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So I've been trying to understand the boy-girl paradox, and I reframed it in terms of coin flips to help it make more sense to me and I get now why the probability is different depending on whether we're considering a specific child vs just either child. But now I'm stuck on how to think about the selection process when we're not starting with a specific child (or coin, or my rewritten version of the problem).

Specifically, I'm trying to figure out where I'm going wrong in thinking about, in a sequence of two coin flips P(both coins are heads|at least one coin is heads). It's easy enough to list out the the three possibilities and see that only one of them has both coins as heads and so the answer is 1/3. But where I'm confused is that I'm reasoning about it in what seems le should be an equivalent way, but getting the wrong answer:

If it's given that at least one coin is heads, then either, the first coin is heads and the probability that the second coin is heads (and hence both are heads) is 1/2. And similarly if we instead know the second coin is heads to start with. The probability that we're in the first situation is 1/2 and vice versa. So the total probability should be 0.5×0.5 + 0.5×0.5 = 0.5. Clearly that's wrong, but what's the specific error? My thought is that I actually should be breaking it up into three cases: the first coin is heads, the second coin is heads, both coins are heads. But it seems like I shouldn't need to do that, because an OR probability is inclusive. If I do P(second coin is heads|first coin is heads) + P(first coin is heads|second coin is heads) shouldn't that have already accounted for the possibility of both coins being heads?

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    $\begingroup$ The specific error is that you are not correctly applying the axioms of probability. Your calculation amounts to two ways of describing the same event and then summing their chances, which is unjustified. When you want to check any elementary argument concerning probabilities, it is always helpful to attempt to justify each step by appealing to a probability axiom. It's easy to do--in most axiomatic systems there are only three simple axioms--and the effort is quickly repaid by identifying potential flaws in the argument. $\endgroup$
    – whuber ♦
    Jul 4 at 20:18

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the first coin is heads and the probability that the second coin is heads (and hence both are heads) is 1/2. And similarly if we instead know the second coin is heads to start with. The probability that we're in the first situation is 1/2 and vice versa. So the total probability should be 0.5×0.5 + 0.5×0.5 = 0.5.

What you are doing here is adding the probabilities of two cases. There are two things wrong with this.

  1. Your two times 0.5x0.5 computation is for the probability that 'the first coin is heads and the second coin is heads' and the probability that 'the second coin is heads and the first coin is heads'. But those are just the same events that you are adding twice!
  2. When adding and multiplying probabilities you should think about whether the events are disjoint or not and for multiplying probabilities you should think about correlation. The addition rule of probabilities is the case when we compute the probability of an event that can be composed of disjoint subevents. E.g. the probability of only a single heads is the probability of HT plus the probability of TH and is 1/3 + 1/3 = 2/3. See also the final Venn diagram here: Multiply, add, or condition on probability?
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