me and cheesepuffly have the same birthday haha, 16th october.
having recently turned 19, i feel absolutely no difference which i tell all the elder family members who ask that question. and i spent the evening making a nice new weapon for the scions hehe
all in all, it was a pretty pants day... im sure birthdays get worse as you get older
You posted about this? Heheh. Huh, interesting, yours was probably better than my 15th birthday, i mean i got kewl stuff, and a cheese cake, but nothing really else.
oh by the way? Chances is 1/364
Actually, that's incorrect. (http://en.wikipedia.org/wiki/Birthday_paradox)
To compute the approximate probability that on a forum of n members, at least two have the same birthday, we disregard variations in the distribution, such as leap years, twins, seasonal or weekday variations, and assume that the 365 possible birthdays are equally likely. Real-life birthday distributions are not uniform since not all dates are equally likely.
It is easier to first calculate the probability p'(n) that all n birthdays are different. (p' being used here instead of a p with a line over it, which is the opposite of p.) If n > 365, by the pigeonhole principle this probability is 0. On the other hand, if n ≤ 365, it is
p'(n) = 1 * (1 - (1/365)) * (1 - (2/365)) * ... * (1 - ((n-1)/365)) = (365!)/((365^n)((365-n)!))
because the second person cannot have the same birthday as the first (364/365), the third cannot have the same birthday as the first two (363/365), etc. This can be represented as a summation but I am having enough trouble writing this on a forum already!
The event of at least two of the n persons having the same birthday is complementary to all n birthdays being different. Therefore, its probability p(n) is
p(n) = 1 - p'(n) (as I said, opposite of p is p')
This board has 387 total members, so that is our n.
First, the probability that no one has the same birthday of 387 members is:
(365!)/((365^387)((365-387)!)) = (365!)/((365^387)((-22)!)) = ERROR
Note that (-22)! is impossible, thus this goes over 100% probability that two people will share the same birthday.
If we want to say the probability that you, and specifically you, and some other have same birthday, that is:
1 - ((365 - 1)/365)^n, and thus: 1 - (364/365)^387 = 0.65414271530895277770031243411142
OR: 65.414271530895277770031243411142%
That might seem inflated but it is influenced by those other than you with the same birthdays.
Statistics are cray aren't they?
EDIT:
To get the probability of specifically you two specifically in this group, we get the probability of selecting you two:
(1/387) * (1/386) = 0.0025906735751295336787564766839378, or 0.25906735751295336787564766839378 %
Multiplying that by the chance of any two having the same birthday, which in a group this large is nearly 100%, you get the same thing, 0.25906735751295336787564766839378%
Additionally, the chances of two specific people on this forum having the same birthday (last thing I said) on a specific day (16th of October), is:
(above) * (1/365) or 0.0025906735751295336787564766839378 * 0.002739726027397260273972602739726 = 7.0977358222726950102917169378451e-6
OR .0000070977358222726950102917169378451 OR .00070977358222726950102917169378451% OR .0007%
Have a nice day!
..................................................................................................wow.
How much of a social life do you have n1?
Quote from: bigbadbogie on October 27, 2008, 03:47:32 AM
..................................................................................................wow.
Quote from: Steeveeo on October 27, 2008, 03:49:04 AM
How much of a social life do you have n1?
My thoughts exactly... I can't understand where people can find the time to figure all of this stuff out and post it all, but its interesting nonetheless. I've always wondered about this though, so thank you for that very, very,
very, specific percentage. If only I could figure out how to work it in other places (giving or taking more people).
This is just a case of the Birthday Paradox - see http://en.wikipedia.org/wiki/Birthday_paradox (http://en.wikipedia.org/wiki/Birthday_paradox) . Once you have a group of 23 people, odds are 50/50 that two of those people share the same birthday. Now, calculate the odds of three people on that day.
-- GSH
I used the Wikipedia page I linked for everything before the edit. The rest was from high school statistics.
And here I was thinking that you had 10 sheets of paper and a calculator. *Smacks own forehead*.
No, BBB, I had to really do the math. No paper though.
lol. interesting...
GAH! Learning! IT HURTS!
Actually, with what i was doing last year, it almost makes sense.
I wasnt too far off lol.