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Mental_floss has a list of ten paradoxes, dealing with math, logic, physics, language, or some other method of cramping your brain. There’s the crocodile who grabs a kid, a race with a tortoise, the dehydrated potatoes, and this one:

3. THE BOY OR GIRL PARADOX

Imagine that a family has two children, one of whom we know to be a boy. What then is the probability that the other child is a boy? The obvious answer is to say that the probability is 1/2—after all, the other child can only be either a boy or a girl, and the chances of a baby being born a boy or a girl are (essentially) equal. In a two-child family, however, there are actually four possible combinations of children: two boys (MM), two girls (FF), an older boy and a younger girl (MF), and an older girl and a younger boy (FM). We already know that one of the children is a boy, meaning we can eliminate the combination FF, but that leaves us with three equally possible combinations of children in which at least one is a boy—namely MM, MF, and FM. This means that the probability that the other child is a boy—MM—must be 1/3, not 1/2.

Wait a minute, I’ve flipped enough coins in statistics class to know that the real answer is still 50%, but how in the world did they come up with 1/3? Wait, wait: just who said these were “equally possible combinations”?

Still, debunking that one was easy compared to some of the other paradoxes in the list at mental_floss.

(Image credit: Flickr user Alex Proimos)

This line of reasoning doesn't work though. Try applying it to the Monty Hall problem:

You pick one of three doors on a game show, trying to find the one of the three with a prize. The game show host then always opens one door that does not have the prize, and asks if you want to switch your choice to the door you didn't choose and he didn't open.

You can demonstrate by straightforward experiments that by switching you have a 2/3 chance of winning, while not switching gives you a 1/3 chance of winning. If you simply ordered the doors by the order they were revealed, you would say that the two remaining doors each have a 50-50 chance of having the prize, and there is no reason to switch. Except both probability reasoning similar to what gives 1/3 for the problem in the story above, and simply trying it out experimentally show it is not 50-50.

The key in both cases is that what you are told gives you information. In the Monty Hall problem, the host always opens a losing door: in the case your first pick was right, it doesn't change anything by switching, but if you were wrong with your first pick, it guarantees your second pick will be right. By saying that one of two child is a boy, you've already said they can't both be girls, but didn't say which of the two kids is a boy.

The genotype vs. phenotype example I previously gave might seem more clear (to me at least). You have two carrier parents (each with a dominant and recessive gene). You look at the offspring, and can see which ones are double recessive, because they express the recessive trait. You grab one that is not expressing the recessive gene, what are the chances of it being a carrier versus double dominant? That is the same as saying you know one of the genes is dominant because it shows the dominant phenotype, but there is still a 2/3 chance of it being a carrier versus a 1/3 chance of it having no recessive genes at all. You can't say "Gene #1 is the one I know is dominant" and "Gene #2 is the one I don't know" because you don't know which of the two you actually know (e.g. if they both were dominant, you couldn't point to one and say that was Gene #1).

In other words, if you were told of two kids: Pat and Chris, and told one, but not which one, is a male, you would say the first child in your order is the first one revealed, a male one. If you found out both Pat and Chris were boys, which one was the one you were labeling as the "first child?" You can't say if you were referring to Pat or Chris in that case, so your original labeling was not referring to a specific child.
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Confidence intervals with the binomial distribution has some issues, but ignoring that and using the crude Wald confidence interval, lets look at your numbers and see what are the chances you missed 50% and above by luck (plus-minus errors below are approx 1 sigma):

Run 1: 13/31 -> 42+/-9% -> about 16% chance you got below half by luck
Run 2: 12/29 -> 41+/-9% -> about the same
Run 3: 10/27 -> 37+/-9% -> about 8% chance you got below half by luck
Run 4: 8/21 -> 38+/-11% -> about 15% chance getting below half by luck
Combined: 43/108 -> 39.8+/-4.7% -> less than a 2.5% chance of it originally being half or above, i.e. a 95% confidence interval would exclude one half but include one third.

If you think the problem is your runs were too short, try the link I give several comments down (not in response to a person, just the the story). You can easily do runs of a million or more. For example, if I do three runs, the percentages I get for males are: 33.36%, 33.29%, and 33.30%.
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Here's another one,
Take two pieces of paper (blank on both side), then choose one piece and put an 'X' on one side.
Now, get a opaque bag, and put both papers (without folding them up) inside the bag, and mix them up (including mixing up the front and back)
I will now play a betting game with you:
you can take out either piece of paper from the bag but without turning the paper over, and then choose how much you want to bet if the other side of the paper has an X; of course if the X is showing when you first take it out , then the betting is off, and the paper is return inside the bag for another turn.

Here's the question, is having the X on the other side a 50% chance?

The answer would be no. The chances are 33%.
Let's say blank=O and 'X'=X; the two pieces are OO and OX
there is 50% chance that you take out OX, within that, 50% chance that O is showing and the bet is on; and 50% chance that X is showing and the bet is off.
There is also 50% chance that you take out OO, and the bet is on.
That means for the condition that the bet is on, there is only 33% chance that the paper is OX.

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Now back to the boy/girl question; both 1/3 and 1/2 could be correct, depending on how the 'boy known' has come to be known.

I believe there is consensus that when having two children, 'one boy one gir'l is double the chances of 'two boys'; in this case where the sex of both children is known beforehand, the answer would be 1/3.

But if the sex of only one child chosen at random is known, and it happens to be a boy, while the sex of the other child is unknown, then the answer would be 1/2.
Referring to the betting game above, OX represents boy/girl, and OO represents boy/boy; but now add in another OX as boy/girl is double the chances of boy/boy.
The doubling of OX cancels out the 50% chance where boy/girl becomes invalid. Therefore the overall chances would become 1/2.
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Random integers between 1 and 2 for both columns, run for forty rows/sets four times - the first two columns are the 'generators,' the other columns are the generated numbers copied over (note that columns M & N match columns A & B.) Colored cells are sets containing at least one male, blue cells are sets containing both males. As you'll see, the numbers aren't even hitting the predicted 50%, but I'll put that down to short runs.
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