Age before beauty

I went out tonight with a bunch of colleagues and ex-colleagues. Five people in total, myself included.

The five of us sat around a square table, and for one reason or another, we soon established that we were sat in age order, clockwise. The second oldest person (me) was sat to the left of the oldest, the next youngest was sat to his left, the next youngest to his left and the youngest to his left, and indeed to the right of the oldest person.

I jokingly asked what the chances were of that happening before calculating the odds of us sitting in age order, either clockwise or anticlockwise, as one in twelve, or 8.333%.

A lot of frustrating discussion ensued, during which I tried to convince my company of the veracity of my claim.

I’ve since sent them an email complete with the full set of 120 possible seating combinations, the ten age-ordered combinatons highligted.

I think I’m owed either an apology or the £10 offered by way of a bet.  Or maybe both.

Comments

4 Responses to “Age before beauty”

  1. Nick Robinson on April 9th, 2009 06:38

    I get the 120 possible seating combinations – that’s 5! isn’t it?

    But where do you go from there – is it nCk or nPk or something else?

    In my defense, nobody ever gets probability, otherwise we couldn’t subsidise opera by taxing all those poor people via the Lottery

  2. admin on April 9th, 2009 08:47

    The best way to think about this is as follows, I think, Nick.

    The order in which people sit down is immaterial, given that we assume they sit down randomly. So let’s assume the oldest person sits down first.

    Where they sit is immaterial to the outcome. Now the probability of the next oldest person sitting to his left is 1 in 4, given that one in four people can opt to sit in that seat. Once that fella’s sat down, there are three comparative youngsters still standing. The chances of the oldest remaining fella sitting in the next seat round is therefore 1 in 3. The odds of the second youngest fella sitting in the next seat is 1 in 2. And if everyone else has already sat down, the youngest fella only has one place to sit.

    1/4 * 1/3 * 1/2 = 1/24. That’s the probability that people will sit in descending order of age, running clockwise. The chances of them sitting in the right order but anti-clockwise is the same.

    So 1/24 + 1/24 = 1/12.

    Or alternatively…

    The 120 seating arrangements is actually inflated by the fact that everyone can shuffle round one and be technically in a different formation even though, relative to one another, they are in the same order. So instead of thinking of 120 seating arrangements, think of 24, where one person is in a fixed seat and the other four people shuffle and move around them.

    Given this, only two seating combinations will result in an acceptable progression. One running clockwise, one running anti-clockwise. 2 in 24 = 1/12 and we’re done.

  3. Jon on April 9th, 2009 08:48

    Dan,

    Looks like you got the simplest part of this wrong:

    “an apology OR the £10 offered by way of a bet.”

    Shouldn’t that be an AND?

  4. Jon on April 11th, 2009 18:03

    When re-posting this article did you add “Or maybe both” or did I miss it when i read it first time round?

    If I missed it, apologies for my previous comment, it looks very picky of me.

    If I didn’t miss it and you’ve just added the “Or maybe both” piece then I’m looking very picky through no fault of my own at best, or a right **** at worst.

    Fair enough, not the first time that either of these things will have been thought about me, but I’m usually quite capable of getting this sort of reaction on my own, I don’t need any help.

    Just saying.

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