# Will the fact that my Italian son is going to attend a primary school change the expected number of Italian children to be present in his class?

This is a question stemming from a real-life situation, for which I have been genuinely puzzled about its answer.

My son is due to start primary school in London. As we are Italian, I was curious to know how many Italian children are already attending the school. I asked this to the Admission Officer while applying, and she told me they have on average 2 Italian children per class (of 30).

I am now at the point in time where I know that my child has been accepted, but I have no other information about the other children. Admission criteria are based on distance, but for the purpose of this question, I believe we could assume it's based on random allocation from a large sample of applicants.

How many Italian children are expected to be in my son's class? Will it be closer to 2 or 3?

• This reminds me of the old joke, "I always carry a bomb when I travel, because what are the odds of two people having a bomb on the same plane?" – Bill the Lizard Sep 23 '15 at 18:39
• That the Admission Officer told you that they have on average 2 Italian children per class makes this data 'suspiscious' to me. If it stemmed from a real computation you would have expected a non-round number. So it's possible that the true value is 1.51 or 2.49, say. Also since the Admission Officer is more likely to try to 'please you' with their answer, they may have rounded up rather than down (if they did think you'd be pleased to have your child among other Italians), suggesting that the probability distribution over values near 2 would be non-symmetrical. Answers below can be adapted. – PatrickT Sep 24 '15 at 7:25
• @PatrickT "Mode" is a valid type of average. – Ian Ringrose Sep 24 '15 at 8:46
• Thanks a lot guys for responding. I have now also posted a similar question, but with a different framing (stats.stackexchange.com/questions/173969/…), which has been triggered by some of your input / answers. – jj90213 Sep 24 '15 at 11:32
• @PatrickT I think that there're a lot more poorly educated people who'd be confused by 1.5 ("How do you have half a kid?") than stats nerds annoyed about excessive rounding seems more likely to me. (Assuming the more precise number isn't actually 1.9 or 2.1 anyway.) – Dan Is Fiddling By Firelight Sep 24 '15 at 17:57

As always you need to consider a probabilistic model that describes how the school distributes children among classes. Possibilities:

1. The school takes care that all classes have the same number of foreign nationals.
2. The school even tries to make certain that each nationality is represented roughly the same in every class.
3. The school doesn't consider nationality at all and just distributes randomly or based on other criteria.

All of these are reasonable. Given strategy 2 the answer to your question is no. When they use strategy 3, the expectation will be close to 3, but a bit smaller. That is because your son takes up a "slot", and you have one less chance for a random Italian.

When the school uses strategy 1 the expectation also goes up; how much depends on the number of foreign nationals per class.

Without knowing your school there is no way to answer this more perfectly. If you have just one class per year and the admission criteria are as described the answer would be the same as for 3 above.

Calculating for 3 in detail:

$$E(X) = 1 + E(B(29, 2/30)) = 1 + 1.9333 = 2.9333.$$

X is the number of Italian children in the class. The 1 comes from the known child, the 29 are the rest of the class and 2/30 is the probability for an unknown kid being Italian given what the school says. B is the binomial distribution.

Note that starting with $E(X|X\geq1)$ does not give the proper answer, as knowing that a specific child is Italian violates the exchangeability assumed by the binomial distribution. Compare this with the boy or girl paradox, where it makes a difference whether you know that one child is a girl vs. knowing that the older child is a girl.

• Let's make the binomial assumption and let $n=30$. It seems that the choice between $E(X\sim B(30,2/30)|X\ge1)$ and $E(B(29,2/30))$ can depend on the assumptions. For example, if I assume that any Italian father in London is very likely to be as puzzled as @user90213 and is then going to post a question here, then seeing this one question does not alter my expectations much. I only learned that one kid is Italian and would compute $E(X|X\ge1)$. Is it what you called "exchangeability"? If on the other hand user90213 is my close friend and I know his son, then I would arrive to your answer. – amoeba Sep 23 '15 at 15:30
• @amoeba Knowing that in a specific school and in a specific class there is the kid of user90213 is enough to distinguish him from the rest, it does not depend on how special your relationship to user90213 is. But it's tricky in that it matters how you learn the information. For example, if you ask by email that the oldest Italian child in a class contact you by name and you get a reply, you would go for the $E(X|X>1)$ approach even if you can distinguish the child afterwards. Try googling for the girl-boy paradox or even make a more general question for that. There is lots of discussion on it. – Erik Sep 23 '15 at 15:49
• That's right, thanks Erik. What I meant in the previous comment is something similar to your e-mail example. If I assume that all Italian parents in a class will post a question here, then seeing this question is exactly like getting contacted by the oldest Italian child. It seems that we are generally in agreement, +1. The wiki link is indeed interesting. – amoeba Sep 23 '15 at 15:57
• (+1) But puzzled why you say "If you have just one class per year [...]". – Scortchi - Reinstate Monica Sep 23 '15 at 16:06
• @Scortchi If the school has just one class per year, then it can use the two strategies denominated 1 and 2, since every child that is accepted at the school this year ends up in the same class. – Erik Sep 24 '15 at 11:09

Another way to look a this is at the level of individual children. Assuming that 30 children drawn randomly from a population (which you've indicated we can), we can work backward to the rough probability of an Italian child being drawn from this population: $2/30$ = $1/15$.

