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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

If there are 30 kids in a class, what's the odd's of

thanks.

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

Just edited it to make it clearer

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## Comments

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.

26 people is the threshold?

Go on then Stu.

for me it's

1/365th X 2 divided by 30.

So it's unlikely statistically, although it's not uncommon in practise as the dates of conceiving aren't randomly distributed.

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.

Lol- i was waiting for that.

Sponsored links: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.

And the probability that birthdays are not spread evenly across the year, you usually get peaks in the birth rate during the year.

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 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

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

Anyway I prefer Shag's answers!

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