Question

Be honest, have you ever met someone, only to be shocked that you share the same birthday? We have all been there at one point or another. In this week's discussion, you will be asked to address this event with a measure of probability.

See, each semester I ask each of my classes of about 32 face-to-face students the question, "What do you believe the probability is that there exists a common birthday(s) among those in this class?" After each person has the opportunity to consult his or her neighbor, I proceed to write down the guesses on the board. Those guesses often look something like this: 1%, 10%, 25%,50%, 67%, 90% 100%. As we all know, only one of these can be correct (if any).

Using your understanding of counting techniques and probability, investigate the answer to this question. What do you calculate to be the true probability of such an event? Provide any and all reasoning. PLEASE RECORD YOUR OWN INTUITION AND NOT ANSWERS YOU FIND ON THE INTERNET!! You are graded on the thoughtfulness of your response.

Answer 1

Solution:

We assume there are 365 days in a year and there are 32 students in the group.

Each student can have birthday on any of the 365 days in the year and it is also possible more than one student can have birthday on the same date – as an extreme case, all 32 students can have their birthdays on the same date. Thus, total number of possible ways 32 students can have their birthdays is 365³²

Out of these possible 365³² ways, the possibilities when no two students have their birthdays on the same date is 365 x 364 x …………. x 334 [i.e.,(365 - 31)]

[Reasoning: The first student can have birthday on any of the 365 dates, then the second student can have birthday on any of the remaining 364 dates only, the third student can have birthday on any of the remaining 363 dates only, and so on, the thirty second student can have birthday on any of the remaining 334 dates only]

So, probability no two students have their birthdays on the same date is:

(365 x 364 x …………. x 334)/365³²    

= 11184/9.84854E+81

= 1.1356E-78 which is negligibly very small – 0.00000 (78 zeros)114

So, then the probability that at least two students will have the same birthday

= 1 - 1.1356E-78

= 1.

Thus, interestingly, and perhaps unbelievably, it is certain that in a group of 32 persons at least two will have the same birthday.