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zeroonetwothree commented on The sisters “paradox” – counter-intuitive probability   blog.engora.com/2025/08/t... · Posted by u/Vermin2000
lightvector · 2 days ago
One of the challenges with puzzles like this it that it gives you "at least one of them is a girl" as a mathematical assertion where you're not supposed to further introspect the context of how/why you're being given that fact.

But that's unrealistic. In real life, the context for how and why there would be a speaker telling you such a thing in the first place can be relevant and affect the probability!

How is this possible? Suppose among all the math-riddle-loving parents of two children who would ask such a puzzle in the first place there are an equal number of parents of B-B, B-G, G-B, G-G, and that each is equally likely to ask you such a riddle when you meet them.

Suppose when asking such a riddle the B-B parents tell you "at least one of them is a boy" (they don't have any girls, so that's the only way they can ask this kind of riddle), the G-G parents tell you "at least one of them is a girl" (same thing but in reverse), while the B-G and G-B parents say one of "at least one of them is a boy" and "at least one of them is a girl" equally at random.

Then, conditioned on being told that "at least one of them is a girl", the probability of another girl is actually 1/2, not 1/3 like the paradox answer claims. To see this, imagine 40 examples of the above puzzle asking taking place. You get 10 B-B parents saying "at least one of them is a boy", 10 G-G parents saying "at least one of them is a girl", and among the 20 (B-G and G-B) parents since they choose randomly, you have 10 saying "at least one of them is a boy" and saying "at least one of them is a girl".

So out of the 20 times where "at least one of them is a girl" is said, there are 10 cases where it's a G-G family and 10 cases where it's a B-G or G-B family, therefore conditioned on being told "at least one of them is a girl", the probability of two girls is actually 1/2.

If there were some gender bias in how the B-G and G-B families might ask the question, or other differences that affect how likely different of these people would be posing the puzzle to you, then the probability could be yet different than either of 1/3 or 1/2.

So there's a difference in being present something as a flat mathematical assertion that you're supposed to take at face value and not supposed to question further (where the probability is 1/3, as the article claims). Versus being told something in real life, where you always need to take into account the context and situation of the speaker, and the probability could be different.

There are real life implications of this too - the big classic one being publication bias / newsworthiness bias. As most people intuitively know by now, it is also often wrong to take the statistical analysis or claims of a particular research study or paper entirely at face value, because there is a bias in the fact that "positive" and "exciting" results are more likely to be reported in the first place, and so statistical outliers that aren't actually replicable are disproportionately likely to be reported (see also https://xkcd.com/882/). And publication bias still occurs with respect to the reporting of results, amplification or not in the media etc, even when the the authors themselves are trustworthy and have done their analysis within the paper in a statistically proper way. So conditioned on you hearing about the result in the first place, it is often less likely to be true (and less likely to replicate in the future, etc) than you would think if you just took the statistical analysis in the paper at face value, even when that analysis was done correctly. The situation in the "sisters paradox" of computing a probability taking a statement entirely at logical face value is rare in real life.

zeroonetwothree · a day ago
It’s not meant to be a real life problem but a math one. Would you like it better if it was phrased instead as: X and Y are random Bernoulli variables with p=0.5. What is P(X+Y=2| X=1 or Y=1)?
zeroonetwothree commented on The sisters “paradox” – counter-intuitive probability   blog.engora.com/2025/08/t... · Posted by u/Vermin2000
JeffJor · 2 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.

zeroonetwothree · a day ago
Well if we frame the question as “what is the probability of winning by always switching” then this doesn’t play into it and the answer is indeed 2/3. Hence as a question about general strategy the standard answer is correct.

You’re right if we are asking about a specific case though.

zeroonetwothree commented on The sisters “paradox” – counter-intuitive probability   blog.engora.com/2025/08/t... · Posted by u/Vermin2000
simonh · a day ago
These two questions are not equivalent.

Q1: I looked at only one of a pair of two randomly selected children and it was a girl. What is the probability the other I didn’t see is a girl?

Q2: I looked at both of two randomly selected children and at least one of the pair of children is a girl. What is the probability the other is also a girl?

zeroonetwothree · a day ago
There is no “the other” in the original question. Introducing that completely changes the meaning.

