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Birthday Probability: CR4 Challenge (09/09/08)

Posted September 07, 2008 5:01 PM

This week's CR4 Challenge Question:

How many people must be in a room in order for the probability to be greater than 1/2 that at least two of them have the same birthday? (By "same birthday", we mean the same day of the year; the year may differ.) Ignore leap years.

Thanks to Maths_Physics_Maniac for this question!

Answer:

Given n people, the probability, Pn, that there is not a common birthday among them is

The first factor is the probability that two given people do not have the same birthday. The second factor is the probability that a third person does not have a birthday in common with either of the first two. This continues until the last factor is the probability that the nth person does not have a birthday in common with any of the other n - 1 people.

We want Pn < 1/2. If we simply multiply out the above product with successive values of n, we find that P22 = 0.524, and P23 = 0.493

Therefore, there must be at least 23 people in a room in order for the odds to favor at least two of them having the same birthday.

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#1

Re: Birthday Probability: CR4 Challenge (08/26/08)

09/08/2008 11:48 AM

366?

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#3
In reply to #1

Re: Birthday Probability: CR4 Challenge (08/26/08)

09/08/2008 2:31 PM

I guess that would be 100% probability. Again, my lack of math skills becomes apparent. Why I try answering the challenge questions, I have no idea...

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#22
In reply to #3

Re: Birthday Probability: CR4 Challenge (08/26/08)

09/09/2008 8:32 AM

I'm with you- I enjoy the questions but it is apparent to me that I should be looking at the answers and not giving them. Anyway, I like your answer. It seems to me like the probability of guarranteeing even one pair of birthdays is only met when you have accounted for all of the days in the year and then add one. I had entertained adding another 182 to the 365 (547) believing that it would take that many for at least 1/2 the birthdays in a year to be covered.

I figured your 100% probability was for 1 pair, maybe not for 1 out of every two (I don't know!)

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#23
In reply to #1

Re: Birthday Probability: CR4 Challenge (08/26/08)

09/09/2008 8:37 AM

12???

In my grade school class, there were 12 people. There were 2 shared birthdays within that sample. 2 on March 15, 1954, and 2 on June xx, 1954.

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#2

Re: Birthday Probability: CR4 Challenge (08/26/08)

09/08/2008 1:49 PM

Assuming that "ignore leap years" means "assume people born on February 29th were systematically excluded from the room", rather than that birthdays in leap-years don't count:

I think it is appreciably less than 30 (that would give 435 pairs of people, but the probability will need to take account of duplications) - I'm out tonight, so I'll do the numbers tomorrow - unless some-one else has done it first.

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#4

Re: Birthday Probability: CR4 Challenge (08/26/08)

09/08/2008 2:40 PM

Twenty eight

Consider if we start in a room with just two people. The probability that the second person has the same birthday as the first is 1/365. If we add one more to the room, the probability that person matches one of the other two birthdays is 2/365. Add a fourth person and their contribution to the birthday match probability is 3/365. For the four person problem, the probability of a birthday match is 6/365 or 1.6 percent. If we keep adding one person at a time and adding in their probability, we get a probability of 0.962 for 27 people and 1.036 for 28 people. If we use a 366 day year, we get a probability of 0.956 for 27 people and 1.033 for 28 people so the effect of leap year don't matter.

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#140
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Re: Birthday Probability: CR4 Challenge (08/26/08)

04/01/2009 3:41 AM

Wrong answer!

P <= 1

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#5

Re: Birthday Probability: CR4 Challenge (08/26/08)

09/08/2008 3:37 PM

This problem is best solved by finding the probability that the condition isn't met.

With 1 person in the room, there is 100% (1.00) chance that no one else in the room has his birthday

When someone enters he has 364/365 (99.7%) chances that his birthday is not "owned" by someone in the room.

As the third person enters he has a 363/365 (99.5%) chance that his birthday is not "owned" by someone already in the room and there was a 99.7% chance that there is no "shared" birthday already present.

At his point we see that as 3 enters the room, the chances of 1 AND 2 AND 3 having distinct birthdays are (365/365)*(364/365)*(363/365) = 99.2%.

Clearly, we can keep this pattern going until the chances of entering the room, in which there are no "shared" birthdays, with a birthday which isn't "owned" are less than 50%

That is (365/365)*(364/365)*(363/365)......*(366-N)/365 < 50%

This happens as the 23rd person walks in and his chances are 49.3%.

This means that with 23 people in a room the chances are greater than 1/2 that there is at least one shared birthday

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#6
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Re: Birthday Probability: CR4 Challenge (08/26/08)

09/08/2008 5:06 PM

Good answer. I knew I had missed something and now I see where I went wrong.

Jim

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#11
In reply to #5

Re: Birthday Probability: CR4 Challenge (08/26/08)

09/09/2008 4:50 AM

Agreed (so G.A.) if the population has a uniform distribution of birthday dates.
And even were we to have a party for people born only in leap years, that answer would not change.

In practice there are seasonal peaks and troughs; these vary by country, but figures for the US* indicate about 12.5% more births in September than in January (in spite of the shorter month). Then there are specific dates where staff shortages are predictable, and there is a tendency for inductions and Caesarians to avoid these. It's possible that taken together these effects will reduce the answer by a day, but I don't enough data to be certain.
. If the gathering covers only a narrow age range, other factors enter the equation.
Inductions tend to be carried out early mid-week, as do non-emergency Caesarians; this will influence the distribution if the gathering covers a narrow age range (as do many).