Given that we know that one of the 30 is Italian, we only have to compute the probability for the remaining children:

$$29 \cdot 1/15 = 29/15 = 1.933\ldots$$

So, knowing that your child is Italian changes the expected number of Italian children in the class to approximately 2.933, which is much closer to 3 than 2.

Here's my thoughts on how to approach this:

Let the random variable $S_n$ denote the number of Italian children in a class that is currently of size $n$. Let $X$ be the indicator for a new child's being Italian. Suppose that we add child $X$ to this class. Then the expected number of Italian children in this augmented class of size $n+1$ is $\mathbb E(S_n + X) = \mathbb E(S_n) + \mathbb E(X) = \mathbb E(S_n) + \mathbb P(X = 1)$. Note that independence doesn't matter here since we're only using the linearity of expectation. If child $X$ is known to be Italian then $X = 1$ with probability 1 so we have increased the expected value by 1.

• So there'll be $n+1$ children in this class after the addition of the Italian child? – Scortchi - Reinstate Monica Sep 23 '15 at 14:44
• Yes. Is there something that I'm missing related to that? – jld Sep 23 '15 at 14:47
• Depends how you read the question. Suppose classes are of exactly 30 children. – Scortchi - Reinstate Monica Sep 23 '15 at 14:49
• Maybe I misunderstood the question. I thought it was asking about how the addition of a known Italian child changes the expectation. – jld Sep 23 '15 at 14:49
• That's a very good point about class sizes possibly being capped – jld Sep 23 '15 at 15:49

Based on the Admission Office info, the number of Italian children follows binomial $\mathrm{Binom}(30, 2/30)$, assuming independence. Now you know in your class, there is at least one Italian child, so the expectation becomes $\mathbb{E}(X|X\geq1)$. For $X\sim \mathrm{Binom}(30, 2/30)$, this evaluates to $2.28$ (if I get my calculation right).

Edit. Evaluation of the expectation: $$E[X|X\geq1]=\sum_{i=0}^{30}iP(X=i|X\geq1)=\sum_0^{30}i\cdot \frac{P(X=i, X\geq1)}{P(X\geq1)}=\sum_1^{30}i\cdot \frac{P(i)}{1-P(0)}$$

(note the change in summation lower bound at last step)

• Can you elaborate on the conditional expectation? – Antoni Parellada Sep 23 '15 at 14:54
• Your answer is incorrect. The proper way to calculate this would be as 1 (the known child) + E(B(29, 2/30)) which turns out as 2.9333. And the assumption of binomial distribution is questionable. – Erik Sep 23 '15 at 15:04
• One more thing I would like to point out: a) your calculation of the conditional expectation is wrong. But b) more importantly, starting with your conditional expectation is incorrect. Knowing that a specific child is Italians breaks the exchangibility assumed by the binormal distribution. It's very similar to the boy-girl paradox (en.wikipedia.org/wiki/Boy_or_Girl_paradox) where it makes a difference whether you know that the older child is a girl or know that one of the two children is a girl. – Erik Sep 23 '15 at 15:16
• Scratch comment a) from above. But b) is more serious anyway ;) – Erik Sep 23 '15 at 15:23
• I agree. For OP, the distribution is no longer binomial(30, 2/30), but indeed 1+binomial(29, 2/30) – jf328 Sep 23 '15 at 15:29

No. Your knowledge of the impending events changes nothing about the school's typical experience.

• -1. This is incorrect, as explained in detail in other answers and comments here. – amoeba Sep 24 '15 at 14:19
• forgive my lack of advanced maths, but what makes this gent's child to be NOT one of the 'typically 2' children?.. such that we end up closer to 3. – Mart Sep 24 '15 at 16:35
• – amoeba Sep 24 '15 at 17:46
• Mart: Imagine I toss a coin ten times and count the heads; nothing odd about the coin or the way I toss it. I repeat that experiment many times, and on average I see almost exactly 5 heads in ten tosses; which results you see (1000 tosses in all, of which 50.3% were heads, well within the expected variation for a fair coin-tossing procedure; we decide agree that the process seems at least practically fair). Now I do the experiment one additional time with you, and you see that the first 4 tosses are all heads. What is the expected number of heads in the complete set of ten tosses? 5? more? – Glen_b Sep 25 '15 at 5:48
• Note that by your earlier argument, the first four "could have been four of the expected five". But then you'd be saying that there's less than a 50% chance on the next six tosses (in fact you're saying there's only a 1/6 chance on average). How would the coin know to come up heads less often? – Glen_b Sep 25 '15 at 5:55