Birthday thread.

[Pando]

Seriously though, with the current amount of active members the test won't have enough people in it

That is my theory to prove. I want to know how many members it would take to have one on each day. Perhaps it will take years to happen, perhaps it never will.

and the mathematic formula would get ugly as some days can be considered more popular than others.

I don't get your point?

Recipe for the world's biggest hangover surely.

You only get a hangover after you actually stop drinking.

[Rockefella]

You only get a hangover after you actually stop drinking.

Right, forgive me for my stupidity.

[Matra et Alpine]

All we need is a mathematic formula and this test.

Back to my University statistics classes

You don't try to work out the probability of 2 HAVING the same day, you work out the odds of them NOT having the same day

P(event happens) + P(event doesn't happen) = 1
P(two people share birthday) + P(no two people share birthday) = 1
P(two people share birthday) = 1 - P(no two people share birthday).

First person can have any birthday. The second person's birthday has to be different. There are 364 different days to choose from, so the chance that two people have different birthdays is 364/365. That leaves 363 birthdays out of 365 open for the third person.

The probability that both the second person and the third person will have different birthday is:
(365/365) * (364/365) * (363/365) = 132132/133225,
which is about 99.18%.
Four people with all have different birthdays is:
(364/365) * (363/365) * (362/365) = 47831784/ 48627125,
or about 98.36%.

A formula for the probability that n people have different birthdays is

(365_P_n)/(365^n)

which is

((365-1)/365) * ((365-2)/365) * ((365-3)/365) * . . . * ((365-n+1)/365).

or

365! / ((365-n)! * 365^n).

So you keep iterating until you get the probability you want

ie 23 people gets you at 50%

[Waugh-terfall]

Bring in the Pi

[Matra et Alpine]

I don't get your point?

aha a birthday on a given day is something else.
THe population are not equally spread over 365 days.
Some days are more populated with newborn than others.
THere are many factors including cultural - Glasgow holidays uised to always be 2nd week in June .... guess when lots of births occurred ? yep, 9 months later
So you have a skewed population to select from and that alters the probability of anyone being on a SPECIFIC day greatly and very complex to determine. You need population distribution curves of your sample population. You need to define your sample population first

[twinspark]

I don't get your point?

If we had the same birth probabilities for every day, it would be possible to calculate the amount of people needed to get one for every day with, for example, 95% probability. But as there tends to be more people born approximately nine months after local celebration days (those obviously vary culturally), it wouldn't be accurate.

Edit: too slow...

[Pando]

Yay, found the formula!

How many people do you need to meet, in order to meet a person for each possible birthday? Intuitively, this number will be much higher than 365, but how much? (This problem is a special case of the Coupon collector problem.)

After having met people with n different birthdays (n < 365), the chance that the next person you meet has a colliding birthday is (365-n)/365. If the birthdays are distributed uniformly, this means that you will have to meet 365/(365-n) people on average to find one with the a new birthday. This means that, on average and if you meet a person per day, it will take a year to find the person with the last missing birthday.

The total number of people to meet is then , or 2364.64 people.

Which means it's doable!

[Pando]

aha a birthday on a given day is something else.
THe population are not equally spread over 365 days.
Some days are more populated with newborn than others.
THere are many factors including cultural - Glasgow holidays uised to always be 2nd week in June .... guess when lots of births occurred ? yep, 9 months later
So you have a skewed population to select from and that alters the probability of anyone being on a SPECIFIC day greatly and very complex to determine. You need population distribution curves of your sample population. You need to define your sample population first

If we had the same birth probabilities for every day, it would be possible to calculate the amount of people needed to get one for every day with, for example, 95% probability. But as there tends to be more people born approximately nine months after local celebration days (those obviously vary culturally), it wouldn't be accurate.

Edit: too slow...

Yes, I see your points and I've though about it myself, but just to keep it simple, lets keep all those factors out. In tests like these there is a factor evens out those popular days as well, when there's only a few days left people are prone to start "looking" for those and oversee other dates.

[Matra et Alpine]

Yay, found the formula!
Which means it's doable!

No
Because

If the birthdays are distributed uniformly

is not feasible in the real world unless you control the population.

[Pando]

No
Because is not feasible in the real world unless you control the population.

Yes, of course but with over 3000 active members, and new ones every day - it is doable. Especially since we have members from all over the world with different popular/unpopular birthdays. I'm optimistic.

[Matra et Alpine]

Yes, of course but with over 3000 active members, and new ones every day - it is doable. Especially since we have members from all over the world with different popular/unpopular birthdays. I'm optimistic.

"do-able" ?
It isn't, not to any reasonable degree of accuracy.
Probability is heavily driven by population.
One simple case -- Feb 29th !!! Ouch
3000 is not a large population when there are 366 possible outcomes and a skewed distribution.
I woudl guesstimate that to within 25% is reasonable and might even be 10% significance. Just an unknown at the moment.

[Pando]

"do-able" ?
It isn't, not to any reasonable degree of accuracy.
Probability is heavily driven by population.
One simple case -- Feb 29th !!! Ouch
3000 is not a large population when there are 366 possible outcomes and a skewed distribution.
I woudl guesstimate that to within 25% is reasonable and might even be 10% significance. Just an unknown at the moment.

I think we can find a person for each day, although that will take time. Accuracy? It is/will be just a slightly manipulated test, which isn't completed until it is (no in betweens). I'm not trying to make a bet that there is one of each birthday among the 3000 active members, all I'm saying is that it will get us a good way. We can search for the last handful if we want.

February 29th will be a bonus level.

[IWantAnAudiRS6]

Who cares? Quit nay-saying, sit back, and enjoy! It's like a census of members, I guess. Check out the triplets, twins, the first 3-in-a-row (that's Fred, me and Tom). Why does everything have to have some smart-arse 'end' to it?

By the way, Matra, I already added yours in- read the thread first...

[IWantAnAudiRS6]

Now Sam, I'm still waiting for a reply to this post:

Should we start putting links to the profiles as well? [Little voice]Yes, let's[/Little voice]

Eh, just search them in the member's thread! I think doing that would put me over the character limit for a post- otherwise, I would.

[Sweeney921]

September 21