Birthday Minimum to cover all 365 days...


by Aquafire
Tags: birthday, days, minimum
Aquafire
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#1
Aug25-06, 12:34 AM
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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.
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CRGreathouse
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Aug25-06, 12:37 AM
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Quote Quote by Aquafire
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?
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
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#3
Aug25-06, 12:56 AM
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Quote Quote by 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 April 29th birthday.
Why April 29th? Don't you mean February?

CRGreathouse
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Aug25-06, 02:33 AM
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Birthday Minimum to cover all 365 days...


Quote Quote by d_leet
Why April 29th? Don't you mean February?
Bah, yes of course. It's clearly too late for my brain to function.
d_leet
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#5
Aug25-06, 02:35 AM
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Quote Quote by CRGreathouse
Bah, yes of course. It's clearly too late for my brain to function.
Yes I know how that is, I'm pretty close to that point myself right now.
CRGreathouse
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Aug25-06, 02:37 AM
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Quote Quote by d_leet
Yes I know how that is, I'm pretty close to that point myself right now.
Yeah. Back pain has kept me up all night, ugh.
Aquafire
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#7
Aug25-06, 04:10 AM
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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
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#8
Aug25-06, 05:51 AM
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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
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Aug25-06, 06:16 AM
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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
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Aug25-06, 07:56 AM
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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
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#11
Aug25-06, 11:20 AM
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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
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#12
Aug25-06, 11:25 AM
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Quote Quote by 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?
This has no correspondence whatsoever to the puzzle being asked. Bzzt.
shmoe
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Aug25-06, 11:30 AM
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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
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Aug25-06, 11:32 AM
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Quote Quote by DaveC426913
This has no correspondence to the puzzle being asked. Bzzt.
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
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Aug25-06, 11:52 AM
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Quote Quote by DaveC426913
This has no correspondence whatsoever to the puzzle being asked. Bzzt.
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
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Aug25-06, 01:35 PM
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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
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Aug25-06, 02:00 PM
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Quote Quote by shmoe
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.
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
Aquafire
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#18
Aug25-06, 08:14 PM
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Goodness me,

I didn't realise this problem was going to be so difficult for some of you to understand.

The point is I DON'T KNOW the birthdates of any of these people ahead of time.

The Idea is to congregate a large mass of strangers...whose 'birthdays' I haven't the foggiest clue about.

In otherwords..I have no preknowledge of their birthdates...so the crowd is utterly random...

And from there...I start calling out dates of the year...in random order...or even in straight calendrical sequence...or backwards...or standing on one leg....it really isn't important.

What is important...is to find out what is the minimal numbers of people...I would have to have in my set to guarantee with 100% certainty that a tleast one person will be present for every calendar birthday...called out. IE 365 calls..an one hand goes up for every one of those 365...days..

Now the chances of gathering 365 strangers together into one place and finding that not a single person in that group shares the same birthday date of the year...Ie 365 seperate birthdays...without knowing their birthdays ahead of time.....would have to be pretty remote.

Likewise, the chance of all 365 strangers...(remember I have no way of knowing ahead of time what their birthdate actually is) having the exact same birthday...lets say 4th of July...seems extremely remote..

So again...I ask with some trepidation....

What is the minimal number required...to guarentee a person will be present for everyone of the 365 potential days of the year that match someones birthdate...

Again I emphasise I don't have anyway of knowing ahead of time...what the birthday is of any single stranger...

Thanks guys..


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