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**Au101****Member**- Registered: 2010-12-01
- Posts: 286

Hi, I've been looking at the birthday problem (which is a statistical problem which aims at finding out the how many people you would need in a random group to be certain that two of them shared a birthday. Obviously the vacuous answer is 367, but as it turns out, there is a probability of 99% that two people will share a birthday in a group of just 57 and 50% in a group of just 23 (see: http://en.wikipedia.org/wiki/Birthday_problem)).

Okay, this is fair enough and very interesting, but I was trying to take the principle further. i wanted to find out how many people you'd need to be confident that two of them shared a birthday in the same week and then how many shared a birthday in the same month, then how many shared a birthday in the same two month period. Unfortunately, I got completely confused, so I came over here. I know this is 3 questions in one, but i imagine it's just number shunting.

I was also playing with the numbers for a friend and trying to work out the probability that in a random group of five people, two of them would share a birthday in the same month and, i suppose the final question i really have is: imagine a person has met 500 people in their life whom they've really had the chance to get to know. what is the probability that seven of those 500 share a birthday in the same two month period?

I know the last question is quite difficult, but those numbers aren't entirely plucked out of thin air, I was discussing with my friend the statistical significance of her friends sharing birthdays in broad ranges like this.

--

I've been thinking about this and I've realised I've not worded this at all clearly. Let me try and make this question intelligible:

Question 1: What is the smallest group of randomly selected people required such that the probability that two of them share a birthday within one week of each other is at least 75%?

Question 2: What is the smallest group of randomly selected people required such that the probability that two of them share a birthday within thirty days of each other is at least 75%?

Question 3: What is the smallest group of randomly selected people required such that the probability that seven of them share a birthday within sixty days of each other is at least 75%?

Question 4: In a group of 30 randomly selected people, what is the probability that seven of them will share a birthday within fifty days of each other?

*Last edited by Au101 (2013-07-15 06:18:36)*

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,776

Hi;

If you do not mind 2 or more then it is relatively easy.

These are done using a well known formula.

Exactly 2 is different.

**In mathematics, you don't understand things. You just get used to them.Of course that result can be rigorously obtained, but who cares?Combinatorics is Algebra and Algebra is Combinatorics.**

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**Au101****Member**- Registered: 2010-12-01
- Posts: 286

Thank you bobbym, that's great. I've been playing with the numbers myself and I'm completely with you on 1 & 2 . 3 & 4 are utterly beyond me, though. Someone suggested the formula in the wikipedia article under collision counting as a possible approach to me, but i've no idea, really, what to do with it

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,776

You want to see how to use the formula?

Is his suggestion about 3 and 4?

**In mathematics, you don't understand things. You just get used to them.Of course that result can be rigorously obtained, but who cares?Combinatorics is Algebra and Algebra is Combinatorics.**

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**Au101****Member**- Registered: 2010-12-01
- Posts: 286

I think his suggestion was about 3 & 4, yes.

I was having a lot of problems with 1 & 2, but then someone pointed me in the direction of the formula:

Where *n* is the number of people in the group, *k* is the range of days (so *k* = 7 in the case where I'm trying to calculate the number of people required in the group for me to be certain that two of them have birthdays within a week of each other) and *m* = 365 (the number of days, excluding the 29th of February.)

Using this, I was able to get the same answers as you

I'm completely stuck on 3 & 4, though. I can understand that there isn't an algebraic solution to this problem, but is there a numerical one? The only advice i've been given is to try and make use of that formula, but I wouldn't know what to do with it!

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,776

That formula is for the expected number of collisions as far as I know, not the probability.

Unless I have the wrong formula. Please tell me which one then.

There is a computer simulation possible that could get the probabilty.

**In mathematics, you don't understand things. You just get used to them.Of course that result can be rigorously obtained, but who cares?Combinatorics is Algebra and Algebra is Combinatorics.**

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**Au101****Member**- Registered: 2010-12-01
- Posts: 286

Sorry, to quote:

"The probability that the *k*th integer randomly chosen from [1, *d*] will repeat at least one previous choice equals *q*(*k* − 1; *d*) above. The expected total number of times a selection will repeat a previous selection as *n* such integers are chosen equals:

I can't see what to do with this, either, but I suppose you could incorporate the expected total number of collisions into your calculation, along with the probability of two collisions already calculated. I just don't know how you would do that. If you know of a computer model, though, I'd be very happy just to have a numerical answer, I can understand that it would be very difficult to calculate this algebraically

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,776

Hi Au101;

As far as I know that formula is useless for your question.

The formula can give you the average or expected number. I do not see how you can get the 75% in there. You can try to get more information from the guy who suggested that. It may have been an offhand suggestion, one without much thought.

I can have a computer solution in a day or two, maybe a little longer.

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**Au101****Member**- Registered: 2010-12-01
- Posts: 286

bobbym you're a star, that'd be fantastic, thanks a lot

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,776

Will post it as soon as I compute them. Please have patience it is a big calculation and will take a long time.

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**Au101****Member**- Registered: 2010-12-01
- Posts: 286

That's no problem, I'm not in a rush

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,776

Hi;

Working on it but in the meantime if you come across any analytical solution please post it.

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**Au101****Member**- Registered: 2010-12-01
- Posts: 286

Definitely Thanks a lot

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,776

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**Au101****Member**- Registered: 2010-12-01
- Posts: 286

Bobbym that's fantastic, thank you!

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,776

Hi;

Each simulation has been checked using two different methods. I used my own method and the method employed by blank which is much faster to check.

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**Au101****Member**- Registered: 2010-12-01
- Posts: 286

Who is blank?

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,776

Someone who supplied me with a different way of solving the problem. I wanted to mention him because he did not get the credit he deserved.

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**Au101****Member**- Registered: 2010-12-01
- Posts: 286

Okay, well, thank you very much, both of you, that's all i needed

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