*I'm not certain how much of this is due to responses to climate and whether any of it is due to planning so children are old in their year at school - which apparently confers an educational advantage. For what it is worth, here are averaged US figures by month starting January (I forgot to check the units. Strangely, one unit looks to be equivalent to 10/year):

26123
26785
26838
26426
26702
27424
28655
28954
29407
27705
26842
26864
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#26
In reply to #5

Re: Birthday Probability: CR4 Challenge (08/26/08)

09/09/2008 9:19 AM

Back in my mis-spent youth, I read a Time-Life book on mathematics and one of the examples was this probability problem. My recollection is the answer was around 23?

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#63
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Re: Birthday Probability: CR4 Challenge (08/26/08)

09/10/2008 12:51 AM

you are right vote you

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#85
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Re: Birthday Probability: CR4 Challenge (08/26/08)

09/13/2008 10:53 PM

Having been many times confused while trying to do statistics, I decided to do a numerical experiment to confirm SlideRuler's answer. I did an Excel spread sheet that used a random generator to generate a whole bunch of ordered lists (1008) each with 42 birthdays. I used a 365 day year. Then starting at the beginning of each list, I calculated which position in the list had the first birthday that matched a previous one in that list. For 95 of the lists there was no match in the 42 birthdays. The plot below shows the results for each of the number of people versus the number of times that was the lowest number for matching birthdays compared with the equation from SlideRuler. While there is considerable scatter in the data, the general shape of the data agree.

When the cumulative probability is calculated, the numerical data shows just over 50% for 23 people in the room.

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#86
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Re: Birthday Probability: CR4 Challenge (08/26/08)

09/14/2008 8:54 PM

sweet cumulative probability curve jim35848!

milo

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#88
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Re: Birthday Probability: CR4 Challenge (08/26/08)

09/15/2008 4:02 AM

All computer generated random data are psedorandom, and pseudorandom data tend to be biased! They therefore have lower probabilities than actual ones!

How biased it is, depends on how the psedorandom is generated!

Cheers,

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#91
In reply to #88

Re: Birthday Probability: CR4 Challenge (08/26/08)

09/15/2008 6:45 AM

You are quite incorrigible!
It seems that you would rather rubbish anything that conflicts with your viewpoint rather than reconsider.

"Pseudorandom data tend to be biased"
A lot of work has gone into making pseudo-random data as unbiased as possible. Of course there is a conflict between obtaining repeatable simulations of random effects and the simulations being truly random, but that is not the same as bias - and wouldn't lead to the sort of effect you imply on the results.

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#95
In reply to #91

Re: Birthday Probability: CR4 Challenge (08/26/08)

09/15/2008 1:33 PM

You sounds like what you are saying!

"as unbiased as possible" is still biased! I wish to find one such computer generated pseudorandom algorithm that are really random and shall be happy to be proven wrong.

Are you saying MS Excel is one of such genuine random stuff?

AC Wing.

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#7

Re: Birthday Probability: CR4 Challenge (08/26/08)

09/08/2008 7:21 PM

22

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#8

Re: Birthday Probability: CR4 Challenge (08/26/08)

09/08/2008 10:54 PM

The last company I worked for, there was me and another (with < 5 employees)

we had the same Initials (RS)

Same birthday (excepting the year, and that was about 3 years different, 24/10 (and for those Yanks 10/24)

Same First name...

Tghe other guy was employed on his merits, not probability of similarities

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#9
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Re: Birthday Probability: CR4 Challenge (08/26/08)

09/09/2008 1:00 AM

Last year I went to UK to attend my eldest sons graduation ceremony with my family (my wife & younger son). We were on an Emirates flight Birmingham-Dubai-Colombo on economy class. While on transit while walking up to the gate someone behind called me by my name to find that was a friend of mine also from Sri Lanka. He studied with me in school in same class then went to the same University (uni of Perdeniya, Sri Lanka) with me and did same degree (Mech. Engg) with me in 1980. After graduation he worked for Mitsui I worked as an Asst.lecturer, but after 1 year we both joined state Petroleum Co. After being there both of us left and when in our own ways.

Funny thing is he was traveling business class and when we walked up to the gate at the counter we were told that for our family seats have been upgraded to business class.

Funniest thing was I have been allocated the seat adjoining my friend.

How do I calculate such a probability!

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#12
In reply to #9

Re: Birthday Probability: CR4 Challenge (08/26/08)

09/09/2008 4:53 AM

You don't, but it's very high that some such thing will happen at least once during your lifetime. It looks significant at the time, but (similar to the birthday problem) the huge number of opportunities over a lifetime means it is a regular occurrence.

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#24
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Re: Birthday Probability: CR4 Challenge (08/26/08)

09/09/2008 8:45 AM

Its called Small world Netwoks or "six degrees of freedom"

It is apparent to me that both you and your friend are what might be called "nodes" and so have a higher probability of being at the airport then say your other 199 friends and neighbors...

http://www.small-world-network.com/

Ifind that there is a better than 50% chance of my finding someone I know at an airport, baseball, or football game, by Apparent chance, even when i'm out of town...

Large numbers, exponential powers, and shortcuts of nodes across networks are the drivers of these "coincidences."