Deleted Comment

zeroonetwothree commented on The sisters “paradox” – counter-intuitive probability   blog.engora.com/2025/08/t... · Posted by u/Vermin2000
zeroonetwothree · a day ago
So we would write down the possibilities as (B-Tue, B-Tue), (B-Tue, G-Any), (B-Tue, B-NotTue) and the inverse for the latter two. This results in 27 cases. Of those 13 have two boys so the answer is 13/27.
zeroonetwothree · a day ago
Similarly if you had the knowledge “at least one is a boy born on May 11” then it would be very close to but slightly less than 50%.

So we can see in the limit as the information becomes more and more specific it turns into the unconditional probability. That is, the case of “the first is a boy, what is the probability both are boys” (50%).

I think this clarifies the situation in the OP pretty well.

zeroonetwothree commented on The sisters “paradox” – counter-intuitive probability   blog.engora.com/2025/08/t... · Posted by u/Vermin2000
jmount · 2 days ago
Peter Winkler shares some great variations of this: "Boy Born on Tuesday" (p. xix) and "Men with Sisters" (p. xxii) in "Mathematical Puzzles".

"Mrs. Chance has two children of different ages. At least one of them is a boy born on Tuesday. What is the probability that both of them are boys?"

(note: it is a puzzle, not a biology or data demography problem. so there are 50/50 independence assumptions on gender and uniform day of week assumptions prior to adding the conditioning.)

zeroonetwothree · a day ago
So we would write down the possibilities as (B-Tue, B-Tue), (B-Tue, G-Any), (B-Tue, B-NotTue) and the inverse for the latter two. This results in 27 cases. Of those 13 have two boys so the answer is 13/27.
zeroonetwothree commented on The sisters “paradox” – counter-intuitive probability   blog.engora.com/2025/08/t... · Posted by u/Vermin2000
layer8 · 2 days ago
Here “on Tuesday” is ambiguous, in my opinion. I first thought it meant “on a Tuesday” and that it was just a diversion. But it is likely intended to mean “last Tuesday” or “this Tuesday” (which excludes the boy-then-girl case). Wording it more clearly would likely reduce the ratio of wrong answers.

Furthermore, “of different ages” is likely intended to exclude the case of twins. However, even with twins, one is generally nominally older than the other. (Not to mention that it’s possible for two non-twin siblings to be the same age in years, at certain points in time.) Why not just say “that aren’t twins”?

I loathe when logic puzzles are obscured by ambiguous language, turning them more into “gotcha” text interpretation riddles than logic puzzles.

zeroonetwothree · a day ago
I think it clearly means “on a Tuesday”, anything else wouldn’t make sense as a puzzle. We are meant to assume each day of the week is equally likely.

Excluding twins is so that we can assume the probability of each day of the week is independent.

zeroonetwothree commented on The sisters “paradox” – counter-intuitive probability   blog.engora.com/2025/08/t... · Posted by u/Vermin2000
taeric · a day ago
So, my problem with how this is modeled is it assumes order doesn't matter in one aspect, but that it does in another.

Simply stated, if you allow the possibility space of "boy-girl" and "girl-boy", you have to also have two "girl-girl" states. Since you don't know which of the kids is known. Why is that not correct?

State it with coins, if I know that you flipped a quarter and a dime and one turned up heads, what are the odds that both are heads?

zeroonetwothree · a day ago
The answer is the same in your coin version. There are four possible outcomes: (Q, D) = HH, HT, TH, TT. Given that one turned up heads that eliminates TT so we see that HH has 1/3 probability.

As you can see there aren’t two HH states just as there aren’t two GG states in the original question.

zeroonetwothree commented on Comet AI browser can get prompt injected from any site, drain your bank account   twitter.com/zack_overflow... · Posted by u/helloplanets
Anon1096 · 6 days ago
You can safeguard against this by having a whitelist of commands that can be run, basically cd, ls, find, grep, the build tool, linter, etc that are only informational and local. Mine is set up like that and it works very well.
zeroonetwothree · 6 days ago
Everything works very well until there is an exploit.

u/zeroonetwothree

KarmaCake day5422November 5, 2011View Original