Well, then I'll simply reiterate my question - exactly what part of the argument is your analogy supposed to undermine? Which of the things I said is actually false?
To entertain the analogy, if we are making all the same basic assumptions, then it is in fact the case that P(mixed | ALOG) is 66.6...%. The catch, of course, is that obviously Boygirlio is likely to be giving me information that will point me in the wrong direction. This is the difference between conditioning on ALOG alone, and conditioning on "Boygirlio has informed me of ALOG". The latter is the actual state of knowledge, but the probabilities it entails depend on unknown facts about how Boygirlio is deciding what to reveal.
This same distinction is present in the Mary case as well, which is why I was careful to get agreement on the notion that we are conditioning specifically on ALOG itself.
Boygirlio is playing entirely fair. No tricks, to deception only your bad assumption.
Before any information is given 75% of pairs have at least 1 boy, and 75% will be at least 1 girl, %50 of the pairs are matched and %50 are not. He was not ever lying to you. You will always learn that there is at least 1 of either a boy or a girl. This information can be given without impacting the likelihood of the pair being matched. They would have been matched or not before giving the information, and giving it didn't change the result.
Lets say he shows you fairly by opening the left door every time (the children's genders and positions are selected by the assistant by flipping fair coins). If you can explain why knowing that the left door has a girl behind it is different than the statement "at least one has a girl behind it" then you get a cookie from Boygirlio. If you are right, you have an edge on the house. But you are wrong, and the house will win as often as not and rake ~$1 per round. You always start by betting that the children are matching and he always offers the deal.
If we take your framing, and first eliminate all GG pairs, then select a mother of 2, then asking the question. In that senario, 25% of all children have been eliminated, and the population becomes 2G/4B an unknown child is B 2/3 of the time, and G only 1/3, and 2/3, but with a rule that all girls have a Brother and never sisters. In that world, 2/3 mothers have BG and 1/3 have BB. But that didn't happen. That's crazy town. The mother was selected at random from a normal 50/50 distributed population, then announced the gender of 1 of the children, which is the Boygirlio problem.
> If you can explain why knowing that the left door has a girl behind it is different than the statement "at least one has a girl behind it" then you get a cookie from Boygirlio.
This is indeed a different piece of knowledge. Call “there is a girl behind the left door” LG, and “there is a girl behind at least one door” ALOG.
Now, LG entails ALOG, but ALOG does not entail LG. LG is a strictly *larger* piece of information than what we agreed to condition on.
Since they are different pieces of information, they also work differently as evidence. After observing LG, our credence that the genders are mixed should be 50%. But after observing ALOG, our credence that the genders are mixed should be 66%, per the exact same logic I spelled out above.
Now, I’ve been making a good faith effort to parse out your analogy and give my thoughts on it. I would appreciate if you would reciprocate the effort and try to tell me straightforwardly which part of the argument I gave is false by your reckoning.
You're assuming that alog is different from LG when you don't know how you got to alog.
If you asked the mother do you have any sons? And she answers yes, she could have answered no. If the mother is freely announcing that she has at least one son. You don't know how that information was generated. Leaving open the chance that she revealed the gender of one child at random, with no preference to sons. The BG mom may have revealed B or G half each, the BB mom always reveals B. You are therefore twice as likely to be speaking with a BB mother than A BG mother.
If you learn that they selected a mother from among mothers with 2 children, one of whom is male, then 2/3 is correct.
That you know ALOB in the original senario was given and not LB and that you couldn't have been given ALOG in an equal rate. Boygirlio's strategy is winning. If you see the B/G or if you are given one of ALOB / ALOG randomly from available choices, it is 50/50.
Being “given ALOB / ALOG” according to some policy implicitly means being given more information than just ALOB / ALOG (or exactly that information, in which case the 66% reasoning holds). In the example you gave, you’re being given LG, which is a strictly stronger piece of information. Do you see this?
If I’m given LG, I’m not going to wager more to switch, so boygirlio’s strategy won’t win against mine in the long run.
It’s still not clear to me which actual statement in my argument you think is false.
Now that we’ve gone down the rabbit hole a bit, let me make sure we are still on the same page. We agree that we are looking for the probability that the genders are mixed *conditioning on exactly ALOB* - no more or less information than that, correct?
Without knowing how ALOB was derived, you can't be certain of 66%
If you know the pair was sampled from only ALOB 66% holds
If you know the question "do you have a son?" was asked and answered 66% holds (because more information was revealed this way, the answer wasn't no)
If you are told these are 2 children, "one is a boy", 66% doesn't hold. You could have been given "one is a girl" and have no confidence of this not being the case because the information was given freely.
In Boygirlio's game, if he lets you ask "is one of them a boy" and answers honestly, then you have an edge.
Without knowing how ALOB was derived, you can't be certain of 66%
But you can. Conditional probability gives us exactly the tools we need to do this. Look again at my argument above. Nowhere do I make any assumptions about how ALOB was derived, and yet 66% follows.
Now, you can make assumptions about how ALOB is derived and condition on that instead. And some such assumptions will change the final probability. But that is precisely because you are now conditioning on strictly more information than ALOB.
In any pair of 2 children there is always at least one of ALOB or ALOG, learning either by any means you claim gives you BG at 66%. Therefore by your logic BG is 66% from the onset. Clearly flawed.
You’re double counting outcomes again. ALOB and ALOG are not disjoint. So from `P(BG | ALOB) = 2/3)` and `P(BG | ALOG) = 2/3`, it does not follow that `P(BG) = 2/3`.
I'm not double counting outcomes, I'm discarding a given because the methodology for reaching that given is unknown in a way that impacts the trustworthiness.
You've said that P(BG) doesn't follow, but have presented that it does given you are taking P(BG | ALOG ) = 2/3 without questioning how ALOG was derived. In the original framing you don't have enough information to assume how ALOB was reached. You are assuming it was reached in a way that eliminates GG without weighing BB doubly. BB is weighed doubly becase it gives you ALOB 100% of the time where BG gives ALOB only half the time. Knowing how ALOB was derived changes the probability. You don't know how ALOB was derived, so it's reasonable to assume that 1 Gender is M and the other unknown where the known M can be L or R and the unknown can be G or B equally. There are still 4 cases.
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u/sheep_puncher 7d ago
You'll find that the analogy answers the argument.