children

A while ago I set some probability puzzles. If you’ve not yet pondered them, give them a whirl now. It’s OK, I’ll wait… All done? Final answer?

1 Girls, boys and doughnuts

We know that Laura has two children. There are four possibilities: two girls ( $\mathrm{GG}$ ), a boy and a girl ( $\mathrm{BG}$ ), a girl and a boy ( $\mathrm{GB}$ ) and two boys ( $\mathrm{BB}$ ). The probability of having a boy is almost identical to having a girl, so let’s keep things simple and assume that all four options have equal probability.

In this case, (i) the probability of having two girls is $P(\mathrm{GG}) = 1/4$ ; (ii) the probability of having a boy and a girl is $P(\mathrm{B,\,G}) = P(\mathrm{BG}) + P(\mathrm{GB}) = 1/2$ , and (iii) the probability of having two boys is $P(\mathrm{BB}) = 1/4$ .

After meeting Laura’s daughter Lucy, we know she doesn’t have two boys. What are the probabilities now? There are three options left ( $\mathrm{GG}$ , $\mathrm{GB}$ and $\mathrm{BG}$ ), but they are not all equally likely. We’ve discussed a similar problem before (it involved water balloons). You can work out the probabilities using Bayes’ Theorem, but let’s see if we can get away without using any maths more complicated than addition. Lucy could either be the elder or the younger child, each is equally likely. There must be four possible outcomes: Lucy and then another girl ( $\mathrm{LG}$ ), another girl and then Lucy ( $\mathrm{GL}$ ), Lucy and then a boy ( $\mathrm{LB}$ ) or a boy and then Lucy ( $\mathrm{BL}$ ). Since the sex of children are not linked (if we ignore the possibility of identical twins), each of these are equally probable. Therefore, (i) $P(\mathrm{GG}) = P(\mathrm{LG}) + P(\mathrm{GL}) = 1/2$ ; (ii) $P(\mathrm{B,\,G}) = P(\mathrm{LB}) + P(\mathrm{BL}) = 1/2$ , and (iii) $P(\mathrm{BB}) = 0$ . We have ruled out one possibility, and changed the probability having two girls.

If we learn that Lucy is the eldest, then we are left with two options, $\mathrm{LG}$ and $\mathrm{LB}$ . This means (i) $P(\mathrm{GG}) = P(\mathrm{LG}) = 1/2$ ; (ii) $P(\mathrm{B,\,G}) = P(\mathrm{LB}) = 1/2$ , and (iii) $P(\mathrm{BB}) = 0$ . The probabilities haven’t changed! This is because the order of birth doesn’t influence the probability of being a boy or a girl.

Hopefully that all makes sense so far. Now let’s move on to Laura’s secret society for people who have two children of which at least one is a girl. There are three possibilities for the children: $\mathrm{GG}$ , $\mathrm{BG}$ or $\mathrm{GB}$ . This time, all three are equally likely as we are just selecting them equally from the total population. Families with two children are equally likely to have each of the four combinations, but those with $\mathrm{BB}$ are turned away at the door, leaving an equal mix of the other three. Hence, (i) $P(\mathrm{GG}) = 1/3$ ; (ii) $P(\mathrm{B,\,G}) = P(\mathrm{BG}) + P(\mathrm{GB}) = 2/3$ , and (iii) $P(\mathrm{BB}) = 0$ .

The probabilities are different in this final case than for Laura’s family! This is because of the difference in the way we picked are sample. With Laura, we knew she had two children, the probability that she would have a daughter with her depends upon how many daughters she has. It’s more likely that she’d have a daughter with her if she has two, than one (or zero). If we’re picking families with at least one girl at random, things are different. This has confused enough people to be known as the boy or girl paradox. However, if you’re careful in writing things down, it’s not too tricky to work things out.

2 Do or do-nut

You’re eating doughnuts, and trying to avoid the one flavour you don’t like. After eating six of twenty-four you’ve not encountered it. The other guests have eaten twelve, but that doesn’t tell you if they’ve eaten it. All you know is that it’s not in the six you’ve eaten, hence it must be one of the other eighteen. The probability that one of the twelve that the others have eaten is the nemesis doughnut is $P(\mathrm{eaten}) = 12/18 = 2/3$ . Hence, the probability it is left is $P(\mathrm{left}) = 1 - P(\mathrm{eaten}) = 1/3$ . Since there are six doughnuts left, the probability you’ll pick the nemesis doughnut next is $P(\mathrm{next}) = P(\mathrm{left}) \times 1/6 = 1/18$ . Equally, you could have figured that out by realising that it’s equally probable that the nemesis doughnut is any of the eighteen that you’ve not eaten.

When twelve have been eaten, Lucy takes one doughnut to feed the birds. You all continue eating until there are four left. At this point, no-one has eaten that one doughnut. There are two possible options: either it’s still lurking or it’s been fed to the birds. Because we didn’t get to use it in the first part, I’ll use Bayes’ Theorem to work out the probabilities for both options.

