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GAINING WEIGHT... posted:I'm not sure about this. I think throwing in order of birth is just a red herring. B G and G B are the same "possibility" for Bill BB BG GB GG And then you eliminate the ones that don't fit the criteria. For Bob that results in: BG GG As there are only two possibilities left, and one has a boy, the odds the other child is a boy is 50%. For Bill on the other hand you get: BG GB GG The two out of three equally likely possibilities means there's a 67% chance that the other child is a boy. e: Of course none of this is actually true in real life, as any given child is more likely to be a boy than a girl. A Buttery Pastry fucked around with this message at 17:53 on Sep 3, 2016 |
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Sep 3, 2016 17:51
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GAINING WEIGHT... posted:Another way to think about it: we have the four combinations - BB, BG, GB, GG - each with a 25% probability, and we have information we can use to eliminate some of our choices. In the first, this is easy, because we are told something about birth order, so we take away BB and GB, leaving us with the other two. In the second, we aren't told order, so it feels like GB and BG are both still on the table, giving us 3 equally likely outcomes. This isn't the case, though; one of those two options is impossible, because there is definitely a girl, we're just not sure where. If the girl is the older sibling, the possibility that the older sibling is a boy is 0, even if we don't know that. True, we can't know which choice between BG and GB is impossible, but knowing that one of them is impossible is enough. We can know that BB and [one of (BG or GB)] are eliminated, giving us 2 options. Phyzzle posted:It is the same. For 99 of the flips, the probability of heads is now 0, which does not change the probability of the other flip. The final flip is heads (Bob scenario): 50% Any combination with only one heads (Fertile Bill scenario): HTT...T THT...T TTH...T and so on until TTT...H Plus a final "no heads combination" TTT...T Leaving you with 100 options where you have exactly one heads, and one option with no heads, resulting in a 100/101 chance of there having been 1 heads if 99 of a 100 are tails, or ever so slightly above 99% chance. Any combination with at least one heads (New scenario): 1-(1/2)^100 = basically 100% chance
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Sep 4, 2016 07:27
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GAINING WEIGHT... posted:[2] Two equal choices: GAINING WEIGHT... posted:To put a finer point on my above reasoning with the 100 coins example, your treating of "no heads" as one option is mistaken. It's actually 100 options: the coin we didn't know about is the first, and it was tails; the coin we didn't know about was the second, and it was tails; etc. There's 100 cases where the "other coin" was tails, and 100 cases where it was heads, giving us a 100/200 overall chance of heads and a 100/200 overall chance of tails. TTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTT While the scenario where there is a single heads can have that head be at any point in the chain, resulting in 100 combinations. Buried alive posted:You're being inconsistent. Yes, if BG is the case then GB is not the case. However, we don't even know if Bill has a boy at all. If GG is the case, then neither BG nor GB are the case. Similarly, if it turns out that BG is the case, then neither GB nor GG are the case. Two of the three options are impossible, depending on what the result is. Which results in this: 
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Sep 4, 2016 15:49
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