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?

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sokoloff 4 days ago | parent | next [-]

There aren't two "girl-girl" states, because of the stated assumption in the problem:

> Assume the family is selected at random because they have at least one girl.

Given that plus "a family has two children" and "Assume that the probability of having a girl or boy is 50%"

That means you're starting from the set of all two child families: BB, BG, GB, and GG, being told that you do not have the BB case, leaving 3 ways in which the family could be composed and being asked about "the one which is not a G".

That's different from the dime and quarter case, and would also be different if you were told "the oldest child is a girl", because being told "the oldest child is a girl" eliminates both BB and BG.

Being told "[at least] one of the coins is heads" or "[at least] one of the kids is a girl" only eliminates one of the four cases, while being told "the quarter is heads"/"the oldest is a girl" eliminates two cases.

	▲	taeric 4 days ago \| parent [-]
		I think where this fails for me, is it seems to change the odds of one of the flips? (And again, to be clear, I fully expect that I am the one that has this modeled incorrectly...) Consider. You flip a dime and a quarter, one of them is heads. What is the odds that the dime is heads? What are the odds that both are heads? If you do as stated in this model, you would say the possible states are HT, TH, HH. Doing that, you conclude that the odds of any single one being heads is 2/3. But, that does not make any sense, as you did nothing to change the expected 1/2 odds of either coin. And if the odds of either being H is 2/3, then the odds of both would be? So, in the original question, what would the expectation be on the question of "what are the odds that the first born is a girl, knowing that at least 1 is a girl?" My intuition would be that it should be 1/2? As the two children are independent events. Just as the two coin flips are independent of each other. Not knowing anything about either coin, the best you can do is the original odds. (Things change if talking about after you observe one, of course. If you see a heads, and I say that at least one is heads, you are then modeling if I am disclosing the one you already know or not. Basically, we have moved into liars dice.) So, how can you model this such that you keep the 50% odds per coin, but also have it so that the "at least one of" could apply to either coin? Conversely, why does knowing "at least 1 is heads" change the odds of either single coin being heads?

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quectophoton 3 days ago | parent | prev | next [-]

> So, my problem with how this is modeled is it assumes order doesn't matter in one aspect, but that it does in another.

Yeah, I'm with you here. Assuming heads=girl and tails=boy:

* If order does NOT matter (GB=BG), then it means Alice-Albert (GB, Heads-Tails) is the same as Albert-Alice (BG, Tails-Heads).

* If order DOES matter (GB!=BG), then it means Alice-Barbara (GG, Heads-Heads) is different than Barbara-Alice (GG, Heads-Heads). Thereforce, GG!=GG.

Either way, the stated problem seems badly defined.

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zeroonetwothree 4 days ago | parent | prev | next [-]

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.

	▲	taeric 4 days ago \| parent [-]
		Well, sorta? You were either in the world where dime was heads, in which case the space has TH, HH, or the Quarter was Heads, in which case you have HT, or again HH. Both with 50% probability, no? You were never in a scenario where HT and TH have the same probability as each other. (Well, again, sorta. Point being that you know one of them is possible.) Edit: I want to be clear that my initial thinking was exactly what you said. I was trying to "steelman" the bad intuition and I think I've trapped myself. :D

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pinoy420 4 days ago | parent | prev [-]

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