Cheeky probability problems to irritate your friends
Suppose a friend of yours has two children. But you cannot remember their genders. One day, he implies that one of the children is male. Now here’s the question: what is the probability that the other child is female? You may assume for this question that the only options are male/female.
Go ahead, have a think.
I got this one completely wrong, the first time I heard it.
I should have learned my lesson from the time I encountered the Monty Hall problem. Because in a sense, this is the Monty Hall problem, wearing a big fluffy wig and a stick-on moustache. It’s a tricky problem of conditional probability.
I’d imagine your initial guess was 1/2, or 50%. It seems like a 50% chance, because in this scenario, the child could be either male or female.
However.
The answer is actually 2/3. Grrrrrr 😠
To leave that issue on a cliff-hanger, let’s talk about the Monty Hall problem. If you haven’t heard of the Monty Hall problem before, you’re in for a treat! The general premise is as follows:
Don’t get a shock, but somehow we’ve ended up on a game show! And we’ve arrived at a challenge. We have three doors in front of us. Behind one door, is a car. And oh we really want to win that snazzy set of wheels.
The host of the game show announces to the crowd: we, the contestant, must choose a door. In the heat of the moment, we go for the middle door. Actually no, the left-hand door! The middle door is far too obvious.
Now the host won’t open the left door for us just yet. Of course not — he needs to drag this on a bit and give the crowd a spectacle. So he announces that he will open one of the other doors, and show us what is behind:
Ahhh, nothing. Of course. And now the host turns back to us. Would you like to change your decision?
We’ve tentatively locked in the left-hand door. The host has opened the middle door to reveal… nothing. Shall we switch to the right-hand door? Or stick with our gut decision?
He’s probably trying to trick us. And it’s a 50–50 chance, anyway. We stick with the left-hand door! The host opens it with a flourish… and behind the door, is nothing.
Damn. No car for us.
Now let’s just rewind a little bit. After the host opened the middle door, we assumed he’d presented us with a 50–50 chance. The car is, of course, behind either the left-hand door, or the right-hand door, with equal probability.
Actually… no. And this is where people get very, very angry about the Monty Hall problem.
Let’s start from the beginning. Before the game gets going, the car sits behind each door with 1/3 probability.
When we selected the left-hand door, the probabilities look like this: a 1/3 chance that we picked the correct door, and 2/3 chance that we got it wrong.
Now here is the key. The host will always open one of the doors we did not pick. He will always open a door which does not have the car behind it. And this actually gives us some information.
Remember: there is a 1/3 chance that our original decision was correct: it’s behind the left door. And a 2/3 chance that it was wrong: centre door or the right door. And Monty has just tossed away the centre door. There is now a 2/3 chance of the car sitting behind the right-hand door alone:
People get very annoyed about this problem, because it’s deeply unintuitive, at least on initial viewing.
If you don’t believe my answer, you can simulate this problem very easily with a dice and a friend. Use the dice to allocate the ‘car’ to a ‘door’ (say if you roll 1–2, it’s left, 3–4 it’s centre and 5–6 it’s right). And then you can play the host, and have your friend play the contestant. Re-run the scenario a number of times, and you’ll quickly notice your friend has far better chances if they switch, than if they remain.
Or perhaps you’ll believe my computer simulation of the problem:
The secret behind this problem lies with information and conditional probabilities.
We picked the left-hand door. And if the car is not behind our door (as it won’t be 2/3 of the time), we have forced the host to reveal to us which door the car is behind. Meaning that 2/3 of the time, we’ll get the car if we switch.
So let’s return to our initial problem — the boy-girl issue. Have you figured it out yet?
We know that there are three categories for the pair of children: MM, MF, or FF. That is: two males, one male and one female, or two females.
Ultimately, the category of interest is MF — one male, and one female. Before the father drops any hints, we have the following situation:
Now, when the father reveals one child to be a male, this actually removes one possibility (FF), leaving us the following scenario, where we have only the categories MF and MM:
And clearly, MF occurs 2/3 of the time, when we know FF is not an option. In other words, given that one child is male, the probability of the other child being female is 2/3.
If you still don’t believe me, you can try and devise a way of simulating this problem.
Aren’t probability problems lovely 😅
Thank you for reading :)
-Marc
