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]

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]

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]

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

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

 

[d_leet]

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]

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]

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]

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]

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]

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]

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]

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]

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 http://www.census.gov/ipc/www/popclockworld.html (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]

Goodness me,

I didn't realize 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 separate 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 guarantee 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..

 

[shmoe]

Goodness me,

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

What part of the responses don't you understand? We have answered:

1. Drawing from the real population of the earth. Total number of people - # of birthdays on feb 29th +1 (this is under the pretty reasonable assumptions that feb 29th is the least frequent and all birthdays are indeed represented)

2. Drawing from some unknown population where birthdays are random- There is NO amount of people you can select to GUARANTEE you hit every date. No matter how many people you select, you cannot guarantee you don't miss some date. You can't guarantee ANYTHING except that you will have one day represented, it will still be possible for 10^1000 people to land on the same day (assuming large enough population, uniform choice of b-days, etc). Even if you somehow allowed infinite number of people selected, the event of missing days is not impossible (though will have probability zero). Maybe you have a different meaning of "Guarantee", but this is how I, and everyone else who has replied, has interpreted it.

3. I have proposed you consider P(N) that I described earlier. This is what you should ask if you want to be able to answer how "remote" the probability of missing days is if you select N people. I really think this is what you are interested in and should abandon this "guarantee" requirement (though we have answered your version)

Now, do you understand what I've said above? If not, where are your issues?

 

[shmoe]

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 separate 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..

"Remote" is still non-zero, and not a "guarantee" that this won't happen. I strongly urge you to consider my #3 suggestion. Then you can ask what kind of threshold you want "remote" to mean, e.g. how large must N be to have P(N)>0.9999?

 

[DaveC426913]

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.

You are misunderstanding the problem being asked.

From a bag with 2 colours of marbles in it, you are guaranteed that the bag actually contains both colours. You MUST eventually draw one of each colour. (in your example of 10 read and 10 blue, by the 11th marble, you MUST have one of each.)

As to birthdays, you could have 20 people or 20 million people. There is no guarantee whatsoever, that if you call out "March 13th" anyone in that population will have that as a birthday.

 

[Aquafire]

What part of the responses don't you understand? We have answered:

2. Drawing from some unknown population where birthdays are random- There is NO amount of people you can select to GUARANTEE you hit every date. No matter how many people you select, you cannot guarantee you don't miss some date. You can't guarantee ANYTHING except that you will have one day represented, it will still be possible for 10^1000 people to land on the same day (assuming large enough population, uniform choice of b-days, etc). Even if you somehow allowed infinite number of people selected, the event of missing days is not impossible (though will have probability zero). Maybe you have a different meaning of "Guarantee", but this is how I, and everyone else who has replied, has interpreted it.

3. I have proposed you consider P(N) that I described earlier. This is what you should ask if you want to be able to answer how "remote" the probability of missing days is if you select N people. I really think this is what you are interested in and should abandon this "guarantee" requirement (though we have answered your version)

Now, do you understand what I've said above? If not, where are your issues?

Thankyou Shmoe..

I did not mean for you to be part of the "difficult to understand group"..

I meant it in reference to the other answers.

What lies at the kernel of my issue is the fact that seemingly no-one can offer a 100% guarantee that all the days of the year will have a hand raised.

It seems strange that we can create a formula that may offer 90% accuracy or even 99.9999999% accuracy..but never 100%

In effect even if I extrapolate the latter percentage as 999,999999 people put them all together...there is a still a remote chance that one calander day will not be represented...

So is there a figure that could ever offer 100%...?

It seems not...

 

[shmoe]

You are misunderstanding the problem being asked.

From a bag with 2 colours of marbles in it, you are guaranteed that the bag actually contains both colours. You MUST eventually draw one of each colour. (in your example of 10 read and 10 blue, by the 11th marble, you MUST have one of each.)

As to birthdays, you could have 20 people or 20 million people. There is no guarantee whatsoever, that if you call out "March 13th" anyone in that population will have that as a birthday.

Please, it was an example of a specific distribution that corresponds to the real world's population and reflects the assumptions in the answers that had been given before my post, but simplified to have less birthdays and less people. Does anyone here believe that there are birthday's on the Earth that are missed?

