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)
No, the answer is one third, and you can check it by doing an experiment with coins (or an Excel spread sheet, or simple program). The key point is they are not giving info about one child and asking for the chance of a specific child being male or female, but saying "one" and "the other" places constraints only on the whole system. If you flip two coins, there are four equal possibilities: TT, TH, HT, HH. Even if not keeping track of order, it is TT, HT, HH with a 25%:50%:25% split. The information given allows you to only eliminate one of those (say TT), and not say which coin is heads, and either way end up with a 1:3 ratio.
You can literally test this with coins. Flip a pair of coins. If the pair is TT, don't count it and reflip. Otherwise record how many times you get HH and how many times you get one of each. With about 25 flips, you should have less than a 5% chance of getting 50% or higher. 100 flips would give you less than a thousand chance of getting 50% or more HH, and about 5% chance of getting more than 40% HH, at which point it would be rather tedious but separated from 50%.
This question comes up a lot in three contexts: lessons about the subtle meanings of words in word problems, basic level combinators/probability, or random "paradox" lists. But even like the Monty Hall probability question, people will argue until against the results of even basic experiments until the day they die.
Notice how, in the boy/girl version, the latter two choices suddenly introduces the older/younger sister issue. This is a red herring. You can have an older or younger sibling, and you can have a male or female sibling. These remain separate and have no bearing on one another, nor is age mentioned as a factor in the question - it only asks about boy or girl. However, if you insist on this age thing, then there is an additional possibility not mentioned: not just two boys as they say, but an older brother or a younger brother, adding another option and once again balancing back to 50%.
It's also extremely simple: a 50/50 chance is not modified by any other factors, as most people rightfully believe. Nor does it change with repetition, as many people get wrong - while repeated coin tosses will increase the probability of getting a certain result among the total, no one toss changes from 50/50.
The coin example from PlasmaGryphon (and many others) includes a similar mistake. The choices are listed as TT, TH, HT, and HH - but this only holds true if you have assigned a particular coin to each position. Otherwise, TH and HT are the same thing. Now, if you've accepted this concept, you cannot arbitrarily choose which coin counts as the 'official' heads. If it is the first coin, then the first two choices are obviously ruled out, since the first coin in both of those is tails - what remains is still heads or tails for the second coin.
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.