Answer to Question #131268 in Statistics and Probability for Raymond

Question

Consider a random group of kpeople who tested positive for COVID-19. Their birthdays have been written down for statistical purposes regarding COVID-19 control measures.
a.Determine the least k such that it is more likely than not that two people or more have the same birthday.
b.Determine the smallest k such that it is more likely than not that two or more people share your birthday.

Answer 1

We know that the sum of the probability that an event will happen and the probability that the event won't happen is always 1. If we can work out the probability that no two people will have the same birthday, we can use this rule to find the probability that two people or more will share a birthday:

• P(event happens) + P(event doesn't happen) = 1
• P(two or more people share birthday) + P(no two people share birthday) = 1
• P(two or more people share birthday) = 1 - P(no two people share birthday).

The first person can have any birthday. The second person's birthday has to be different. There are 364 different days to choose from, so the chance that two people have different birthdays is 364/365. That leaves 363 birthdays out of 365 open for the third person.

To find the probability that both the second person and the third person will have different birthdays, we have to multiply:

   (365/365) * (364/365) * (363/365) = 132,132/133,225, which is about 99.18%.

If we want to know the probability that four people will all have different birthdays, we multiply again: (364/365) * (363/365) * (362/365) = 47,831,784/ 48,627,125, or about 98.36%.

We can keep on going the same way as long as we want. A formula for the probability that n people have different birthdays is:

$\frac{365-1}{365} * \frac{365-2}{365} * \frac{365-3}{365} * . . . * \frac{365-n+1}{365}$

We could multiply a term $\frac{365}{365}$ to the above equation (in effect it's equivalent to multiplying the above equation by 1)

So we now have, $\frac{365}{365}*\frac{365-1}{365} * \frac{365-2}{365} * \frac{365-3}{365} * . . . * \frac{365-n+1}{365}$

This is equivalent to $\frac{_{365}P_n}{365^n}$

That's the same as $\frac{365!}{(365-n)! * 365^n}$

We have to determine the least k such that it is more likely than not that two people or more have the same birthday. This is equivalent to saying that we have to determine a k such that: the probability of having two or more people with the same birthday is at least 50%

We know the probability that no two people out of k people have birthdays on the same day is $\frac{365!}{(365-k)! * 365^k}$

Thus probability that two or more people out of k people have birthdays on the same day is $1-\frac{365!}{(365-k)! * 365^k}$

We can find the desired k by trial and error:

When k = 10: $1-\frac{365!}{(365-k)! * 365^k} = 1-\frac{365!}{(365-10)! * 365^{10}}=0.1169$

We need the desired probability to be just above 50%. So we try with a higher number

When k = 20: $1-\frac{365!}{(365-k)! * 365^k} = 1-\frac{365!}{(365-20)! * 365^{20}}=0.4114$

When k = 22: $1-\frac{365!}{(365-k)! * 365^k} = 1-\frac{365!}{(365-22)! * 365^{22}}=0.4757$

When k = 23: $1-\frac{365!}{(365-k)! * 365^k} = 1-\frac{365!}{(365-23)! * 365^{23}}=0.5073$

Thus k=23, we get the probability that two or more people will have birthdays on the same day just above 50%.

So our desired value of k is 23.