 

[shmoe]

It seems strange that we can create a formula that may offer 90% accuracy or even 99.9999999% accuracy..but never 100%

With a totally random population, no matter how many people you pick, there is always a non-zero probability that you have missed Aug. 26th (or any other day).

Anyway, have you tried to find an exact formula?

The other responses were fine, you hadn't specified all your assumptions (random population, how you are picking the people, etc) and they did outline the assumptions they made.

 

[daveb]

Actually, you can guarantee it, because it is a fact (though not a mathematical fact) that eventually you can find a person born on each of the 366 days (365 plus Feb 29). The problem with answering your question is that you would need to know beforehand exactly how many people are born on each day. You calculate the minimum to guarantee the "calling out dates" scenario as follows: What is the number of people born on the date that has the least number of people born that day (probably Feb 29)? Then the minimum number of people you need is the population of earth, minus the number of those people, plus 1.

Now mathematically, we cannot guarantee with 100% certainty that a person was born on every day. We can statistically guarantee with 100% certainty that this is the case since there are records out there that track birth dates.

 

[shmoe]

Have we not yet beaten to death the difference between taking your population to be the Earth's and taking your population to being some infinitly large one (or at least "sufficiently large") with some kind of distribution on the birthdays? The exact circumstances you propose were dealt with in the second response to this thread.

Another question that Aquafire might want to consider, under the random population assumption, what is the expected number of people you have to select to get one from each day? This might end up nasty, I haven't tried to work it out.

 

[Aquafire]

Another question that Aquafire might want to consider, under the random population assumption, what is the expected number of people you have to select to get one from each day? This might end up nasty, I haven't tried to work it out.

Nor have I schmoe.

Although I have tinkered with the answer since you posted the above...I have to say I am probably not bright enough to come up with a satistactory answer to your sub~ text of my original question..

That said, from my rudimantary mathematical knowledge...I am inclined to believe one needs to multiply...365x364x363x362x361x360x359 and on...till ~ 2x1......

But in all honesty Schmoe...I have no other probable answer...

As I keep reminding you all, (here I am not embarrassed to say I want to learn (but the reality is...I genuinely don't have the answer...)

That's why I thought to pass it over to posters like you Shmoe, who clearly have a greater mathematical grasp than me..

My job/pleasure (perhaps some will find it annoying) ...is just to ask questions that are (at least for me_ baffling and perplexing.

Beyond that I want to learn what I can..

I hope I won't be condemmed for such honesty..

Respectfully

Aquafire

 

[shmoe]

The question of P(N) isn't so bad. Under the assumption the birthdays are uniformly distributed and independant of one another, you can use the principle of inclusion/exclusion. Google will get plenty of hits for inclusion/exclusion, many will be phrased in terms of counting problems but there's really not much difference. We'll get:

$$P(N)=\sum_{i=0}^{365}(-1)^{i}\binom{365}{i}\left(1-\frac{i}{365}\right)^N$$

If you want to find some exact values here, the computer is the way to go, and entering it as:

$$P(N)=\frac{-1}{365^N}\sum_{i=0}^{365}(-1)^{i}\binom{365}{i}i^N$$

will end up substantially faster (though none of what's below will take very long either way).

Some values:

P(500)=9.8...*10^(-70)
P(1000)=1.7...*10^(-17)
P(1500)=0.001978...
P(2000)=0.2161...
P(2500)=0.6804...
P(3000)=0.9072...
P(4000)=0.9756...
P(4500)= 0.9984...
P(5000)=0.99959...
P(10000)=0.999999999555...

You could also approximate this with the first few terms of the first sum I gave (Or the last few terms of the second sum, they are reversed). How many terms you need depends on how large N is and how many decimals you want.

 

[gnomedt]

See the "collectors" problem.

Basically, the problem is that if you're a toy collector, and you want to collect all possible 10 Burger King toys, on average how many times do you have to eat at Burger King? What is the probability, after n runs, that only m toys are missing?

You're collecting people and categorizing them into "toys" according to their birthdays, so it's the same problem.