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Au101
2013-07-18 09:58:02

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

bobbym
2013-07-18 09:40:00

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.

Au101
2013-07-18 07:02:54

Who is blank?

bobbym
2013-07-17 22:28:56

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.

Au101
2013-07-17 22:16:06

Bobbym that's fantastic, thank you!

bobbym
2013-07-17 10:34:19

Au101
2013-07-17 09:15:00

Definitely Thanks a lot

bobbym
2013-07-17 08:33:14

Hi;

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

Au101
2013-07-17 05:41:48

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

bobbym
2013-07-17 05:35:25

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

Au101
2013-07-17 05:27:32

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

bobbym
2013-07-17 05:26:14

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.

Au101
2013-07-17 05:17:33

Sorry, to quote:

"The probability that the kth 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

bobbym
2013-07-17 05:13:11

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.

Au101
2013-07-17 05:09:26

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!