Here is a brief overview from a review on Amazon :

Coincidence is the current focus of modish mathematical investigation. Kicked off, according to Buchanan, by a 1998 paper published in Nature, research on the nature of coincidence posits that deep-seated principles order huge, seemingly inchoate assemblies of objects. According to these conjectured principles, any member of a gigantic assembly of similar members (say of six billion human beings) can connect with any other member in astonishingly few steps. The idea seems ubiquitous, cropping up in food chains, the cell, neural networks, disease propagation, or electrical power grids--all arenas explored by Buchanan. This connection of objects in a set, dubbed "small worlds," comes in two "flavors": egalitarian networks and aristocratic networks, an example of the latter being the Internet. for this book:

http://www.amazon.com/Nexus-Worlds-Groundbreaking-Theory-Networks/dp/0393324427

It is a very worthwhile read.

http://www.ncbi.nlm.nih.gov/pubmed/9623998

milo

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#30
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Re: Birthday Probability: CR4 Challenge (08/26/08)

09/09/2008 9:54 AM

Addendum Milonum

Will a network science emerge that helps us understand a variety of complex organizational systems by describing the puzzles of human behavior and connections in mathematical terms? So argues Buchanan, former editor of Nature and New Scientist. Buchanan, who holds a Ph.D. in physics, delivers a good introduction to theoretical physics and the "small worlds" theory of networks. He sees biology, computer science, physics, and sociology as intimately connected. Buchanan illustrates social and physical networks with examples ranging from the infamous "six degrees of separation" theories, to the spread of the AIDS virus, to the mapping of the nervous system of the nematode worm. Are the similarities among these networks merely a coincidence or the result of some underlying physics? Only further research will tell, but in the meantime this book is a good primer to basic network concepts and contains references to key journal articles and studies for further reading. The subject will be of particular interest to mathematicians, physicists, and computer scientists and of general interest to those in most other disciplines. Recommended for academic and larger public libraries. Colleen Cuddy, Ehrman Medical Lib., NYU Sch. of Medicine

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#31
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Re: Birthday Probability: CR4 Challenge (08/26/08)

09/09/2008 10:16 AM

Thanks Duckinthepond.

Its worth the read!

But the network science doesn't explain a darn thing about the Female mind...

milo

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Re: Birthday Probability: CR4 Challenge (08/26/08)

09/09/2008 11:27 AM

I've been meaning to read this book and never got around to it.

Am setting sail to Parry Sound library where a single copy is available. Running low on single malt also.

Thanks for reminder!

ps....my wife calls herself Mrs. Stepford whenever she doesn't have her way (which has yet to happen). I prudently blend in with the workshop woodwork at those times.

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Re: Birthday Probability: CR4 Challenge (08/26/08)

09/09/2008 12:01 PM

Thanks Milo you opened my eyes to a completely new world that I did not know ever existed (perhaps does not exist at all!).

Since in Mechanic of Machine lectures my lecturer got confused while confusing the rest of us students trying to explain Nth degree of freedom and tried to extrapolate that to infinite number of degrees of freedom I thought ' 4 degrees of freedom" is enough for me!

That was way back in 1978 or 1979

How silly I was!

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#77
In reply to #9

Re: Birthday Probability: CR4 Challenge (08/26/08)

09/11/2008 11:40 PM

astrologically you have uncanny resemblance

planetary influences may drive you to commonalities and your choices for attaining your needs...perspectives may be more family influenced

probabilities? i think we all have shadowed counterparts, but what is the likelihood we meet them face to face? calculate planetary environment to familial drives

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#21
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Re: Birthday Probability: CR4 Challenge (08/26/08)

09/09/2008 8:07 AM

"...other guy was employed on his merits, not probability of similarities..."

Odd how that makes it sound like you were employed on probability of similarities... Nah, I'm sure that was merit-based as well!

The job I had immediately before this current one, there were 4 of us in the department, and I shared the same date of birth (including the year) with one of them. But what amazed people the most is that the shared birthday is 31 December. Why do people think it's strange for some of us to be tacked on at the end of the year for tax purposes?

OBTW, I heard a similar challenge question years ago, and if I remember correctly, the probability approaches 100% at about 50 people.

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#10

Re: Birthday Probability: CR4 Challenge (08/26/08)

09/09/2008 4:18 AM

ignore leap years, there is 365 days in a year,

so the correct answer is

(364/365)n < 0.5

where n is people number who have the same birthday!!!!

who like to calculate this number n ?

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#15
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Re: Birthday Probability: CR4 Challenge (08/26/08)

09/09/2008 7:20 AM

oh, I forget to add one item. the equation should be :

(364/365)n + n(364/365)(1/365)n-1 < 0.5 who can solve this equation?

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#20
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Re: Birthday Probability: CR4 Challenge (08/26/08)

09/09/2008 8:05 AM

sum mn=2 Cmn (1/365)m (364/365)n-m

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#141
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Re: Birthday Probability: CR4 Challenge (08/26/08)

04/01/2009 3:46 AM

Wrong answer!

Wrong analysis!!!

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#13

Re: Birthday Probability: CR4 Challenge (08/26/08)

09/09/2008 5:30 AM

none likes to calculate it? let me calculate myself,

its 250.8--250.9 , so that the least number will be 251.

