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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...

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  • $\begingroup$ 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
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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.

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  • $\begingroup$ 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

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