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Yes it's another birthday question, no I cannot find this variation anywhere else! Here goes:

In a class, 20 students sit at the front row and another 280 sit in the seats behind them. The instructor plays the classic birthday problem; he asks the students one by one about their birthdays (not the year) and lets any student in the room interrupt if they hear their birthdays heard. Find the probability that no coinciding birthdays occur (within the first row and between the first row and the rest of the classroom) by the time all 20 students in the first row have been asked. Assume nobody is born on February 29th and instructor does not reveal his own birthday.

MY PROGRESS:

So no Feb 29th means using 365 days. Also assuming a person actually does shout out if they hear their birthday announced.

The issue I am having is that the 280 students behind the first row CAN have overlapping birthdays with each other, just not with anyone in the front row.

Breaking the question down:

If we simply had the 20 in the front row only then the probability would be $\frac{365}{365}$ x $\frac{364}{365}$ x ... x $\frac{346}{365}$ which would give us 0.589 (3 d.p.)

Then if nobody in the front row shares a birthday, we have 20 of the 365 possible days 'used up,' and the remaining 280 students can have birthdays on any of the remaining 345 days, and ARE allowed to overlap with each other. What I am not sure is how to proceed from here. Any help much appreciated!

UPDATE: I think I have the solution but not feeling confident about it so comments welcome:

We need BOTH that the 20 on the front row don't share birthdays and also that given the front row don't share birthdays that the remaining 280 students don't share birthdays with anyone in the front row.

If nobody in the front row shares a birthday, then there are 20 days in the year that now have someone whose birthday is not on that day. We need to fit the other 280 students on to the remaining 365-20=345 days that have no birthdays so far.

For 365 days and 280 students there are 365*280=102,200 ways in which these people can have birthdays. The good cases are where they have birthdays on the 345 remaining days which gives 345*280=96,600. So the probability of the good cases occurring = 96,600/102,200 = 0.945 (3 d.p.).

Finally we multiply this result by our earlier result (probability) that the 20 in the front row did not share birthdays, so 0.589*0.945 gives us 0.557 (3 d.p.)

Does this seem right?!

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  • $\begingroup$ Not following the last bit. the condition that nobody in the back $280$ matches any of the front $20$ means, as you say, that they have $345$ days to choose from. So, any of those rear folk can be any of those days, making a whopping $345^{280}$ cases. $\endgroup$ – lulu Nov 3 '16 at 23:50
  • $\begingroup$ Ok thank you, so does that mean the probability for none of the back 280 matching the front $345^{280}$ / $365^{280}$ ? $\endgroup$ – Fkins Nov 3 '16 at 23:54
  • $\begingroup$ Well, not quite. That fraction is the conditional probability for the event you want given that no two people in the front row match. (that's what you actually want anyway). $\endgroup$ – lulu Nov 3 '16 at 23:57
  • $\begingroup$ Ok thanks so the final answer would then be 0.589 * $345^{280}$ / $365^{280}$ (with the 0.589 part being the probability the 20 at the front don't share birthdays) ??? $\endgroup$ – Fkins Nov 3 '16 at 23:59
  • $\begingroup$ I didn't check the first part but that number looks about right. the algebra looks good. $\endgroup$ – lulu Nov 4 '16 at 0:03
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You were doing great!

The probability that the first $20$ students don't have repeated birthdays is $\frac{365!}{345!365^{20}}\approx 0.557$.

The second thing that needs to happen is that none of the birthdays of the remaining $280$ students is one of those $20$ birthdays, and the probability of this is $(\frac{345}{365})^{280}\approx 1.4\times 10^{-7}$.

So the probability that both of these happen is $\frac{365!}{345!365^{20}}(\frac{345}{365})^{280}\approx 1.4\times 10^{-7}\approx 8.26297 × 10^{-8}$

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  • $\begingroup$ Ok makes perfect sense, thanks for this! $\endgroup$ – Fkins Nov 4 '16 at 0:23

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