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#14

Re: Birthday Probability: CR4 Challenge (08/26/08)

09/09/2008 7:00 AM

NC2×1/365 = 1/2

where n is the number of people in the room

n * (n-1)(n-2)!/ 2!*(n-2)! *1/365= 1/2

n2-n-365=0

n = 19.6 ?

so 20 people....

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#16
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Re: Birthday Probability: CR4 Challenge (08/26/08)

09/09/2008 7:28 AM

G.A.

This is the way we estimate also the match of birthdays of the 2nd person to the third and the third to the 2nd, the 4th to the 3rd and 3rd to the 4th,... so on and so forth.

Surely, we assume the population gives birth to children uniformly throughout the year, or if birth rate is higher seasonally or due to other factors, then the answer would be different.

BTW, shouldn't it be NP2×1/365 = 1/2 that give your calculation?

AC Wing.

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#17
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Re: Birthday Probability: CR4 Challenge (08/26/08)

09/09/2008 7:33 AM

Well the reason i used combination was because the order doesnt matter. I thought about that point too.

I wonde rif the order does matter, i guess it would then it would be 14.01 people. Ya youre right, its people, matters which one you pick first.

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#18
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Re: Birthday Probability: CR4 Challenge (08/26/08)

09/09/2008 7:35 AM

My fault. Yes, it should be combination, NOT permutation! I was thinking something else!

Cheers.

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#19
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Re: Birthday Probability: CR4 Challenge (08/26/08)

09/09/2008 7:53 AM

I would appreciate another opinion on this, namely yours . Combination sounds right? Unordered set of elements, if its distinct elements then shouldnt the order matter? i'm such an idiot, if it is distinct elements then the order doesnt matter because well one set is the same as the other other in reverse or some other order.

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#34
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Re: Birthday Probability: CR4 Challenge (08/26/08)

09/09/2008 11:11 AM

How does order matter, in this case?

We don't care if its the first and second people whose birthdays match. We just want to know how large must the set be in which we have a 50% probability of any 2 matching.

Order is not an issue.

milo

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#46
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Re: Birthday Probability: CR4 Challenge (08/26/08)

09/09/2008 1:56 PM

Sticking strictly to the original question and assuming there is NO difference between the person entering the room having same birthday to one inside the room and a person inside the room have same birthday to one who is entering, I think order does not matter; i.e. A has same birthday to that of B is the same as B has the same birthday to A.

On the order hand, SlideRuler actually calculated cases or 2, 3, 4, 5,...N people having same birthdays, excluding only NO match case, hence his/her answer was larger, 23.

In your case, you used NC2 (meaning you were concentrating on only 2 people having same birthdays)×1/365(each has an their own birthday of the year), you were calculating only the probability of have only 2 person of shared birthday, hence you have a smaller answer, 20.

Who is right, depending on how you interpret the question, with the key words summarized below:

1. At least

2. 2 shared birthdays

I am inviting you all to read the question again and comments please!

AC Wing.

I don't think my answer is off topic! It should be strictly sticking to the question! Can anyone in CR4 rectify this?

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#51
In reply to #46

Re: Birthday Probability: CR4 Challenge (08/26/08)

09/09/2008 2:07 PM

I agree not off topic.

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#53
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Re: Birthday Probability: CR4 Challenge (08/26/08)

09/09/2008 2:09 PM

What about my reasoning?

AC Wing.

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#55
In reply to #53

Re: Birthday Probability: CR4 Challenge (08/26/08)

09/09/2008 2:11 PM

Reasoning sounds reasonable. I voted non off topic too.

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#54
In reply to #46

Re: Birthday Probability: CR4 Challenge (08/26/08)

09/09/2008 2:10 PM

you got a not off topic from me.

milo

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#57
In reply to #54

Re: Birthday Probability: CR4 Challenge (08/26/08)

09/09/2008 2:23 PM

Thanks for you 2.

Cheers,

AC Wing.

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#58
In reply to #46

Re: Birthday Probability: CR4 Challenge (08/26/08)

09/09/2008 5:47 PM

There is of course no difference between everyone being there together and entering one at a time. Any order would do, but description via ordering clarifies the calculation.

The reason for the difference in count is that one (23) gives the probability of at least two people sharing a single birthday on that particular occasion (the situation that I understand is specified in the challenge).
The other (20) appears to be the average number of days on which more than one person was born - although I haven't checked the calculations.

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#72
In reply to #58

Re: Birthday Probability: CR4 Challenge (08/26/08)

09/11/2008 1:56 PM

My understanding is different as I've explained previously.

23 actually includes 2 and more than 2 matched birthdays,

20 estimated only 2 matches birthdays,

all at chances of 1/2.

AC Wing.

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#75
In reply to #72

Re: Birthday Probability: CR4 Challenge (08/26/08)

09/11/2008 3:52 PM

The challenge was specified as the probability of at least one coincidence;
There is a different statistical specification for the meaning that I believe you desire: "the expected number of days on which two or more people share a birthday".
(Another clue -we all know that the definition of probability means that the probability can never exceed unity - what with your calculation is the probability after 28 days?)

By the way, even for the second meaning the method of calculation is at best oversimplified. The maximum number of days on which a coincidences can occur cannot exceed 365, which according to my calculation of NC2/365 would be exceeded once there are 517 people in the room.

