# Birthday probability of $k$ people and $n$ days a year such that at least $2$ people have the same bday

A certain planet has n days in one year. What is the probability that among $k$ people on that planet there are (at least) two who share their birthday?

My answer to this practice question is: There are $N^k$ probabilities/cases in total. We now have to count the favorable/conditional cases. There are ${n}\choose{k}$ ways to select no two people having birthday on the same day. The probability is then $p=1-\frac{{n}\choose{k}}{N^k}$.

But I'm sure the problem is more complicated than that...

• Hi James, please use MathJax: math.meta.stackexchange.com/questions/5020/… – 0rka Jan 28 '18 at 9:20
• got it thankss! – james black Jan 28 '18 at 9:42
• Have you tried applying pigeon-hole principle in this problem ? – spkakkar Jan 28 '18 at 9:49
• Apart from your mixing $N$ and $n$, it is no more complicated than that. Test it with $n=365$ and $k=23$ and see if you get a probability just over $0.5$. If $0 \lt n\lt k$ you can take ${n \choose k}=0$ so the probability is $1$ in that case – Henry Jan 28 '18 at 12:48
• what is the difference between big N and small n? – james black Jan 30 '18 at 8:57

While $k>n$, it is obviously a $100$ percent chance, so we shall not consider a case. So, as the answer says, the number of undesired outcomes is $({n\atop k})$, because we are choosing $k$ different days from $n$ to ensure that there is no overlapping of birthdays. Note the importance of the world different, as it means that there are $k$, and only $k$ days to have the birthdays, or else there would be overlaps.