# What are Belgium's chances on getting a medal at the 400m finals, given that they have 2 of the 8 athletes?

This is a question my wife and I pondered over at the breakfast table this morning: Belgium has two athletes in the finals of the 400m at the World Championship athletics this year.

What are Belgium's chances of getting a medal, assuming every runner has the same capabilities? Plus, what's the formula/explanation for calculating this?

I know/think that, if we'd have 1 athlete, it would be 3/8, because there are 3 possible medals, and 8 athletes competing. (I hope this step is correct?)

But what if you have 2 athletes? Is it then just 6/8? Intuitively, that feels incorrect. But I would love to get a decent explanation on how to calculate this.

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Since they are brothers, we can hope that they won't quibble over political issues, therefore that rises their chances. :D –  Raskolnikov Aug 30 '11 at 7:39
I don't really understand what kind of assumptions you're making. Are you assuming that all of the athletes are equally good, so ranking in any given event will just be random? Why would you do that? –  Qiaochu Yuan Aug 30 '11 at 7:50
Guys, we know that the problem description doesn't take all the variables into account. But give us citizens of small nations some slack, please. The OP is to some extent gloating about the rare occasion of Belgium having two athletes in top 8 in any event. On any given year (but, alas, this one is a likely exception) I would do the same thing about men's javelin :-). –  Jyrki Lahtonen Aug 30 '11 at 8:02
Ok ok, I'm no mathematician, so indeed, it's assuming everyone has equal chances. Also assuming every lane on the track has the exact same resistance, wind conditions,... ;) I thought the question would be clear enough. It's about the calculation of the probability. –  Peter Aug 30 '11 at 8:06
The intuitive 6/8 falls down because there is a chance that they both win medals. The expected number of medals between them is 6/8, but $\frac{3\cdot 2}{8 \cdot 7}=\frac{3}{28}$ of the time they both medal. –  Ross Millikan Aug 30 '11 at 13:12

http://en.wikipedia.org/wiki/Combination

There are $\binom82$ possibilites for the ranks those athletes finish in.

In $\binom52$ of those, neither of them gets a medal,
so in $\binom82-\binom52$ of those, at least one of the athletes gets a medal.

Therefore the chance is $\frac{\binom82-\binom52}{\binom82} = \frac9{14}$ .

(A simple way to convince yourself that 6/8 should be wrong is that when
you apply that reasoning to 3 athletes, you get a 'probability' greater than 1.)

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Haha, great answer. Never thought about expanding the reasoning to 3 athletes. –  Peter Aug 30 '11 at 8:07
Usually, calculating the inverse and substracting it is easier. Same with dice: what is the chance that you roll an odd die? 3/6 -> 1/2. Now what is the chance, with two dice, that you roll at least one odd die? Well, take the chance you roll none (1/2*/1/2=1/4) and inverse it (1-1/4 = 3/4). –  Konerak Aug 30 '11 at 8:14
@UncleZeiv: Eight rankings in total, of which Beligians can occupy two; five non-medal-winning rankings, also of which Belgians must occupy two if neither of them is to win a medal. –  Niel de Beaudrap Aug 30 '11 at 11:07
@Niel: I see. I was reasoning the other way round, which comes more intuitive to me: 8 players can occupy the podium in (8 3) ways, of which (6 3) don't contain any of the Belgians. I end up with the same result so I guess it's correct as well? –  UncleZeiv Aug 30 '11 at 15:31
@UncleZeiv: right-o; both different ways of exploring the space of possibilities and filtering for the condition of there existing (or failing to exist) a Belgian on the podium. –  Niel de Beaudrap Aug 30 '11 at 15:35

It is easier to compute the probability of them not winning a medal. There are ${5\choose 2}=10$ different ways for the two Borlées to be placed in position fourth to eighth (5 positions, two brothers). Altogether there are ${8\choose 2}=28$ ways to put the two athletes in the 8 slots (1st to 8th). Assuming that positions are random (obviously a stretch of facts, but we ignore that), the chances for them coming out empty is thus $$\frac{10}{28}=\frac{5}{14}.$$ So the chances for a Belgian medal are $$1-\frac{5}{14}=\frac{9}{14}\approx0{,}643$$ or about 64%.