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#79
In reply to #75

Re: Birthday Probability: CR4 Challenge (08/26/08)

09/12/2008 2:33 PM

The original challenge:

"How many people must be in a room in order for the probability to be greater than 1/2 that at least two of them have the same birthday?"

and it was clarified that no leap years should be considered.

I desire nothing more than this question and I believe the answer provided by SlideRuler and Sahasushank were clearly explained. Although I was thinking something else when I said about the permutation, but nothing like what you said!

Actually I intended to choose 20 as the answer because the challenge "How many people" implied at least how many people should be there in order to have at least 2 matched birthday! 20 is what I am sticking to!

Any confusion?

Period.

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#80
In reply to #79

Re: Birthday Probability: CR4 Challenge (08/26/08)

09/12/2008 3:38 PM

As you make no answer to any of my specific points, I'm totally confused whether your statement is because:
a) You don't agree the definition of probability, or
b) You believe that Sahasushank's initial answer in post #14 was a mathematically correct derivation of the probability as usually defined.

If a), you can join either the ever-expanding club of people who redefine standard terms, or you can obtain a book on probability and work on the standard basis (much better for communication, as there are proper terms for the alternative meanings).

If b), you are ignoring the constraint that probability cannot be greater than unity. You are also also ignoring Sahakushank's revised answer in post #73.

Plus: if you are taking Sahakushank's formula from #14 to represent what is usually known as the "expected number" of days on which there might be coincidences, how do you explain the possibilty that the expected number of coincidences eventually becomes more than the number of days in a year?

BTW, are you aware that if you have three birthdays of Arthur, Brenda, and Charles on the same day, the formula of #14 would count the coincidences in the following way:
Brenda and Arthur is one event,
Charles and Arthur is a second event, and
Charles and Brenda is a third event.

I'm sorry you are suffering from a period.

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#81
In reply to #80

Re: Birthday Probability: CR4 Challenge (08/26/08)

09/12/2008 3:54 PM

P.S. I don't understand or agree the equation in #73 either - but at least it shows second thoughts

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#84
In reply to #81

Re: Birthday Probability: CR4 Challenge (08/26/08)

09/12/2008 10:05 PM

I'd chucked all those books ages ago!

Look at the question again, and read the meaning of each of the equations, you won't have such confusions!

Equation #73 is just the same as that of SlideRuler's!

AC Wing.

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#90
In reply to #84

Re: Birthday Probability: CR4 Challenge (08/26/08)

09/15/2008 6:29 AM

Communication error:

Equation in #73 is due to Lendog:
. probability(n) = (1/2 x n x (n-1))/365
In concise from, SlideRuler's is:
. probability(n) = 1-365!/365n/(365-n)!

Not only is the form different, but so are the results
I list them from 20 to 29:

200.4114380.520547945
210.4436880.575342466
220.4756950.632876712
230.5072970.693150685
240.5383440.756164384
250.56870.821917808
260.5982410.890410959
270.6268590.961643836
280.6544611.035616438
290.6809691.1123287

Not only are there differences in detail, but the results of #73 clearly give "probabilities" exceeding unity!

Sahkushank's number in post #74 (= 23) is the same as SlideRuler's, but there seems to be a problem with the posting, as there is no relation between the equation that appeared and SlideRuler's. My point was that the originator of the answer "= 20" (that you use as a reference for your viewpoint) has since changed his mind.

P.S. It's never a good idea to chuck good technical books until you've assimilated the content.

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#94
In reply to #90

Re: Birthday Probability: CR4 Challenge (08/26/08)

09/15/2008 1:28 PM

Whoever changed his opinon, I stick to mine.

I tend to prove all equations I used and make sure I understand the meaning of each term in them, therefore would not be confused! Trying to hang on to books or equations that we don't understand is more confusing than ever, not a good idea neither!

I stick to 20, the MINIMUM number that would give the required probability under the said conditions!

AC Wing.

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#96
In reply to #94

Re: Birthday Probability: CR4 Challenge (08/26/08)

09/15/2008 2:32 PM

Please restate the equation you believe in your own posting - then maybe we can talk.

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#132
In reply to #72

Re: Birthday Probability: CR4 Challenge (08/26/08)

09/24/2008 6:34 AM

Even if we accept your interpretation of the question, there seems to be a problem here.

Two matched birthdays or more includes the situation of exactly two matched birthdays. So the probability of exactly 2 matched birthdays must be less than (or possibly equal to) the probability of two matched birthdays or more.

So, if you have a probability of ∏(2+) for two matched birthdays or more, the probability ∏(2) of exactly 2 will not be greater than ∏(2+).

That means that, if 23 people are required for ∏(2+)>1/2, at least 23 will be required for ∏(2)>1/2.

In short, it is not possible for 23 to be correct for the standard interpretation and for 20 to be correct for "exactly 2". The answer 23 is supported by spreadsheet simulation and also by experiment.

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#65
In reply to #17

Re: Birthday Probability: CR4 Challenge (08/26/08)

09/10/2008 7:34 AM

Personally, I usually count things one at a time - typically the order of counting is whatever is convenient to ensure I don't double-count. It does not make any difference what order I choose. The same applies here - an order allows the calculation process to be structured, but the order chosen makes no difference to the result

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#59
In reply to #14

Re: Birthday Probability: CR4 Challenge (08/26/08)

09/09/2008 6:57 PM

Had to throw in my two cents.... I also calculated 20 people, but I did not have your formula. My calculations based on the number of single non-repeating pairs in a room of 'n' people with a 1/365 chance that any one pair would match. 20 people exceeds a 182.5/365 chance. (just over 19 w/ 182 non-repeating pairs).

