# Puzzle procrastination: perplexing probabilities part II

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.