Analyze the Famous 'Birthday Problem' and Discuss the Statistical Implications (Math Problem Sample)

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21st January 2016
The Birthday Problem
Everyone loves their birthdays and also their kith and kin’s, best friend’s and so on. What if a bizarre coincidence happens and there comes a situation wherein two persons share the same birthday, say in a group? And they all share the same room? It will be interesting to experience such a scenario. This situation, also known as the birthday problem, till now remains as one of the problems most discussed and most debated at. This problem completely relies on the beauty of the concept of Probability, to arrive at a plausible conclusion and we shall now see in detail, how the problem unfurls.
To achieve this goal, I shall employ the concepts of the Poisson distribution and Combinations, in detail. Also, while discussing this problem, there also lies the fact that how many people should be there in the room in order to make the above situation happen? Let us find the answers now.
The Problem and some of its Illustrations the problem can be roughly said as: how many persons should be present in a room so that there is at least 50% chance of any two persons sharing the same birthday? (IB Maths Resources from British International School Phuket)
The above situation is somewhat analogous to the situation of flipping a coin. There is an equal chance of getting either a head or a tail. Now, let us consider the situation of a classroom. How many students must be compulsorily there in a classroom, so that the above said concept holds valid?
Suppose, let us assume that only two students are there in a classroom and that the year of the birthday has 365 days and that the two students share the same birthday. Now, the first student can have any of the day in a year, as his/her birthday. So, there are 365 distinct possibilities in 365 days. So, the probability here is P(X) = 365/365 = 1, a sure event where X denotes the event of having a birthday in 365 days.
Now, the second person also sharing the same birthday is a mutually occurring event. Hence, to find its probability, we have to multiply the probability of getting a birthday in 365 days (i.e.) 1, with the probability of having the same birthday (i.e.) 1/365. Thus, (1/365)*(1) = 1/365 (Whitcher).
Now, consider the situation where in three persons are there in a room. What is the probability that all the three persons share the same birthday?
This question cannot be solved directly as did previously. Instead, we are going to consider the formula, P(X) + P(X’) = 1, where X denotes a sure event and X’ denotes an impossible event. So, we shall now find the probability of these three persons having different birthdays.
Let x, y, z be the three different persons. As we have said above, these three persons have three different birthdays in the same year. Then, the probability of x having a birthday in that year is 365/365 = 1, obviously. Now, the probability of y having a birthday, apart from x, is 364/365 which can be taken as, (364/365)*(365/365) = 364/365 itself (as we have considered the possibility that all the three persons have unique birthdays, the birth date of x, which is already considered is ruled out. Hence, out of 365 days, x’s birthday is omitted, which means we have 364 days for consideration only). Now, considering the case of z, there are only 363 days left, in the similar fashion explained above. Thus, probability of z having a birthday, apart from x and y is (365/365)*(364/365)*(363/365) [mutually exclusive events] = (364*363/365*365) = 0.99179 ---------- (1)
Now, if it is the case of four persons having different birthdays, then the above calculation will be remade as, (365/365)*(364/365)*(363/365)*(362/365) = 0.99179*(362/365) [using (1)] = 0.9836. thus, in the same fashion we can calculate any number of distinct birthday possibilities. But, extending for many number of persons by manual calculations, like we have done above is quite difficult. It will be far more easier if a particular pattern hass been found for easy calculation purposes. Hence, for this purpose the method of mathematical induction comes into play. Mathematical induction method enables us to take all the above mentioned criteria into consideration and allows us to form suitable patterns for larger cases. By the term ‘induction’, we are inducing some specific pre stated conditions, to obtain larger results. Here, in our case the prestated conditions are the probabilities found above. These probabilities hold for all the natural numbers like n = 1,2,.. hence, with the aid of this method we are fitting the probabilities in the form of a pattern which holds true in all the cases, with the golden rule of induction (i.e.) if a rule holds good for (n-1) numbers, then it also holds good for n numbers.
This method holds when n=1, 2, 3, 4... Hence, using the method of mathematical induction, we can extend this pattern to n persons having different persons. The probability in that case will be, 1*(364/365)*(363/365)*(362/365)....*(365-(n-1)/365)
The above notation is nothing but 365Pn / (365)n where nPr = n!n-r!
Now, when there is more than one person in a room, the situation of comparison arises. That is, when two persons are there in a room, one can compare themselves with the other. If there are 3 persons, then the first person can compare with the second and the third and the second person with the third and so on. (Here, the comparison of the second person with the first and the comparison of the first person with the second is one and the same). In this way, there are 3 combinations possible. When four persons are there, (say) P, Q, R, S, then the combinations of PQ, PR, PS, QR, QS, and RS are unique and are thus taken into considerations, which make the total possible combinations as 6 (Floor). Extending this pattern like what we did above, it is obvious that for n persons in a room, there are n(n-1)/2 combinations and this is obtained by nCr = n!n-r!r! where the number of potential combinations of at least two persons sharing the same birthday is nC2 = n!n-2!2! which is n n-1n-2!n-2!2! = [n(n-1)]/2.
Let us now tabulate our findings
Number of persons having the same birthday

Corresponding Probability

1

1, a sure event

2

1/365

3

0.99179

4

0.9836

In general, for n persons

365Pn / (365)n

Number of persons having the same birthday

Corresponding Combinations

1

No comparison needed

2

1

3

3

4

6

In general, for n

[n(n-1)] / 2

We haven’t answered our actual problem till now. That is, we have to yet find out, how many people ought to be there in a room, so that there is at least 50% chance of two persons sharing the same birthday? This can only be found out by repeated calculations and the answer turns out to be 23 as, when 23 persons are present in a room, using the above said method, the first person has 365 days to choose from, the second person has 364 days to choose from and going on, we have the twenty third person to choose from 343 days to choose from. Hence, the probability of all the 23 persons having different birthdays is, 1*(364/365)*(363/365)*(362/365)....*(343/365) = 0.4927.
Thus, P(shared birthdays among the 23 persons) = 1-0.4927 = 0.5073 ≈ 0.51
Let us ascertain the above fact using, Poisson distribution and before that, we will show that our usage of the Poisson distribution is valid, as our ‘Birthday Paradox’ case, can be treated as a Poisson experiment.
Poisson Experiment Here, we are particularly using Poisson distribution because it gives the probability of any number of independent events occurring in a fixed time (Weisstein). First, let us ensure that all the attributes of the Poisson experiment are satisfied in our case.
Here, the shared birthdays are treated as successes and distinct birthdays are treated as failures. Our region, in which the successes are calculated, is a room and is measured here, in terms of the capacity, (i.e.) number of persons present and the average number of successes (λ) is unknown here. Also, the probability of getting a shared birthday is proportional to the capacity of the room. That is, more the number of persons present, higher are the chances of success. Suppose, the region of consideration, is smaller than a room. Then, the probability of sharing of birthdays by some number of persons is zero obviously. Thus, our case satisfies all the points needed to become a Poisson Experiment and hence Poisson distribution can very well be applied here (Stattrek.com).
Let X denotes the number of shared birthdays and let k = 0, 1, 2,... and let λ be the parameter. Then,
When k = 0, we have P(X=0) as e-λ0! which is e-λ itself as 0! is 1. Hence, P(X>0) = 1-P(X=0) = 1- e-λ. Also, the problem says that at least 50% chance of sharing the birthdays should be possible. Hence, our inequality must be framed exceeding the probability of 0.5. Thus, P(X>0) > 0.5, is the necessary requirement.
That is, (1- e-λ ) > 0.5, this implies - e-λ...