I'm rooting for them anyway :-)

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@Theo Buehler: Thanks for the corrected spelling. I really should have known better. I'm afraid I know next to no French, but that is, of course, a lame excuse to spell a name incorrectly. –  Jyrki Lahtonen Aug 31 '11 at 7:24
No problem at all and no need to apologize :) I should have added a link to their homepage which contains a few nice pictures of the three Borlées. –  t.b. Aug 31 '11 at 7:38
@Theo: Thanks for the link. I watched the final live, but the screen on my cell phone is too small for me to make out the letters, let alone to spot an accent. –  Jyrki Lahtonen Aug 31 '11 at 8:01

The first brother has a 3/8 chance of winning, the questioner was correct.

The tricky bit is why the second brother doesn't add another 3/8, and that's because the question was asking what was the chance that at least one brother got a medal. Therefore, if the first brother got a medal, we just don't care what the second brother does.

So the only way the second brother can help is if both the first brother doesn't get a medal, and the second brother does.

To find the chance of 2 independent things both happening, we multiply. The chance the first brother doesn't get a medal is 5/8, and the chance the second brother does is 3/7. (O.K., 3 out of 7 is another tricky bit. We have assumed the first brother lost, so with him "out of the running" there are only 7 people who have a chance of being in the 3 winning spots). Multiplying gives 15/56, which is the chance that only the second brother wins a medal.

So adding the 3/8 from the first brother to the 15/56 of help from the second brother, we get 36/56 (the same as everyone else, whew!).

The calculations other people used are based on generalizations of this simple case to any number of successes out of any number of tries, and on the realization that the easier way to think about this in general is to count the number of possible successes out of the number of possible outcomes.

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Very clear answer for people bad at math indeed. Deserves more upvotes if you ask me. –  Peter Aug 31 '11 at 8:42

So, there are 3 medals. You have 8 finalists. You are interested in only the winning chances of two of them. You can compute it as follows: count how many ways you can pick 3 winners out of 8 such that none of them is Belgian. (So, for the gold medal, you have 6 choices, for the silver 5 and for the bronze 4.) Divide this by the total amount of possible combinations of 3 winners, regardless of nationality. (Now you have 8 choices for gold, 7 for silver and 6 for bronze.) You then get the probability that no Belgian wins a medal. The complement of that probability is the one you are looking after.

The probability of no Belgian getting a medal is:

$$\frac{6 \times 5 \times 4}{8 \times 7 \times 6} = 0.3571 \; ,$$

the complement thus being

$$\text{Probability of a Belgian getting a medal} = 1- 0.3571 = 0.6429 \; .$$

This is making the quite unrealistic assumption though that all athletes are equally fit, that conditions are equal for them, etc...

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There are $\binom83$ ways to pick three winners out of eight. If the Belgians do not win, there are $\binom63$ ways to pick three winners from the non-Belgians. Hence the probability that some Belgian would get a medel is $1-\binom63/\binom83 = 1-\frac{6!5!3!}{3!3!8!} = 1-\frac{5}{14}=\frac{9}{14}.$

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First, write down all possible outcomes: $\Omega = \{(1,2), (1,3), \ldots, (1,8), (2,1), (2,3), (2,4), \ldots, (2,8), (3,1), \ldots, \ldots, (8,7)\}$, where, e.g., (2,4) means that the first athlete gets the 2nd place and the second athlete gets 4th place. Remember to distinguish, e.g., between (1,2) and (2,1).

Now write down all outcomes where they get at least one medal, i.e. $A = \{(1,2), (1,3), \ldots, (1,8), (2,1), (2,3), (2,4), \ldots, (2,8), \ldots, \ldots, (8,3)\}$.

Now just use the Laplace formula $$P("\text{they get at least one medal}") = \frac{\text{number of "good" outcomes}}{\text{number of all outcomes}} = \frac{\# A}{\# \Omega}=\frac{36}{56}\approx 0.64.$$

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Since their mother won't care which son stands (higher) on the podium, you can actually ignore the distinction between $(1,2)$ and $(2,1)$ this time. –  Jyrki Lahtonen Aug 30 '11 at 8:07