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...

  • $\begingroup$ Hi James, please use MathJax: math.meta.stackexchange.com/questions/5020/… $\endgroup$ – 0rka Jan 28 '18 at 9:20
  • $\begingroup$ got it thankss! $\endgroup$ – james black Jan 28 '18 at 9:42
  • $\begingroup$ Have you tried applying pigeon-hole principle in this problem ? $\endgroup$ – spkakkar Jan 28 '18 at 9:49
  • $\begingroup$ 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 $\endgroup$ – Henry Jan 28 '18 at 12:48
  • $\begingroup$ what is the difference between big N and small n? $\endgroup$ – 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.

Then I presume the rest would be easily understandable, because it is just finding the fraction of possible ways over the total ways.

  • $\begingroup$ so is my answer incorrect? $\endgroup$ – james black Jan 30 '18 at 8:56
  • $\begingroup$ Yes, you can accept this answer and upvote $\endgroup$ – QuIcKmAtHs Jan 30 '18 at 9:29
  • $\begingroup$ I mean, the answer above is correct @jamesblack $\endgroup$ – QuIcKmAtHs Jan 30 '18 at 10:58
  • $\begingroup$ That was my answer and I was just asking for a confirmation (because i thoguht that the answer is too easy) haha so my answer is good $\endgroup$ – james black Jan 31 '18 at 5:50
  • $\begingroup$ you lied this was wring...not thee answer $\endgroup$ – james black Feb 4 '18 at 10:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.