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Oxbridge interview question (combinatorics/probability) watch

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    Here's a question from the oxbridge-admissions report:

    If we have 25 people, what is the likelyhood that at least one of them is born in each month of the year?

    My answer: (this is not a complete answer, just gives some idea of what I did and my final probability)
    Spoiler:
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    First I considered the binomial distribution and a few other things, but eventually I sort of gave up. I defined a function f(m, p) to be the probability of success with m out of 12 months still without any people in them and p people who do not have months assigned to them (i.e. whose months are undetermined).
    I then created a recursive formulation of f...
    Spoiler:
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    with each of these cases handled seperately:
    m = 0 (also p < m as an optimisation)
    p = m
    p > m

    I created a Scheme program to calculate the value of f(12, 25). It took a minute to run in the interpreter, and gave me an exact answer.


    My answer was 0.1819 (to 4 dp) but I could never have done it without a computer. Any insights?
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    P(one in jan) = 1 - (11/12)^25
    P(another in feb) = 1 - (11/12)^24
    ...
    P(one in dec too) = 1 - (11/12)^14

    P = [1-(11/12)^25][1 - (11/12)^24]...[1 - (11/12)^14]

    Would that work? A little simpler, still don't think it's easily solvable without a calculator.
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    I make it 9.
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    O_o
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    (Original post by Speleo)
    P(one in jan) = 1 - (11/12)^25
    P(another in feb) = 1 - (11/12)^24
    ...
    P(one in dec too) = 1 - (11/12)^14

    P = [1-(11/12)^25][1 - (11/12)^24]...[1 - (11/12)^14]

    Would that work? A little simpler, still don't think it's easily solvable without a calculator.
    That's a much smaller probability than my one. It looks like you don't consider the 2nd person going into the same month as the 1st, but the next 23 people sorting it out.

    Master621: that isn't a probability
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    I will take your word for it, I am not very good at probability

    On second thoughts it does look a little off.
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    Excel (Monte Carlo method) confirms my result. The experimental result already matches my calculated one to the fourth decimal place and as I add more trials the error is drifting down. However I still feel as though I've missed something.
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    I think I get (24C11)/(36C11) which seems less than your answer.

    I got this thinking of allowed possibilities

    JJJFFMAAAM...D

    as choosing where to put 11 jumps into 24 gaps

    and all possibilities - such as

    JJJJMAAAAAAM....N

    as

    ****||*|******|*...*|

    where there are 11 jumps (|) in amongst 25 months and they can be repeated.

    My method seems to give the right answer for small cases.
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    If you wanted to work it out by hand, you could do 1- P(one month missing) - (two months missing) -... - (11 months missing ie all born in the same month)
    = 1 - 12C1*(11/12)^25 - 12C2*(10/12)^25 - ... - 12C11*(1/12)^25. I'm pretty sure this is the right answer though it might take a while to work out!
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    (Original post by dijon199)
    If you wanted to work it out by hand, you could do 1- P(one month missing) - (two months missing) -... - (11 months missing ie all born in the same month)
    = 1 - 12C1*(11/12)^25 - 12C2*(10/12)^25 - ... - 12C11*(1/12)^25. I'm pretty sure this is the right answer though it might take a while to work out!
    Think about the bold bit. The 12C1 = 12 is for the twelve months. So
    (11/12)^25 needs to be the probability that the 25 are born in Jan-Nov say - which it is - but with no guarantee that all 11 months are included.

    So not convinced you can go down this route.
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    dijon199: The problem is that your "one month missing" includes many from "two months missing" and "three months missing" etc...

    I found out the answer (a math master at my school gave me an answer which works out exactly to my previous one, but with far less calculation).

    Exact and complete formula below.
    Spoiler:
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    \Sigma_{n=0}^{12} (-1)^n \cdot C^{12}_n \cdot (\frac{n}{12})^{25}
    Note: C^n_r is notation for "n choose r" or "nCr"


    I'm going to investigate this one. Maybe I should read some online stuff on combinatorics, as it seems to be a favourite of Oxbridge admissions.
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    (Original post by RichE)
    Think about the bold bit. The 12C1 = 12 is for the twelve months. So
    (11/12)^25 needs to be the probability that the 25 are born in Jan-Nov say - which it is - but with no guarantee that all 11 months are included.

    So not convinced you can go down this route.
    yes, you're right (oops!). I'll have a think about how to get that answer you gave - it looks like the type of answer where you're starting off with a large number, then subtracting, then adding, etc...until you get the answer ie you're getting closer and closer to the answer with each term. But I could be completely wrong! So the 12th term would be the largest and 1-term the smallest as the 0-term is zero.
 
 
 

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