The problem sounds trivial enough, and comes in the form of a statement for which we have to predict the probability. It reads ‘I have two children. One is a boy born on a Tuesday. What is the probability that I have two boys?’
It sounds trivial. The Tuesday bit is just window dressing, so we are looking at ‘I have two children, one a boy. What is the probability I have two boys?’ So with one child a boy, surely there is 50 per cent chance that the other child is a boy and a 50 per cent chance it’s a girl. Which makes the probability of having two boys 0.5, or 50 per cent. There’s a one in two chance.
But unfortunately that is not correct.
The reason we get confused is that when trying to imagine the situation we think of the ‘first’ child we come as a boy, then look at the options for the second child being a boy. However the description of the situation would also work if the first child is a girl and the second child is a boy. The only way to be absolutely certain is to work through every possible combination:
These are the four possible combinations, each equally possible. Of these, three are situations that match my initial statement ‘I have two children, one is a boy’. In all but case 4, one of the children is a boy. But only one of those three combinations with a boy also makes the second child a boy. So the answer to ‘I have two children, one a boy. What is the probability I have two boys?’ is not 50 per cent, or one in two, it is one in three. This part of the problem is probably on a par with Monty Hall in the difficulty of getting your head around it. But there is a more fiendish part. We were wrong to discard the Tuesday. Saying the boy was born on a Tuesday changes the probability.
To see this we need a much bigger table. It starts like this:
In total we have 196 entries in this table. We go through every single sex/day combination in the first column combined with a girl born on Monday (fourteen of them in all), then every single sex/day combination in the first column combined with a girl born on Tuesday (fourteen of these too) and so on until we have cycled through every option for the second child.
Now we need to know two things. How many of those pairs feature a boy born on a Tuesday (like option 2 above) and how many of those have a second boy? We are going to have one combination of child A as a boy born on Tuesday with every possible child B – fourteen of those, plus thirteen other combinations where child B was a boy born on Tuesday, but child A wasn’t (we have already counted the instance were both child A and child B are a boy born on Tuesday). So there are 27 rows that match our circumstance of having a boy born on Tuesday.
We now need to pin down how many of those rows had two boys. The first set of fourteen all had a boy as child 1, and half of those – seven also had a boy as child 2. Of the thirteen additional rows where child B was a boy born on a Tuesday, six would have child A also a boy. So of the 27 rows with a boy born on a Tuesday, thirteen of them have a second boy. The answer to ‘I have two children. One is a boy born on a Tuesday. What is the probability I have two boys?’ is 13 in 27 – almost, but not quite, one in two.
This really upsets common sense. Just by specifying the day on which one of the children was born we change the probability of both children being boys from one in three to 13 in 27. Yet our minds rebel at this. Surely we could have chosen any day? The only way I can see to make some sense of this is to point out that in any particular real circumstance, you can’t choose that day at random; it is extra information that depends on the circumstance. The boy will have been born on a particular day and the result of that is that it cuts down the options, just as Monty Hall did when he opened a door and showed a goat. The reality is even harder to accept in this example.
There's lot's more on the beguiling nature of probability in my book, Dice World.
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Actually, this is wrong. Just look at the first problem, which is simpler. In the two scenarios that there is a boy and a girl, that are 1) and 3), they could have revealed either the sex of boy or the sex of the girl and then ask for the other, each with 50% chance. Nothing establishes that they must necesssarily reveal the boy first.
ReplyDeleteTherefore, after the revelation of the boy, both cases 1) and 3) lose half of their probabilities, meaning that they are currently half as likely as case 2), in which a boy was guaranteed to be revealed. If you take that into account, you will get that each gender is 50% likely.
Ironically, this is the same principle that applies in Monty Hall, it's just that this time it plays in favor of the 1/2. In Monty Hall, when the player's choice has a goat, given that the player's choice is not allowed to be discarded, the host is only left with one goat available to reveal from the rest, being 100% forced to show it, but when the player's choice has the car, the other two have goats so the host is 50% likely to reveal any of them, neither guaranteed.
The analysis would be correct if the person just told you that they have two children and you asked: "Do you have a boy?", in which case you would force them to always reveal the boy whenever they have at least one.
Now fix the same mistake in the second problem and you get 50% chance too.
I think your explanation works only if the first boy reveal was done for a specific birth-order, meaning you start with all 4 possibilities, and that the first reveal has to be that of a first-born or last-born. When the "boy" reveal is not associated with a birth position, Mr. Clegg/vos Savant's explanation would be correct. Work it backwards. Start with the given of two kids and no gender, you would have 4 unknowns. Draw a grid and put each unknown into a box. Now someone tells you box 1 has a boy in it. The question becomes what's the chance that box 1 has 2 boys? Since you can only eliminate the one choice with 2 girls, the chance of box 1 having 2 boys is now 1 of the 3 remaining boxes.
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