After the positive response to the recent brain teaser on medical tests, I thought I’d recycle another statistical problem from my university days. This one kept most of us guessing for a while, but once the lecturer pointed us in the right direction most of us got there in the end.
A university course is promoted across different disciplines, with up to 400 students being scheduled to attend the lecture that is held every Tuesday morning. Because the university is very strict on entry criteria, each of the 400 students in the year was born between 1 September 1992 and 31 August 1993.
Assuming the distribution of birthdays across the year is uniform, how many people need to attend the lecture before the probability of two of them being born on the same day exceeds 50%?
Please put your answers in the comments box below and I will come back with the answer next week.
Best stab I can come up with is 183.
Now, I came across this one ages ago and someone told me that it was 23 (funny what you remember) , but no idea how that could possibly be right.
Apologies for the really delayed response here, holidays etc.
Jeff is absolutely right, it's 23. The best way to think about it is from the reverse of the question – ie, with X people in a room, what is the probability they will all have different birthdays. With 2 people, the probabilty will be 364/365, with three we will have to knock another day off the year and combine it with the result from two people (so the probability will be 363/365*364/365), etc etc. The resulting probabilty collapses quite quickly so that, with 23 people in the room, there is a less than 50% chance that all of them will have different birthdays. And, hence, a greater than 50% chance that two of them will have the same one.