Use Poisson approximations to investigate the following types of coincidences. T
ID: 3435244 • Letter: U
Question
Use Poisson approximations to investigate the following types of coincidences. The usual assumptions of the birthday problem apply, such as that there are 365 days in a year, with all days equally likely.
(a) How many people are needed to have a 50% chance that at least one of them has the same birthday as you?
(b) How many people are needed to have a 50% chance that there are two people who not only were born on the same day, but also were born at the same hour (e.g., two people born between 2 pm and 3 pm are considered to have been born at the same hour).
Explanation / Answer
a. Let I be the first person and a second person comes. The probability that the second person has a different birthday from me is 364/365. When a third person comes, the probability that he has a different birthday to the first two is 363/365. Thus, the probability that there are now no duplicate birthdays is (364/365)*(363/365). After the fourth one comes in, the probability that he has a different birthday to the first three is 362/365. The probability that there are now no duplicate birthdays is (364/365)*(363/365)*(362/365). and so on. It can be seen that the probability that there are no duplicate birthdays falls below 50% when the 23rd person arrives. Therefore, 23 people are needed to have a 50% chance that at least one of them has the same birthday as I.
b. There are 365 x 24 = 8760 hours in an year. Now same as above, Let a second person comes in. The probability that the second person has a different birth hour from me is 8759/8760. When a third person comes, the probability that he has a different birth hour to the first two is 8758/8760. Thus, the probability that there are now no duplicate birth hours is (8759/8760)*(8758/8760). After the fourth one comes in, the probability that he has a different birthday to the first three is 8757/8760. The probability that there are now no duplicate birth hours is (8759/8760)*(8758/8760)*(8757/8760). and so on. It can be seen that the probability that there are no duplicate birth hour falls below 50% when the 111th person arrives. Therefore, 111 people are needed to have a 50% chance that at least one of them has the same birth hour as I.
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