Birthday Puzzle

[badmephisto]

Going in the spirit of the previous puzzle thread, I was going to post a puzzle which is both interesting to me mathematically, and it also comes up with an answer that is non-trivial and cool. (Many people don't actually even believe the answer when I tell them). Here it is:

There are 30 students in a class, each of them is born in the same year. What is the probability that there is at least one pair of students that have the same birthday?

 

[shelley]

Similar to this one is the one that asks how big a group of people has to be before the probability of two people in the group sharing a birthday is greater than 1/2.

 

[immortalcube]

highlight to read ---> it's actually easier (for me at least) to figure out the probability of nobody having the same birthday as anyone else, and then subtract that from 100%. So to figure that out let's start small. What about with two people? the first birthday is automatically the 'right' one so that probability is 365/365. the probability that the second person will not have the same birthday is (364/365)*(365/365) which equals 364/365, or .997. So the probability that they will have the same birthday is .003 (.3 %). For three people, (363/365)*(364/365)*(365/365) = .992, and conversely the probability of two of them having the same birthday is .008 (.8 %). So we can sum this up as (365 * 364 * 363 --> all the way to the 30th number down from 365)/ 365^30 = .2936837573. So the probability that two of them have the same birthday is 1 - ANS, which equals .7063162427 ~= 71%. As for the second answer, after working on it for a bit I have come up with 23, which is a .507 or 50.7% chance.
God I hope I did that math right
EDIT: (highlight) I just realized that it might've been more accurate had I used 366 days (because we don't know for sure that the unspecified year isn't a leap year). However, that shouldn't throw off my calculations by very much (by almost nothing, in fact).

 

[Karthik]

Highlight to see my answer:
Required Probability= 1-P(No two people have the same bithday)
Assuming the year is non-leap.
Now P(No two people have the same birthday)=
364/365*363/365*362/365*....*336/365
So Required Prob=1-(364*363*362*....*336)/(365^29)
The number is pretty unexpected.

 

[badmephisto]

i also tried a different approach today but i cant seem to be able to really solve it that way. I tried doing 30C2 to choose a single pair of people out of all 30 people, and then multiply that by probability that the two share a birthday. so it would be 30C2 * (1/365) = ~1.19 . But obviously thats wrong answer because I think (1/365) is oversimplified. It always assumes one person to automatically have the correct birthday, whatever that is, but that is not correct because to every person there is one single unique birthday. So it should be more complicated, but I cant really think of a simple way to fix this. Can anyone think of a way to do it this way?

 

[Brian Le]

I just took a test on probabilility...

 

[Robert]

This is known as the birthday paradox, I had a class on this.

Answer:

Spoiler

70%

Explanation:

Spoiler

You gotta do the probability of a pair having the same birthday, not somebody having the same birthday of one specific person, so in a 30 class room there are more than 250 pairs. First calculate the probability of the birthdays being different:
(Weird line means its a fraccion)


365!
------------------
365^30(365-30)!

So the chance is 30%

The answer to this is 70%