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#25

Re: Birthday Probability: CR4 Challenge (08/26/08)

09/09/2008 9:01 AM

The last correct answer would be

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#27
In reply to #25

Re: Birthday Probability: CR4 Challenge (08/26/08)

09/09/2008 9:25 AM

613.

then the possibility will be 0.5006

and when 612, the possibility will be 0.4998

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#61
In reply to #27

Re: Birthday Probability: CR4 Challenge (08/26/08)

09/09/2008 9:02 PM

I made a wrong result.

this value would be less than 366 in the least.

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#28

Re: Birthday Probability: CR4 Challenge (08/26/08)

09/09/2008 9:26 AM

When I'm in the same room as my sister there's 100% probability.

I'm exactly 16 years older then my sister and my step-mother couldn't take the time to stop and make me a birthday cake before running off to the hospital but she did have enough time to wake me at 3 in the morning to tell me they were leaving.

That was 30 years ago.

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#29

Re: Birthday Probability: CR4 Challenge (08/26/08)

09/09/2008 9:43 AM

Maybe I am over simplifying, I have been accused of that before.

The question is basically.

Drawing Random numbers between 1 and 365, when does the odds of a repeating number become greater than 50%

Isn't it simply 186 ?

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#32

Re: Birthday Probability: CR4 Challenge (08/26/08)

09/09/2008 10:18 AM

That depends. If you had my father, his father, my sister, a school mate, an ex-girlfriend, and myself in one room, you would have that many people with the same birthday plus whoever else happened to be present. I am not sure that this was what was meant by family planning, but that is the way it worked out.

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#33

Re: Birthday Probability: CR4 Challenge (08/26/08)

09/09/2008 11:05 AM

How many people must be in a room in order for the probability to be greater than 1/2 that at least two of them have the same birthday? (By "same birthday", we mean the same day of the year; the year may differ.) Ignore leap years.

I believe the answer is 23 people for 50% likelihood and 58 people for 99% chance of same birthday.

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#35

Re: Birthday Probability: CR4 Challenge (08/26/08)

09/09/2008 11:27 AM

23 seems to be a common answer, So therefore

Hypothetical

I generate 22 random numbers from a group of 365, lets say for arguments sake the result is;

1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22

Your saying there is a 50% chance that I will draw one of these numbers again????

I just don't see it?????

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#37
In reply to #35

Re: Birthday Probability: CR4 Challenge (08/26/08)

09/09/2008 11:33 AM

Actually thats now how probability works. It is not defined for the set number of tries or a maximum number of tries, etc. Instead if you were to conduct the test infinite number of times then yes the ratio would equal out to 50%, i.e 50% of the time it would be the same number

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#40
In reply to #37

Re: Birthday Probability: CR4 Challenge (08/26/08)

09/09/2008 11:57 AM

Quite true. That is how casinos and all gambling games take your money away!

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#45
In reply to #35

Re: Birthday Probability: CR4 Challenge (08/26/08)

09/09/2008 1:28 PM

I don't understand these smaller numbers either- your example being a good one.

I figure that the second person that walks into the room has a 1 in 365 chance of duplicating the birthday. The second person has a 1 in 364 and so on. Could it be as easy as adding all those up until we get to above .5? At person 145 the sum becomes .500843, but at that number there could still be 222 dates left open (far above 1/2 the dates available). Its so hard trying to remember the stuff they tried to teach me 30 years ago!

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#47
In reply to #45

Re: Birthday Probability: CR4 Challenge (08/26/08)

09/09/2008 1:59 PM

Nope. See slide ruler's post 5.

Also, your analysis leaves the numerator at 1??? The second guy has only no degree of freedom to match he either matches guy number one or not; guy number three has two degrees of freedom he can match guy number one OR guy number two; guy or gal number four has the possibility of matching mssr's 1,2,or 3 this goes on for a long time... your fractionation does not seem to account for this.

Slideruler nailed it by solving for the complement of the event.

milo

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#48
In reply to #45

Re: Birthday Probability: CR4 Challenge (08/26/08)

09/09/2008 2:01 PM

Actually you're going about it the wrong way.

See, what you are calculating is simply the probability of picking up 2 numbers out of 365. if you have to pick a number out of 365 number what is the probability that you will pick a number out, 1/365, probability that you will pick 2 numbers out ... 2/365 and so on and so forth.

On the other hand, in the current problem, you are not given the number of people. That is what you need to calculate. Therefore, suppose there are N people in a room and you pick 2 people out of that N people and then what is the probability that those 2 have the birthday on the same date. so you have to first take into account that you picked up 2 people then you need to multiply it by the requirement that both of them have it on the same day i.e. 1/365

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#76
In reply to #45

Re: Birthday Probability: CR4 Challenge (08/26/08)

09/11/2008 4:24 PM

Just a note-- I guess it is 144 people that the sum becomes .500843

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#70
In reply to #35

Re: Birthday Probability: CR4 Challenge (08/26/08)

09/10/2008 6:26 PM

Let me try to explain where you are going wrong:-

Although the number sequence you give is a possibility, it is not typical. You have chosen to draw 22 distinct numbers and you are now asking about the probability that the 23rd number would match one of your first distinct 22 - this is not the problem given.

