What is the probability that of 25 randomly selected students, no two share the same birthday?
The probability is (approximately) 43.13%.
Ah, this question! It's an old one but a good one!
(We ignore leap years and use a 365-day year.)
Let's start with one student in the group. Since there's nobody else's birthday to avoid, all 365 days are allowed. The probability that we haven't yet found two people with the same birthday is
Now move up to two students. The probability that this 2nd student's birthday is not already one of the group's is
Up to 3 students. The probability of a third unique birthday is now
#"P"("1st unique") xx "P"("2nd unique")xx"P"("3rd unique")#
#= 365/365 xx 364/365 xx 363/365#
When we add a 4th person, you might start to see the pattern. Assuming all previous conditions have been met, the probability of 4 people having unique birthdays is
#"P"("1st") xx "P"("2nd") xx "P"("3rd") xx "P"("4th")#
#=365/365 xx 364/365 xx 363/365 xx 362/365#
Adding another student has caused the probability to go down, as we should expect.
The fraction above can be concisely written as
#"P"("4 unique b-days")=(365!//361!)/(365^4)#
#color(white)("P"("4 unique b-days"))=(365!)/((365-4)!xx365^4)#
This gives us a general form for the probability of having a group of
#"P"(n" unique b-days")=(365!)/((365-n)!xx365^n)#
To find the probability of having 25 students with unique birthdays, we plug in
#"P"("25 unique b-days")=(365!)/((365-25)!xx365^25)#
If you want to do the calculation by hand, you can. With computer software, we get an answer of about
This means that by the time a group reaches 25 students, there's already more than a 50% chance that two of them have the same birthday. (The minimum number of students required to make the chance greater than 50% is 23.)
Pretty neat how that's all it takes!