Probability and the Birthday problem Part-I

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Offline Mohammad Hassan Murad

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Probability and the Birthday problem Part-I
« on: June 02, 2012, 06:47:49 PM »
THE BIRTHDAY PROBLEM

In this current semester I take the course Mathematics-I in the department of EEE. There are two sections. In each the number of students are 43 and 42.
Now one day I ask myself what is the probability for sharing my birthday to any other students?
Or, I can ask what is the probability that any two or more students share the same birthday?

It is a possibly surprising fact that if you have 23 students in your class or at a party, there is a better than 50% chance that two people share the same birthday (same month and day, not necessarily in the same year).

Let us solve this problem. If we ignore the complication of leap-days there are 365 days on which a birthday can occur. A moment of thought tells us that if there are 366 people in a room then it is certain (probability = 1) that at least one pair of them must share a birthday (think about it!!). If the number of people were reduced to 365 or 364, then while it is not certain that at least one pair share a birthday, our intuition tells us that the probability of this must be very close to 1 — indeed we should be astonished if no two people shared a birthday with such numbers. If the number of people were reduced further then, clearly, the probability would reduce but would still remain high down to 200 people or so.
Let us now start at the other end with two people in a room. The probability that they have the same birthday is low and is easy to calculate. One way of thinking about it is to consider that one of them tells you his birthday. Then you ask the other one for his birthday. There are 365 possible answers he can give but only one of them will match the birthday given by the first person. Hence the probability of the birthdays being the same is 1/365.
The other possible way to look at this problem is to first ask the question “In how many possible ways can two people have birthdays without any restriction?” Since there are 365 possibilities for each of them then the answer to this is 365×365—every date in the calendar for one combined with every date in the calendar for the other. Now, we are going to ask a question that is not, perhaps, the obvious one to ask — “In how many ways can the two people have different birthdays?”.Well, the first one has 365 possibilities but once his birthday is fixed then, for the birthdays to be different, the second has only 364 possibilities so the answer is 365 × 364. Now, we can find the probability that they have different birthdays as
pdiff = (Number of ways of having different birthdays)/(Total number of ways of having birthdays)
= (365 × 364)/(365 × 365)
= 364/365
= 0.997260
≈ 99.72%
Now, either the two individuals have the same birthday or they do not and these are the only possibilities and they are mutually exclusive so that
pdiff + psame = 1,
or, psame = 1 − pdiff = 1 – 364/365 = 1/365
the answer obtained previously. The reason we have introduced this apparently convoluted way of getting the answer is because it better lends itself to dealing with the problem of when there are three or more people in the room. Now consider three people in a room. The number of ways they can have birthdays, without any restriction, is
365 × 365 × 365.
The number of ways they can have different birthdays is found by noting that the first person can have a birthday in 365 ways, for the second to be different from the first there are 364 ways and for the third to be different from the other two there are 363 ways. Hence the total number of ways they can have different birthdays is
365×364×363.
Hence the probability that they have different birthdays is
pdiff = number of ways birthdays can be different number of ways they can have birthdays
= (365 × 364 × 363)/(365 × 365 × 365) = 0.9917958.
If the birthdays are not all different then at least two of them must be the same and the probability of this is
pnot diff = 1 − pdiff = 0.0082042.
We should now be able to see by extending the principle that the probability that at least two people out of five people in a room would share a birthday is
pnot diff = 1 – (365 × 364 × 363 × 362 × 361)/ (365 × 365 × 365 × 365 × 365) = 0.027136.
The advantage of calculating the probability in the way that we do, by finding the probability that all birthdays are different and then subtracting from 1, is now evident. Suppose that, we tried to directly find the probability that two or more people in a set of five shared a birthday. We would have to consider one pair the same and the others all different, three with the same birthday with the other two different, two pairs with the same birthday, and many other combinations. All birthdays being different is a single outcome, the probability of which is comparatively easy to find.
Now, we ask the question that is really the basis of this section—“How many people must there be in a room for there to be a probability greater than 0.5 (or, 50% chance) that at least two of them would share a birthday?” This can be done with a calculator. We work out the following product term-by-term until the answer we get is less than 0.5.
(365/365) × (364/365) × (363/365) × (362/365) × (361/365) × (360/365) × • • •
The value of n for which the answer first falls below 0.5 is when the probability of all birthdays being different is less than 0.5 and hence when there is a probability greater than 0.5 that two or more people have the same birthday. This is the number of people that fulfills our requirement. What is your intuitive assessment of n at this point? In the following table, we give the probability of having at least two people with the same birthday with increasing n. We see that the probability just exceeds 0.5 when 23 individuals are present — a result that most people find surprisingly low.


There is greater than 50% chance that two of these will share a birthday!

References
Woolfson, M. M. Every day Probability and Statistics, Imperial College Press, 2008.
Deep, R. Probability and Statistics, Academic Press, 2006.
Blinder, S. M. Guide to Essential Math A Review for Physics Chemistry and Engineering Students, Academic Press, 2008.
« Last Edit: June 02, 2012, 07:09:40 PM by Mohammad Hassan Murad »
Senior Lecturer (Mathematics)
Department of Natural Sciences,
Daffodil International University,
Faculty of Science and Information Technology.

Offline asitrony

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Re: Probability and the Birthday problem Part-I
« Reply #1 on: June 03, 2012, 05:20:22 PM »
Sweet problem!

Feel happy to know such a problem solving method! It's really a new thing to me and I am very much interested to know more about it!

Thanks

Offline Mohammad Hassan Murad

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Re: Probability and the Birthday problem Part-I
« Reply #2 on: June 03, 2012, 07:15:59 PM »
Dear Asit Sir I hope the other parts of this post will make you more interested on this subject.
« Last Edit: June 04, 2012, 01:37:18 PM by Mohammad Hassan Murad »
Senior Lecturer (Mathematics)
Department of Natural Sciences,
Daffodil International University,
Faculty of Science and Information Technology.

Offline sumon_acce

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Re: Probability and the Birthday problem Part-I
« Reply #3 on: June 03, 2012, 08:58:46 PM »
Thanks Murad Sir.......really good post.