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.

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

]]>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.

]]>"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

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

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

]]>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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