
3) 36 combinations have two boys, but not a Tuesday Boy. 169 have no Tuesday Boy. Probability of the information is 0%.
According to Bayes Law, the probability of two boys, given “at least one Tuesday Boy,” is:
(1*1 + 12*0.5 + 36*0)/(1*1 + 26*0.5 + 169*0) = (1+6+0)/(1+13+0) = 7/14 = 1/2.
+++++
A “real sample space” contains all combinations of the random factors, regardless of how they may or may not be observed. Beauty’s “world” exists in only one day at a time, and there are two possibilities for that day. H&Tue is one of those combinations, whether or not Beauty can observe it.
Most halfers can’t explain why they don’t except the day as a random element. They want it to be true because it gives the answer they want to reach. But some resort to calling it an “indexical.” For example, if a probability textbook were to ask “What is the probability that today is Thursday,” it can;t be answered because the context is not defined. This doesn’t apply to the Sleeping Beauty Problem, because the problem establishes the necessary context.
But if that is your objection, Beauty is told that:
1) On Monday, and again on Tuesday after Tails, Procedure A will happen.
2) On Tuesday after Heads, Procedure B will happen.
Procedure A is “You will be woken, and asked for your probability that the coin landed Heads. Procedure B is not described, except that Beauty is assured that she will know which procedure is happening when it does.
Q1: Beauty is woken, and taken to DisneyWorld. She is never asked about the coin. This must be Procedure B. Still, she is curious, and says (to herself) that the probability of both Heads and Tuesday is 100%.
Q2: Beauty is woken, and asked about the coin. This must be Procedure A. Still, she knows that to allow for her curiosity in Procedure B, she would need to separate H&Mon from H&Tue, and T&Tue from H&Tue, in her “real sample space.” With equal probability. BECAUSE THEY REQUIRE DIFFERENT ANSWERS. There is no justification for not doing the same for T&Mon and T&Tue. So her sample space has four equiprobable outcomes, and three of them are consistent with Procedure A.
Note that Q2 does not depend, in any way, on her waking up on H&Tue. Just knowing that it is a possibility, and quite distinct from the others.
Thanks for writing out the Bayes version. Your arithmetic is right, and it matches what I wrote in the article: if one child is picked at random and announced as a Tuesday boy, the answer is 1/2, and if the family is selected for having a Tuesday boy, it is 13/27. Both are Bayes with different likelihoods. They differ only in the process that produced the information.
So I don’t think 1/2 is the correct solution to every version. It is the correct solution to the random-child version. The usual reading of the puzzle is the selection version, where the likelihood is 1 for any family with a Tuesday boy.
One point on the symmetry argument: the 14 statements “at least one [gender] born on [day]” each hold for 27/196 of families, which adds up to about 1.93. That is more than 1, so they overlap and can’t be treated as separate cases. That is why the trick works for Bertrand’s coins, where gold and silver are exclusive, but not for the selection version.
On Sleeping Beauty, I think the four-outcome argument is a good case for the thirder side, but it relies on equal probability for H&Mon and H&Tue, and halfers deny that. It isn’t forced by the setup. I’d treat it as a modelling choice, and I’m happy to read your follow-up.
“So I don’t think 1/2 is the correct solution to every version. It is the correct solution to the random-child version. The usual reading of the puzzle is the selection version, where the likelihood is 1 for any family with a Tuesday boy.”
The “usual interpretation” of the problem is indeed the selection version. But the “usual wording” is that there is a random family, and you somehow know that there are two children with at least one boy (maybe born on a Tuesday). THIS IS NOT THE “SELECTION VERSION.” For it to be, you need an explicit statement that says “boy (maybe born on Tuesday)” was determined before the family was selected, and used as a filter for that selection. That very rarely happens.
Halfers deny equal probability for H&Mon and H&Tue because they either deny that H&Tue happens. That only the possibility of observation allows for the day to exist. There is never a justification provided, just the implication that it cannot happen since Beauty will sleep through it.
Did you known that, in the paper that first presented the problem as “Always Monday But Only Tuesday If Tails,” that was not the question it claimed to answer? Here is what it posed: “Some researchers are going to put you to sleep. During the two days that your sleep will last, they will briefly wake you up either once or twice, depending on the toss of a fair coin (Heads: once; Tails: twice). After each waking, they will put you to back to sleep with a drug that makes you forget that waking. When you are first awakened, to what degree ought you believe that the outcome of the coin toss is Heads?”
The “two days” is not really a part of this, but it presages how Adam Elga defined the experiment to make the “once or twice” part happen. Here’s another way:
Replace the coin with a six-sided die (I can make this use the coin, but it gets convoluted), with sides labeled {H1, H2, H3, T12, T13, T23}. On day 0 the die is rolled. Over the next three days, Beauty is woken if, and only if, the day number appears on that roll. Each tine she is woken, she is asked for the probability that the die roll has an “H.” Then she is put back to sleep with amnesia.
Beauty can call the current day, whatever it is, day X. (Actually, we can tell her the day, but I don’t want to get into justifying that.) The other two days are day Y and day Z. BY HALFER LOGIC, each die roll in the sample space {HX, HY, HZ, TXY, TXZ, TYZ} had a prior probability of 1/6. Burt since she is awake, she knows the sample space is reduced to {HX, TXY, TXZ}. Bayesian probability now say sthe the probability of HX is 1/3.
Thanks, I think we mostly agree. The answer depends on the protocol: random child gives 1/2, selection gives 13/27, and the standard wording doesn’t say which one you’re in. That’s the point I was making in the article.
Two small notes on Sleeping Beauty. Halfers don’t deny that Tuesday exists. They argue that waking up isn’t new evidence, since Beauty knew she’d be woken at least once anyway. And your die version works out to 1/3, but it assumes “today” is equally likely to be any of the days. That self-locating step is exactly what the halfer-thirder debate is about, so I’d call it a thirder argument rather than a proof against halfers.
Either way, I kept Sleeping Beauty out of the Bertrand family on purpose, since it’s a different kind of puzzle. Thanks for the thoughtful discussion!
“The standard wording doesn’t say which one you’re in.”
Exactly. And by not saying, it cannot be what you called the “selection version.” By Bertrand’s Box Paradox. Make three versions of the wording; one with “boy,” one with “girl,” and one with “written down but hidden.” All three have to have the same answer, and the last one has to have the answer 1/2.
“Two small notes on Sleeping Beauty. Halfers don’t deny that Tuesday exists.”
It’s a blind spot. They do so by not allowing H&Tue to have a non-zero probability that is disjoint from H&Mon. The two are distinguishable outcomes, yet H&Tue has no probability to occur anywhere in their solution.
“They argue that waking up isn’t new evidence, since Beauty knew she’d be woken at least once anyway.”
They don’t define, or even comprehend, what “new evidence” means. They just assert that because you know you would wake, you can’t have it.
In fact, it isn’t a defined term in the Mathematics of Probability. It is used, informally, to mean information that eliminates members of the sample space. Which happens.
“And your die version works out to 1/3,”
Yep. And it is a faithful representation of the problem Adam Elga posed. So is the “Always Monday, Tuesday after Tails” one. They can’t have different answers, either. And that one is 1/3.
“[My six-sided die version] assumes ‘today’ is equally likely to be any of the days”
Again, exactly. But it isn’t an assumption, it is a fact. The setup is completely symmetric about which day it is, so “today” is equally likely to be any of them. That’s why we could tell Beauty what day it is; each day functions the same way as any other if we do.