If you want to rephrase the problem in terms of numbers, you can say:-

How many numbers must I draw at random from the integers 1 to 365 inclusive so as to have a greater than 50:50 chance of having at least one matching pair?

N.B. Each draw will contain all 365 numbers, so each time you pick a number( Except for your first pick obviously) there is a chance you will pick a match.

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#71
In reply to #35

Re: Birthday Probability: CR4 Challenge (08/26/08)

09/11/2008 5:34 AM

Another way of looking at the flaw in this approach is:

You have artificially selected your first 22 numbers so that none of them are the same. If you had selected these 22 numbers at random, there could have been some coincidences.

Let us take a partial example - suppose you initially select the numbers 0...21. The 22nd number you select will have a 21/365 probability of coinciding with one of the preceding numbers. If the 22nd number is not one of 1...21, the 23rd number will have a 22/365 chance of coinciding with one of the preceding numbers.

But it is easiest to work your way forwards from an arbitrarily selected first number more-or-less as described by SlideRuler.

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#38

Re: Birthday Probability: CR4 Challenge (08/26/08)

09/09/2008 11:41 AM

23 for 50%

57 for 99%

My wife (an Educator) actually surveyed hundreds of classes and confirmed it many times.

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#39

Re: Birthday Probability: CR4 Challenge (08/26/08)

09/09/2008 11:48 AM

http://omgwtf.superlime.com/f7e89dc20e65cf181996442a22217955/psychic.swf

some really cool 'rithmetic here.................

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#42

Re: Birthday Probability: CR4 Challenge (08/26/08)

09/09/2008 12:02 PM

Can you guess what is the most popular birthday in the US?

it's October 4th...

why?

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#44
In reply to #42

Re: Birthday Probability: CR4 Challenge (08/26/08)

09/09/2008 12:28 PM

Because on average the gestational period for humans in america is 278 days per your data, resulting in a bunch of simultaneous births resulting from the furious copulation from New Years festivities January 1? October 4 is 278 days subsequent to january 1; same as human gestational period.

How about a pandering request for a good answer on that one?

milo

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#43

Re: Birthday Probability: CR4 Challenge (08/26/08)

09/09/2008 12:23 PM

The solution is 22

while P<.5
P=prod(1:n)/(2*prod(1:(n-2)))*(1/365)^2*365
P=P*((365-(n-1))/365)
n=n+1
end

P(22)=0.5438

Chazl

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#66
In reply to #43

Re: Birthday Probability: CR4 Challenge (08/26/08)

09/10/2008 7:37 AM

You forgot that you need two for the first possible coincidence - so add one to give 23.

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#49

Re: Birthday Probability: CR4 Challenge (08/26/08)

09/09/2008 2:03 PM

3 people would come to exactly 50% probability (this would be the least but highly improbable at 1 out of 2 chances to match), so my answer is 2 which would be greater than 50% probability. If there are only 2 this would either be 100% or 0% but still probable.

Highly Highly, unlikely but still probable.

If your looking for a guaranteed match for 50% chance 183. Super Guaranteed 366.

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#50

Re: Birthday Probability: CR4 Challenge (08/26/08)

09/09/2008 2:05 PM

I don't know what the probability is associated with this question, but in my last job, in a group of 150 researchers, there were 3 of us had the exact same birthday. By that I mean we were all born on the exact same day. My technician was, in fact, <8 hrs older than I am (I called him the "old man"), while the General Manager was the youngest. In addition, there were also 8 other people working our of the group of 150 that shared the same birthday (born on the same day of the year). Oh, BTW, the date is March 13.

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#52
In reply to #50

Re: Birthday Probability: CR4 Challenge (08/26/08)

09/09/2008 2:09 PM

My dad's birth date. Mines 14th march, and that is the only distinction i have in my till now non Porsche and mclaren with a wicked v8 500 hp something monster I share it with Einstein.

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#56
In reply to #50

Re: Birthday Probability: CR4 Challenge (08/26/08)

09/09/2008 2:22 PM

"My technician was, in fact, <8 hrs older than I am (I called him the "old man"), while the General Manager was the youngest."

Does this mean the younger people are, the better they are educated?

Or, it actually require less experience to fill a top job?

A joke only!

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#64
In reply to #50

Re: Birthday Probability: CR4 Challenge (08/26/08)

09/10/2008 1:13 AM

March 13? Human Gestation period approx 280 days?

Weddings in June I suppose..!

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#60

Re: Birthday Probability: CR4 Challenge (08/26/08)

09/09/2008 8:41 PM

Its really depressing seeing all you guyz (spelt with a Z so as to make it more of a general calling to males and females than just males) playing with numbers and mathematics, and I'm still struggling trying to find the workings of the first few equations ;o(

Mathematics I have a problem with, Electronics I love and can play with easily, the problem being, Electronics and Mech Eng involves maths ;o(

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#62

Re: Birthday Probability: CR4 Challenge (08/26/08)

09/09/2008 9:03 PM

So how many people researched this answer on the web and got sliderule's answer?

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#67

Re: Birthday Probability: CR4 Challenge (08/26/08)

09/10/2008 8:25 AM

23

also can be done by looking through any "who's who" or similar almanc ofpersonalities.

