
#19
Aug2506, 09:27 PM

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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 bdays, 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? 



#20
Aug2506, 09:57 PM

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#21
Aug2606, 06:46 AM

P: 15,325

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. 



#22
Aug2606, 07:20 AM

P: 48

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 noone 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... 



#23
Aug2606, 07:32 AM

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#24
Aug2606, 07:50 AM

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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. 



#25
Aug2806, 09:33 AM

P: 927

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. 



#26
Aug2806, 09:55 AM

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Have we not yet beaten to death the difference between taking your population to be the earths 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. 



#27
Aug2906, 03:53 AM

P: 48

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 



#28
Aug2906, 07:24 AM

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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:
[tex]P(N)=\sum_{i=0}^{365}(1)^{i}\binom{365}{i}\left(1\frac{i}{365}\right)^N[/tex] If you want to find some exact values here, the computer is the way to go, and entering it as: [tex]P(N)=\frac{1}{365^N}\sum_{i=0}^{365}(1)^{i}\binom{365}{i}i^N[/tex] 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. 



#29
Sep1006, 10:00 AM

P: 38

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. 


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