Question

Probability question

calculate the probability that, in a group of 4 people, at least two have the same birthday.

Answer 1

one method is 1-(no one has the same birthday) = 1 - (354* 364*363*362)/365^4

however, when I tried to use another method to calculate all the cases directly, the result is different.. How to use another method to prove that the first method is correct?

Thank you

Answer 2

probability that exactly 2 have the same bday: = (4choose2)*365*364*363/365^4

probability that exactly 3 have the same bday: = (4choose3)*365*364/365^4

probability that all 4 have the same bday: = 365/365^4

probability that 2 share one birthday and the other 2 share another: = (4choose2)*365*364/365^4

Answer 3

oh! thank you!! I didn't count the 2/2 birthday situation.!

Answer 4

still might be something wrong... adds up to more than 100 %

Answer 5

yah the number is a little bit off....

Answer 6

the problem can be corrected if we get rid (4choose3) for 3 people having the same birthday .. but don't know how we can.

Answer 7

P[at least one] = 1 - P[none]

What is the probability that none of them have the same birthday? Suppose we have four places, each of which a birthday is assigned to it. What is the probability that the first person selected has a different birthday than anyone that came before? 1.0, since he's the first. What about the second one? Well, one of the days is assigned to the first person, so (365 - 1) / 365. Third person? Well, two of the days are already assigned, so his day must be on a different day than both of those days. (365 - 2) / 365. Finally, the fourth one is (365 - 3) / 365.

Now, take the product of this and subtract it from 1 => 1 - (365 * 364 * 363 * 362) / 365^4. That is your answer.

In general, if given n people:
\[\prod_{i=0}^{n-1} \frac{ 365 - i }{ 365 } = \frac{365!} {365^n(365-n+1)!}\]

Answer 8

Correction: take that general formula above, call it F(n), and the correct answer is 1 - F(n).

Answer 9

If we consider two people, the probability that one of the two doesn't have the same birthday as the other is 1 - 1/365 = 364/365. The probability that those two have different birthdays and a third person doesn't have the same birthday as either of them is:
\[\large \frac{364}{365}\times\frac{363}{365}\]
Moreover, the probability that those three don't share a birthday and a fourth person does not have the same birthday as any of the first three is:
\[\large P(0\ of\ 4\ share)=\frac{364}{365}\times\frac{363}{365}\times\frac{362}{365}\]
Therefore the probability that, in a group of 4 people, at least two have the same birthday is given by:
\[\large P(at\ least\ 2\ share)=1-P(0\ of\ 4\ share)\]