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**Au101****Member**- Registered: 2010-12-01
- Posts: 353

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.

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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: 104,771

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.****If it ain't broke, fix it until it is.**

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**Au101****Member**- Registered: 2010-12-01
- Posts: 353

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: 104,771

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.****If it ain't broke, fix it until it is.**

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**Au101****Member**- Registered: 2010-12-01
- Posts: 353

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: 104,771

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.****If it ain't broke, fix it until it is.**

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**Au101****Member**- Registered: 2010-12-01
- Posts: 353

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: 104,771

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.

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**

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**Au101****Member**- Registered: 2010-12-01
- Posts: 353

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: 104,771

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

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**

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**Au101****Member**- Registered: 2010-12-01
- Posts: 353

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

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 104,771

Hi;

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

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**

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**Au101****Member**- Registered: 2010-12-01
- Posts: 353

Definitely Thanks a lot

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 104,771

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**

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**Au101****Member**- Registered: 2010-12-01
- Posts: 353

Bobbym that's fantastic, thank you!

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 104,771

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.

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**

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**Au101****Member**- Registered: 2010-12-01
- Posts: 353

Who is blank?

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 104,771

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.

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**

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**Au101****Member**- Registered: 2010-12-01
- Posts: 353

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

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