Birthday Minimum to cover all 365 days...

Aquafire

A Birthday Puzzle of sorts...

What is the minimum number of people I would have to have gathered in the one place to guarantee with 100% certainty, that no matter what day of the year I call out (from Jan 1st to Dec 31 in any random order)...it will be the birthday of at ONE person in that group?

Does anyone have a definite minimum (calculated number) in mind ?
Is such a number even calculable...?

If so, is it possible to calculate such a figure using the "Birthday Paradox" asa the basis for an answer

http://en.wikipedia.org/wiki/Birthday_paradox

Respectfully

Aquafire

Ps: and no this is NOT homework...I am too olde for that sort of caper.

CRGreathouse

Assuming you mean "at least one person in that group":
If you can choose people for their birthdays, 366 would suffice (choose one with each birthday). If not, the number is W-A+1, where W is the world population and A is the number of people in the world with an February 29th birthday.

d_leet

Why April 29th? Don't you mean February?

CRGreathouse

Bah, yes of course. It's clearly too late for my brain to function.

d_leet

Yes I know how that is, I'm pretty close to that point myself right now.

CRGreathouse

Yeah. Back pain has kept me up all night, ugh.

Aquafire

366...?

Perhaps I am not being clear enough...

How can I guarantee that any day I call out of the 365 days available.. (and called out in any random order) will be covered by having one person in that group put their hand to say it is their birth date...?

(irrespective of which actual year)

Remember, I have only 365 potential days to call...One per each calander day and each call is irrepeatable.

So how many people as a minimum do I need to cover with 100% surity every single day of the 365 days available and called...to make sure one hand matches the day as it is called out?

LENIN

If you selected the people corectly (none have birthdays on the same day) then one for each day would be enogh. If you selected people for your group at rendom then I don't think that you can ever be 100% certine (no mether how large the group is).

matt grime

It is prefectly possible for everyone to have the same birthday i.e. there is no guarantee in that respect. Of course, if you knew the precise distribution of birthdays for everyone in the world you could work it out (it is possibly impossible. It might well be that no one in the world was born on the 3rd of May. Of course it is extremely unlikely, and bound to be false, but there is still a non-zero chance of that being the case).

shmoe

If the answers are still troubling you, think of a smaller problem. If you have 10 red marbles and 10 blue marbles in a bag, how many do you need to pull out to guarantee a red and a blue?

The "366" answer is analagous to you being allowed to look in the bag. You need only 2 picks then.

The massive "all the people in the world less the number whose birthday is on Feb. 29th plus 1" will correspond to the situation where you pull out a bunch of marbles in a dark room, then turn on the lights and examine the colours you've picked. In this case you need more than 10, to ensure you don't draw all red or all blue.

Not being allowed to look at the marbles is the more interesting problem, and I think closer to the version you intended. So what if we now had 10 reds and 10000 blues? You'd need to have pulled out more than 10000 to make sure that hadn't just been unlucky (or lucky depending on your view) and picked all blues.

Now add in some greens, and see how this affects the answer.

DaveC426913

Aquafire:

Since there is no claim that the group of people were hand-picked...

The number of people you need,
to guarantee with 100% certainty,
that no matter what date you call out,
will have someone born on that date,
is:

infinite.

DaveC426913

This has no correspondence whatsoever to the puzzle being asked. Bzzt.

shmoe

If you're taking the assumption that peoples birthdays are randomly distributed, then sure there's no way to guarantee anything.

If you're drawing people from planet earth, then you can do it with the pretty reasonable assumption that every possible day has a representative and assuming that you have a large enough "place" to gather the required number of people in. The minimum required depends on which day has the least number of people, that it's feb 29th is a reasonable assumption as well. You can determine this exactly if you are very fast and run around the world asking peoples birthdays, make sure to account for people who die and are born while you are conducting this survey.

shmoe

Sure it does. Reduced to 2 possible birthdays (or classify them as jan-june, july-dec) and I've given an explicit distribution of a small population.

matt grime

Yes it does. As would an analogy with socks in a drawer.

Since there are not infinitely many people in the world, I wouldn't be too sure of your reasoning.

shmoe

You could also do t-shirts from a closet.

I'd say we've beaten to death the "guaranteed" part of the assumption? Let's remove 'guarantee':

Assume we have an infinite population and their birthdays are uniformly distributed amongst 365 days (ignore leap years). Let P(N) be the probability that if we select N people at random, we have a person on each and every birthday. eg. P(1)=P(2)=...=P(364)=0. What's P(365)? How to find P(N) for N>365? What value of N is needed to get P(N)>0.9 or other threshold?

Can P(N) be approximated easily? An exact answer isn't too bad to get, but the one I have in mind has many terms, and you could approximate well with quite a few less.

CRGreathouse

Exactly. Just for kicks, here are some numbers based on the US Census data for world population (September estimates):

Estimated world population: 6,541,161,782
Estimated number of Feb 29 birthdays: 4,477,181
Estimated people needed: 6,536,684,602