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The Birthday Problem

  • 05-05-2008 6:27pm
    #1
    IrishKnight
    Registered Users, Registered Users 2 Posts: 922 ✭✭✭


    I'm working with the ol' birthday problem.

    I know that, P(at least two people share the same bday)
    =1-P(no one shares the same bday)

    Now, for n people would that make it...

    1-(365-P-n)/365^n

    Where -P- means permutation.

    Am I right or way off?


Comments

  • #2
    Michael Collins
    Moderators, Science, Health & Environment Moderators Posts: 1,852 Mod ✭✭✭✭Michael Collins


    That is spot on. So the probability that

    P(out of 23 people at least two have the same birthday) = 1 - (365P23)/365^23 ~= 50.7 %, which is true!


  • #3
    IrishKnight
    Registered Users, Registered Users 2 Posts: 922 ✭✭✭IrishKnight


    Coolio, that's going to save a hell of alot of time in my exam.


  • #4
    Michael Collins
    Moderators, Science, Health & Environment Moderators Posts: 1,852 Mod ✭✭✭✭Michael Collins


    No bother. It's an interesting question really, people often find it very odd that if you have 23 people in a room, it's more likely that at least two have the same birthday than not. With 60 people it's almost a sure thing, at 99.4%.

    Be careful with computations using that formula though as for a large number of people a lot of standard scientific calculators will have trouble since the intermediate values are so large.


  • #5
    IrishKnight
    Registered Users, Registered Users 2 Posts: 922 ✭✭✭IrishKnight


    Michael Collins wrote: »
    people often find it very odd that if you have 23 people in a room, it's more likely that at least two have the same birthday than not. With 60 people it's almost a sure thing, at 99.4%.

    Somewhat like the Monty Hall Problem, tis hard to think that by changing your mind you have a better chance at wining that wonderful new car!


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