Also applies to deaths

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#68
In reply to #67

Re: Birthday Probability: CR4 Challenge (08/26/08)

09/10/2008 9:31 AM

What a cheerful so&so you must be, to be sure.

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#69
In reply to #67

Re: Birthday Probability: CR4 Challenge (08/26/08)

09/10/2008 11:18 AM

Welcome aboard - good point, too. It doesn't matter any more which end of the lifespan you consider than it does which order the people file into (or out of) the room!

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#73

Re: Birthday Probability: CR4 Challenge (08/26/08)

09/11/2008 3:41 PM

Probability is not my strong suit, but I still think the answer is 20. If you look at "n" number of people in a room, the number of possible non-repeating birthday pairs is

Total(n) = 1/2 x n x (n-1)

If any one pair has a 1/365 chance of matching, then shouldn't the additive probability be

Prob(n) = (1/2 x n x (n-1))/365 ?

If so, then n = 20 yields the first probability greater than 50%, with a probability of 52%

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#74
In reply to #73

Re: Birthday Probability: CR4 Challenge (08/26/08)

09/11/2008 3:46 PM

Nope thats not right actually. The answer would be

(NC2 + NC3 + NC4 + NC5 + .... + NCN-1 + NCN) ×1/365 ≈ 23.

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#78
In reply to #74

Re: Birthday Probability: CR4 Challenge (08/26/08)

09/12/2008 11:50 AM

sahasushank, would you explain your 'C' terms a little more? I would like to understand were I went wrong. Thanks

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#82
In reply to #74

Re: Birthday Probability: CR4 Challenge (08/26/08)

09/12/2008 4:12 PM

What did you actually mean?

Apart from the fact that I couldn't detect a basis for this,
there is no value of N for which 23 is even the closest integral answer.

The values versus N are approximately:
N ...... Equation_sum
2 ...... 0.0027
3 ...... 0.011
4 ...... 0.03
5 ...... 0.071
6 ...... 0.16
7 ...... 0.33
8 ...... 0.68
9 ...... 1.38
10 .... 2.78
11 .... 5.58
12 .. 11.19
13 .. 22.41
14 .. 44.85
(and increasing)

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#92
In reply to #82

Re: Birthday Probability: CR4 Challenge (08/26/08)

09/15/2008 7:10 AM

As is obvious by Physicists reply, my answer is wrong. The equation is not giving the possibility. In fact it is not even dividing the right things to get the answer.

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#93
In reply to #92

Re: Birthday Probability: CR4 Challenge (08/26/08)

09/15/2008 10:30 AM

As I mentioned before, probability is not my strong suit, but I know more now than when I started. I wish I could say I understand what the right answer is, but I know now that my answer is incorrect because my formula can generate a probability greater that one. Learn something new everyday.....

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#83

Re: Birthday Probability: CR4 Challenge (08/26/08)

09/12/2008 9:09 PM

Can anyone explain, whats wrong with this formula? [ in otder that the probability is greater than 1/2 that at the least two of them have the same birthday]

sum mn=2 Cmn (1/365)m (364/365)n - m

n .............n

Σ = [sum]

m=2 ........m=2

where n is totle people should be stay in a room; C means combination.

m is 2,3,...n; means just 2 persons have same birthday, just 3 persons and so on...

Slideruler's condition is one by one enters the room, not in room together. if they are ll in the room and you hope to select at the least two of them have same bithday, I think it should be the above fomula.

but after calculate, I think the sum is too large to right. whats matter?

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#89
In reply to #83

Re: Birthday Probability: CR4 Challenge (08/26/08)

09/15/2008 6:01 AM

Slide ruler's use of "entering the room" is just a convenient way of describing his method. It is just the same principle as counting - one practical way to do this is to institute an order, but the order in which you perform the count is irrelevant.

As to where you have gone wrong with your formula: although I had trouble with the format, it is clear that you will never get near the correct answer if you try to handle the permutations of the group of elements independently of the number of days in the year; the reason for this is that every replication r (2<r<m) reduces by one the number of days on which additional replications of size r can be created.

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#87

Re: Birthday Probability: CR4 Challenge (08/26/08)

09/15/2008 12:43 AM

A thought just occurred to me, so I haven't had time to check it out.

Twins usually have the same birthday.* What if the question were about the probability of someone being a twin, or that two people in the room were twins to each other?** Would the answer come out the same, or would it be reasonably close? What percentage of the population do twins account for? How does that compare to the answer to this question?

*Twins do not have to have the same birthday. Most twin births are separated by a few minutes. One could be born before midnight, and the other after midnight. I don't know of any cases, but with the current world population, I wouldn't be surprised that it has happened.

**I said "twins to each other" to differentiate from people that are twins that have come into the room, but they are not siblings to each other. For example, let's say that Harry and Larry are twins, but Harry's twin is Jerry, Larry's twin is Barry, and neither Jerry or Barry is in the room.

Well, I'm off to Wiki or Google in search of statistics...

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#97
In reply to #87

Re: Birthday Probability: CR4 Challenge (08/26/08)

09/16/2008 10:42 PM

Well, the presence of twins wouldn't skew things at all, because the highest number I've seen is that twins account for about 3% of the population. Now, if there were a twins convention in town...

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