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.

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 “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 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.

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.

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

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.

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.