The probability that Lucy luckily picked that one doughnut to feed to the birds is $P(\mathrm{lucky}) = 1/12$ , the probability that she unluckily picked a different flavour is $P(\mathrm{unlucky}) = 1 - P(\mathrm{lucky}) = 11/12$ . If we were lucky, the probability that we managed to get down to there being four left is $P(\mathrm{four}|\mathrm{lucky}) = 1$ , we were guaranteed not to eat it! If we were unlucky, that the bad one is amongst the remaining eleven, the probability of getting down to four is $P(\mathrm{four}|\mathrm{unlucky}) = 4/11$ . The total probability of getting down to four is

$P(\mathrm{four}) = P(\mathrm{four}|\mathrm{lucky})P(\mathrm{lucky}) + P(\mathrm{four}|\mathrm{unlucky})P(\mathrm{unlucky})$ .

Substituting in gives

$\displaystyle P(\mathrm{four}) = 1 \times \frac{1}{12} + \frac{4}{11} \times \frac{11}{12} = \frac{5}{12}$ .

The probability that the doughnut is not left is when there are four left is

$\displaystyle P(\mathrm{lucky}|\mathrm{four}) = \frac{P(\mathrm{four}|\mathrm{lucky})P(\mathrm{lucky})}{P(\mathrm{four})}$ ,

putting in the numbers gives

$\displaystyle P(\mathrm{lucky}|\mathrm{four}) = 1 \times \frac{1}{12} \times \frac{12}{5} = \frac{1}{5}$ .

The probability that it’s left must be

$\displaystyle P(\mathrm{unlucky}|\mathrm{four}) = \frac{4}{5}$ .

We could’ve worked this out more quickly by realised that there are five doughnuts that could potential be the one: the four left and the one fed to the birds. Each one is equally probable, so that gives $P(\mathrm{lucky}|\mathrm{four}) = 1/5$ and $P(\mathrm{unlucky}|\mathrm{four}) = 4/5$ .

If you take one doughnut each, one after another, does it matter when you pick? You have an equal probability of each being the one. The probability that it’s the first is

$\displaystyle P(\mathrm{first}) = \frac{1}{4} \times P(\mathrm{unlucky}|\mathrm{four}) = \frac{1}{5}$ ;

the probability that it’s the second is

$\displaystyle P(\mathrm{second}) = \frac{1}{3} \times \frac{3}{4} \times P(\mathrm{unlucky}|\mathrm{four}) = \frac{1}{5}$ ;

the probability that it’s the third is

$\displaystyle P(\mathrm{third}) = \frac{1}{2} \times \frac{2}{3} \times \frac{3}{4} \times P(\mathrm{unlucky}|\mathrm{four}) = \frac{1}{5}$ ,

and the probability that it’s the fourth (last) is

$\displaystyle P(\mathrm{third}) = 1 \times \frac{1}{2} \times \frac{2}{3} \times \frac{3}{4} \times P(\mathrm{unlucky}|\mathrm{four}) = \frac{1}{5}$ .

That doesn’t necessarily mean it doesn’t matter when you pick though! That really depends how you feel when taking an uncertain bite, how much you value the knowledge that you can safely eat your doughnut, and how you’d feel about skipping your doughnut rather than eating one you hate.

I enjoy pondering over puzzles. Figuring out correct probabilities can be confusing, but it is a good exercise to practise logical reasoning. Previously, we have seen how to update probabilities when given new information; let’s see if we use this to solve some puzzles!

1 Girls, boys and doughnuts

As an example, we’ve previously calculated the probabilities for the boy–girl distribution of our office-mate Iris’ children. Let’s imagine that we’ve popped over to Iris’ for doughnuts (this time while her children are out), and there we meet her sister Laura. Laura tells us that she has two children. What are the probabilities that Laura has: (i) two girls, (ii) a girl and a boy, or (iii) two boys?

It turns out that Laura has one of her children with her. After you finish your second doughnut (a chocolatey, custardy one), Laura introduces you to her daughter Lucy. Lucy loves LEGO, but that is unimportant for the current discussion. How does Lucy being a girl change the probabilities?

While you are finishing up your third doughnut (with plum and apple jam), you discover that Lucy is the eldest child. What are the probabilities now—have they changed?

Laura is a member of an extremely selective club for mothers with two children of which at least one is a girl. They might fight crime at the weekends, Laura gets a little evasive about what they actually do. What are the probabilities that a random member of this club has (i) two girls, (ii) a girl and a boy, or (iii) two boys?

The answers to similar questions have been the subject to lots of argument, even though they aren’t about anything too complicated. If you figure out the answers, you might see how the way you phrase the question is important.

2 Do or do-nut

You are continuing to munch through the doughnuts at Iris’. You are working your way through a box of 24. There is one of each flavour and you know there is one you do not like (which we won’t mention for liable reasons). There’s no way of telling what flavour a doughnut is before biting into it. You have now eaten six, not one of which was the bad one. The others have eaten twelve between them. What is the probability that your nemesis doughnut is left? What is the probability that you will pick it up next?

You continue munching, as do the others. You discover that Iris, Laura and Lucy all hate the same flavour that you do, but that none of them have eaten it. There are now just four doughnuts left. Lucy admits that she did take one of the doughnuts to feed the birds in the garden (although they didn’t actually eat it as they are trying to stick to a balanced diet). She took the doughnut while there were still 12 left. What is the probability that the accursed flavour is still lurking amongst the final four?

You are agree to take one each, one after another. Does it matter when you pick yours?

Happy pondering! I shall post the solutions later.