08-02-2002 09:36									#91
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quote: Originally posted by Urban Ranger KrazyCat, The difference seems to lie in whether you look at an entire population or an individual. If you look at an entire population, sure, you are correct. However, for an individual, the answer should be 1/2 since there is indeed degeneracy.

Please explain. You appear to be engaging in the fine art of pulling things out of your ass.

	08-02-2002 10:20									#92
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quote: If I have a random number generator that randomly pulls a number between 1 and 365. And I choose any number, every time I run the number generator will give me a 1 in 365 chance of hitting my number. Every subsequent pull would be additive. I don't see the difference.

That isn't quite true. First of all, the probability that person one doesn't have the same birthday as you and the probility that person 2 doesn't have the same birthday as you aren't distinct events; that is, there's some intersection between the two events that must be taken into account.

For an example, let's say if we call the probability that Chow has the same birthday as you E₁ and the probability that Rex has the same birthday as you E₂, and so forth up till we get to all member so 'Poly, the last being E_n. If we go on your theory that P(E₁ U E₂U....U E_n) is simply the sum of individual the probabilities, we obviously end up with counterintuitive result that the probability well exceeds 1.

Another misconception is that you're mathematically introducing the union of the sets, which is the wrong operator in this case. Think of a Venn diagram - the union of two sets includes areas in which the events don't intersect; that is, E1 can be true, while all other events being false. So we get a pretty useless result.

It should indeed be the intresection of two sets, which, as previously indicated, is represented by the product of the two probabilities. Think of one coin. If you flip it, there are two possibilities - {H, T}. So, the probability of either event is 1/2. Now, think of two coins. If you flip both coins, there are four possible results - {HH, HT, TH, TT}. The probability that both are heads is obviously 1/4, which is the product of the individual probabilities, (1/2)(1/2). This situation is analagous to Chow's second problem.

quote: Anyone care to guess the mathematically expected number of different birthdays which would be covered by a random group of 183 people? I don't know the answer; I might be able to figure it out, or hopefully Chowlett can enlighten us.

Like in the previous problem, there are 365^183 total possibilities.

If we assume there are n (n is an integer such that 0 <= n, <= 183) conflicting birthdays, the first person has 365 possible choices. The second has 365 OR 364 possible choices. The next has 365 OR 364 OR 363 possible chioces. The nth has 365 OR 364 OR 363 OR.... 365-n possible choices.

That was pretty futile , let's look at it another way. There is one situation in which there 365^n*365!/(182 + n)! possibilities. There is another such that there are 365^(n-1)364*365!/(182 + n)!. Hell, this may seem hasty given my small expansion but it stands to reason that there are 365!/(182 + n)!(365^n + 364^n +... (182 + n)^n + 365^(n-1)364^n +... [all of those various combinations]). Set this function equal to 1/2, and solve for n, and that would give you the answer.

I don't know how to elegantly write the function, but I gather it shouldn't be too hard just to use a simple program to approximate the solution.

	08-02-2002 18:47									#93
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Problem two: Forget the maths, here was my intuition.

The probability of two people having the same birthday is ~50% for a group of ~25 people. The more people you add the more likely it is that people will share birthdays. Hence with more people the more chance of more birthdays being shared. If we have 183 people as suggested, chances are will have less that 183 distinct birthdays. I figured that for around 250 people we would have around 180 distinct birthdays. With 183 birthdays there is a 1 in 2 chance of my birthday being one of them.

The same logic applies, if we have 365 people chances of them all being different is remote, much closer to having ~260 distinct birthdays (estimate).

	08-02-2002 18:49									#94
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UR, brush up on your conditional probability, I find tree diagrams useful for these types of things.

	08-02-2002 22:41									#95
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To answer my own question (how many different birthdays are expected in a group of 183 people), I used the following reasoning. I'm not certain it's valid; if not, I hope someone will correct me.

Definition: B(N) is the number of different birthdays expected in a group of N people. Obviously, B(1) = 1.

Assumption: If you add 1 person to a group of N, his chance of matching one of the existing birthdays in the group is (B(N)/365). Subtract this number from 1 and you get the increase in B resulting from the addition of that person to the group. Thus, for B > 1, B(N+1) = B(N) + 1 - (B(N)/365). (Making the standard assumptions of evenly-distributed birthdays and no leap year.)

Running this equation through an Excel spreadsheet gives the result that B(183) = 144 (rounded off).

Some interesting results for higher values of N. For N=253, B=183, which agrees with Chowlett's answer on his second problem. For N=730--twice the number of days in a year--B=316, which means you'd still expect to have 49 birthdays uncovered in a group that size.

Last edited by Rex Little on 08-02-2002 at 22:52

	09-02-2002 00:06									#96
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Rex.

That is what I would have expected, the progression is exponential, the number of birthdays that are not covered decays exponentially with number of people.

It is comparable to radio-active decay in this manner, if you use the half life value of 253 people, and that you require around 8.5 half-lives => ~2,150 people before you are "expectant" of all birthdays being covered.

	09-02-2002 00:15									#97
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If you want a formula for the number of uncovered:

Number of unbirthdays = 365 e<sup>-N/365

where N= Number of people</sup>

	09-02-2002 00:26									#98
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That's less of an equation and more of an approximation.

	09-02-2002 00:32									#99
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It is only ever out by a small fraction of a day, and considering you will be rounding to the whole day what is the harm in a fudge

Ok to keep Kitty happy :

Number of birthdays = 365 - 364ⁿ/365^n-1

Last edited by Dauphin on 09-02-2002 at 00:47

	09-02-2002 00:33									#100
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Still an approximation...

You're welcome to kep trying, though.

I doubt the exact formula looks anything like that simple.

	09-02-2002 00:36									#101
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That is exact.

	09-02-2002 00:41									#102
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oops, on second reading.

	09-02-2002 00:48									#103
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For n=1, SD's formula gives 365 - (364/1) = 1. So it's actually a formula for birthdays rather than non-birthdays, but other than that it's correct.

Edit: this was in response to Krazy's post directly above, which he then edited. I think we're all now in agreement.

	09-02-2002 00:50									#104
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Good job I corrected that before you posted.

	09-02-2002 00:51									#105
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I can edit faster than a speeding bullet!