JeffJor (u/JeffJor) - Readit News

JeffJor commented on The sisters “paradox” – counter-intuitive probability blog.engora.com/2025/08/t... · Posted by u/Vermin2000

bitwize · 3 days ago

Related to the Monty Hall "paradox". Spoiler: You'll get the car if you switch doors with 2/3 probability.

https://en.m.wikipedia.org/wiki/Monty_Hall_problem

JeffJor · 3 days ago

Bertrand's Box Paradox, which I wrote about in my own comment, applies to it. The upshot is that probability is not based on which prize placements _could_ lead the current game state, it is the set of all possible game states. Lets assume that the contestant starts off with door #3.

Case 1: The prize is behind door #1, and the host must open door #2. Probability 1/3.

Case 2: The prize is behind door #2, and the host must open door #1. Probability 1/3.

Case 3: The prize is behind door #3, and the host has a choice. Case 3A: The host opens door #1. Probability Q/3. Case 3B: The host opens door #2. Probability (1-Q)/3.

If the host actually opens door #1, the probability that door #2 has the prize is (Case 2)/(Case 2 + Case 3A) = (1/3)/(1/3+Q/3) = 1/(1+Q).

If the host actually opens door #2, the probability that door #1 has the prize is (Case 1)/(Case 1 + Case 3B) = (1/3)/(1/3+(1-Q)/3) = 1/(2-Q).

My point is that, since you get to see which door is opened, 2/3 is correct only if you assume Q=1/2. We aren't told what Q is, but we must assume it is 1/2 because otherwise the answer is different depending on which door is chosen.

JeffJor commented on The sisters “paradox” – counter-intuitive probability blog.engora.com/2025/08/t... · Posted by u/Vermin2000

JeffJor · 3 days ago

Q1: "A family has two children. You're told that at least one of them is a girl. What's the probability both are girls?"

Q2: "A family has two children. You're told that at least one of them is a boy. What's the probability both are boys?"

Note that these are symmetric problems, and must have the same answer.

Q3: "A family has two children. You're told that a gender, that applies to at least one, is written inside a sealed envelope. What's the probability both have that gender?"

In Q3, we have no information. So the answer is the proportion of two-child families that are single gendered. That is, 1/2.

But if we open the envelope, and read what is written inside, the problem becomes either Q1 or Q2. Which have the same answer. So we don't have to open it; whatever the answer to Q1 and Q2 is, opening the envelope in Q3 make its answer the same. If that answer is 1/3, we have a paradox. The answer has to be 1/2 of we don't look.

This is what is known as "Bertrand's Box Paradox." Well, if we add a fourth box to his problem, with one gold and one silver coin. I realize that in modern times the problem itself is called the paradox, but what Bertrand actually wrote (edited to this problem) was "How can it be that opening the envelope suffices to change the probability from 1/2 to 1/3?"

The resolution is that probability must be based on the full set of possibilities, not the possibilities that _could_ result from the full set of _states._ These are the possibilities for this problem:

1) BB and you are told that there is at least one boy. 2A) BG and you are told that there is at least one boy. 2B) BG and you are told that there is at least one girl. 3A) GB and you are told that there is at least one boy. 3B) GB and you are told that there is at least one girl. 4) GG and you are told that there is at least one girl.

Each numbered case has a prior probability of 1/4. Let's say the "A" subcases have a probability of Q/4, so the "B" subcases have a probability of (1-Q)/4.

The answer to the first problem is the probability of case 1, which is 1/4, divided by the total probability of cases 1, 2A, which is (1+2Q)/4. That's 1/(1+2Q).

The answer to the second problem is the probability of case 4, divided by the total probability of cases 4, 2B, and 3B. Which is (3-2Q)/4.

Bertrand's paradox, stated another way, is that these must be equal, but can only be equal if Q=1/2 and both answers are 1/2.

JeffJor commented on Probability Can Bite (2010) maa.org/external_archive/... · Posted by u/behnamoh

movpasd · 2 years ago

#2 is not actually equivalent to "at least one child is a boy". It is rather equivalent to "the first child is a boy". The difference may seem trivial, but one implies the other without the converse being true. This changes the probabilities — it's not an issue with underspecification.

I think your example #1 makes it much clearer why the 1/3 arises, at least in a frequentist analysis.

I would like to offer a similar interpretation but from a Bayesian lens. The 1/3 as rises due to the artificiality of the knowledge condition. Given real-world constraints, we expect any information collected to cleave neatly between the two children in our imagined information gathering scenario. So we implicitly translate "at least one child is a boy" to "we've checked one child, it's a boy".

Consider the following related problem: I have two faucets next to each other, each has a 50% chance of dripping overnight. I leave one shared bucket under both of them. The next day, the bucket is wet. What's the odds that _both_ faucets dripped?

This setup makes the correlative nature of the information much clearer, and I think most people would be less likely to jump to 1/2 as an answer.

JeffJor · 2 years ago

The problem is ambiguous, due to under specification. That means that neither #1 nor #2 is "actually equivalent" to "at least one child is a boy," and more information is needed to construct a probability space.

#1 is "When both genders are known, and boys are preferred in the description, at least one is a boy." The preference is what makes the answer 1/3, and assuming it adds information to the problem.

#2 can be "When only one gender is known, and how we know it is uncorrelated with either possibility, at least one is a boy." But it can also be "When both genders are known, and the description reflects the probability of that gender being chosen at random from the two, at least one is a boy." In both cases, the answer is 1/2.

But being under-specified does not mean the question can't be answered, it just requires applying a reasonable assumption instead of an unreasonable one. #1 is very unreasonable since it adds information, #2 is close, but #3 is best.

And the proof is Bertrand's Box Paradox. That name does not properly refer to a probability problem, it applies to how to make this reasonable assumption.

"Mr. Jones has exactly two children. I have written the gender, of at least one, inside this sealed envelope. What is the probability that both children have that gender?"

If you were to open the envelope, and see the word "boy," the problem becomes the same as the one under discussion. If it can be answered, that answer is correct here as well. But it is an equivalent problem if you see the word "girl," and again the answer must be the same. If 1/3 is an acceptable answer, it means that 1/3 of all two-child families have two of the same gender, and 2/3 have mixed genders.

But that is a contradiction. We know that the split is 1/2:1/2. So the assumption, that 1/3 is a reasonable answer, is disproven. Now, that does not mean that the information came to us via #2 or #3, it just means we can't assume that it was #1.

Most often, the same logic is used for the Monty Hall Problem, it is just applied backwards.