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Any budding Carol Vorderman's out there.

Help sort out a office debate????????

If there are 30 kids in a class, what's the odd's of any two of them having the same birthday, not necassarily birthdate.

thanks.


At least it's taking my mind off the match tommorrow (ish)


Just edited it to make it clearer

Comments

  • Depends how in depth you want the answer, if you want it based purely on the chances of a child being born on any given day being the same as any other day its quite simple.

    However in reality this is not the case, some days (parts of the year) do have a higher chance of being a birthday than others, it also changes from year to year, old jokes about cold winters etc etc being mainly true.
  • edited April 2007
    just over 50% probably

    26 people is the threshold?
  • From what I remember, it's fairly high, something like 14/1, as long as it doesn't have to be a specific date or 2 specific children.
  • edited April 2007
    [cite]Posted By: Stu of SE7[/cite]Depends how in depth you want the answer, if you want it based purely on the chances of a child being born on any given day being the same as any other day its quite simple.

    Go on then Stu.

    for me it's

    1/365th X 2 divided by 30.
  • edited April 2007
    Using a binomial distribution, I got 0.003. (when everyone in the class is born in the same year with an even spread of dates.)

    So it's unlikely statistically, although it's not uncommon in practise as the dates of conceiving aren't randomly distributed.
  • I reckon the odds of no-one having the same birthday = (365-29)/365 * (365-28)/365 * (365-27)/365 …… * (365-2)/365 * (365-1)/365 * 365/365

    Feel free to work it out - ought to come to somewhere round about 0.45 !!!

    then the odds that two or more do is 1 minus that.
  • No one is 0.92
  • What about leap years?
  • I reckon its the twins
  • [cite]Posted By: Shag[/cite]I reckon its the twins

    Lol- i was waiting for that.
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  • [quote][cite]Posted By: Medders[/cite]What about leap years?[/quote]best ignored!

    Found this:
    the total combination of birthdays is 365 x 365 x 365 x....(30 times), or more succinctly 36530. (Again we ignore leap years.) That's our denominator. Without calculating we can quickly see that's a huge number.

    Now, as before, let's calculate the number of possibilities that do NOT provide a matching birthday. The first person states his birthday. For the next person to NOT have the same birthday, she has to pick from 364 days. The next person must choose from 363 days. And so on. Again were trying to calculate the number of combinations in which no one has a matching birthday. For the dice, the calculation was 6 x 5. Here we do the same thing:

    365 x 364 x 363 x 362 x ... x 339 x 338 x 337 x 336

    We can write this more concisely as follows:

    365 x 364 x 363 .... (365 - n+1)

    where n equals the number of kids in the classroom, in this case 30. Calculate and you have our numerator.

    Put the two together to find the odds of NOT finding two kids with the same birthday in a group of 30 kids:
    (365 x 364 x 363 ... x 337 x 336) / 365 to the power 30 = 0.292

    There's about a 30% chance that you will NOT find two kids who share the same birthday in a group of 30. Therefore the opposite situation, that you WILL find two kids with the same birthday in a group of 30, is a whopping 70% (1 - .3 = .7). In fact, we find that in a group of 23 kids, your odds are better than 50% to find two people with the same birthday.

    And that's the Birthday Paradox. It doesn't seem possible that the odds should be so good to find two people with the same birthday in such a relatively small group. But they are. Remember, these are not the odds of finding someone with the same birthday as YOU in a group of 30 people. These are the odds of finding ANY two people out of 30 who share a birthday.
  • What LTGTR wants to know is have they all got their blackburn tickets yet ?
  • [cite]Posted By: Medders[/cite]What about leap years?

    And the probability that birthdays are not spread evenly across the year, you usually get peaks in the birth rate during the year.
  • Cap duly dothed SS.
  • it's about 70%. Look up something called 'The Birthday Paradox'. it came up in a cryptography book I was reading a while back.

    The key thing is not to look at the number of children but more to look at the number of different pairs of children. Which in this case is 435.
  • And to Shag for making me laugh to the power of two!
  • [quote][cite]Posted By: Alex Wright[/cite][quote][cite]Posted By: Medders[/cite]What about leap years?[/quote]

    And the probability that birthdays are not spread evenly across the year, you usually get peaks in the birth rate during the year.[/quote]well that would just make it more likely
  • I added it up and came out at like 68% so decided not to post, as I figured that *had* to be wrong, thanks for clearing is up SS, I don't feel like a complete idiot anymore.
  • edited April 2007
    [cite]Posted By: StanmoreAddick[/cite]Cap duly dothed SS.
    that was lifted, but it is the same as I said above anyway ......

    .... well except the actual calculation which I said though would be about .45

    Anyway I prefer Shag's answers!
  • edited April 2007
    Me too :)
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  • 9 months after Diss, there'll be 13 babies sharing a birthday if Shag gets his way....
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Roland Out!