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I've encountered an interesting probability problem that my little amount of knowledge does not help me to solve.

Select 1000 people from the past, and ignore the year. Also assume that the 365 days of the year is uniformly distributed. I think the probability that two people have the same birthday is $1-\prod^{1000-1}_{j=1} (1-\frac{j}{365}) $. This probability should be just $1$ I think.

But how about two people sharing the same birthday and death date? Should I calculate the two possibility separately and then multiply them?

Thanks

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  • $\begingroup$ The probability that at least $2$ people out of $1000$ share birthdays is exactly $1$. $\endgroup$ – AvZ Feb 26 '15 at 13:56
  • $\begingroup$ @AvZ thanks I just changed that. $\endgroup$ – verticese Feb 26 '15 at 13:56
  • $\begingroup$ I think the probability for this should be $\frac{200}{73\times 365}$ for this $\endgroup$ – AvZ Feb 26 '15 at 13:59
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As it was already pointed out, the probability that at least two people out of 1000 have birthday in one day is exactly 1, since there are only 366 possible dates. Your solution $$1-\prod^{1000-1}_{j=1} \left(1-\frac{j}{366}\right)$$ is not technically correct because even the product is $0$, the term $ \left(1-\frac{j}{366}\right)$ in not a probability for $j > 366$ any more.

If birthday and death day are independent and both uniformly distributed then joint distribution of these two dates is uniform with $366^2$ possible values. Now you can use the same approach so the probability that there are not two people having same birth and death day is: $$1-\prod^{1000-1}_{j=1} \left(1-\frac{j}{366^2}\right)$$

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  • $\begingroup$ $365$ days in a year? $\endgroup$ – graydad Feb 26 '15 at 16:03
  • $\begingroup$ Don't forget the leap years. $\endgroup$ – iiivooo Feb 26 '15 at 16:05
  • $\begingroup$ Right, which makes the average length of a year $\approx 365.25$ days. I suppose it is up to the mathematician whether they want to round up or down. But if you read the wiki page about the birthday problem, every single equation uses $365$. en.wikipedia.org/wiki/Birthday_problem $\endgroup$ – graydad Feb 26 '15 at 16:09
  • $\begingroup$ Interesting, wiki page doesn't round it, they just ignore Feb. 29 :-) You are right, thanks. I hope author can count it now only using 365 dates. $\endgroup$ – iiivooo Feb 26 '15 at 16:20
  • $\begingroup$ Nah your work is still fine :) The wiki page implies that using $366$ is statistically insignificant by saying "...there are 366 possible birthdays, including February 29. However, 99.9% probability is reached with just 70 people... " $\endgroup$ – graydad Feb 26 '15 at 16:25

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