What is the probability of 3 people in a room of 7 having the same birthday ?
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2$\begingroup$ Is that "exactly 3" or "at least 3" $\endgroup$– Graham KempCommented Nov 18, 2014 at 3:22
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4 Answers
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The first answer is ${7 \choose 3}\frac 1{365^2}=\frac 7{26645}\approx \frac 1{3807}$ You pick the three that will match, let the first have any birthday, then force the other two to match that. The actual probability is lower because this counts cases with four matching birthdays four times (and those with more than four even more) and cases with two sets of three twice. Since the starting probability is small, these corrections are very small. I leave the corrections as an exercise.
answered Nov 18, 2014 at 3:26
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$\begingroup$ I'm writing up the "correction" terms right now, and they're a bit annoying to explicitly find, because of so many cases. $\endgroup$ Commented Nov 18, 2014 at 3:27
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$\begingroup$ @ZubinMukerjee: I agree, that's why I left them as an exercise. The logic is not too difficult, but it is a lot of work. The basic approach is here and the numeric impact is quite small. $\endgroup$ Commented Nov 18, 2014 at 3:29
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$\begingroup$ I wonder if there's an elegant way to find the exact answer ... I have no idea $\endgroup$ Commented Nov 18, 2014 at 3:33
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It depends on whether you mean exactly three people or at least three people. I'll answer the former. The probability is going to be a large, messy fraction, whose numerator is going to be the number of "successes" and whose denominator will be the total number of possibilities.
The probability of exactly three people in a room having the same birthday is equal to the number of $7$-tuples of integers between $1$ and $365$ (ignoring leap years) with the property that exactly $3$ of them are the same, divided by the number of possible $7$-tuples.
The number of possible $7$-tuples is simply $$365^7$$
and this is our denominator.
There are $$\binom{7}{3}=35$$ ways to choose the three people who have identical birthdays. Once we've chosen them, there are $365$ ways to choose which day is the special day. Then, having chosen that, there are $$364^4$$ ways to choose the birthdays of the remaining $4$ people. Note the $364$ since we don't want any of those four to have the same birthday as the selected three.
This gives us a preliminary number of successes (our numerator) as $$\binom{7}{3}\cdot 365 \cdot 364^4$$
But we've counted a few more possibilities than we should. The remaining four people cannot all have the same birthday, since that would lead to there being more than three with the same birthday (as opposed to exactly three). We've counted those possibilities once when we shouldn't have at all. Also, we've doubly counted cases in which exactly three of the remaining four people have identical birthdays.
Case 1: Remaining four people all have the same birthday.
There are exactly $364$ overcounts from this case, one corresponding to each of the remaining days of the year.
Case 2: Of the remaining four people, exactly three have the same birthday.
There are four ways to choose the one with the different birthday. Given this, there are $364$ ways to choose his/her birthday, and then $363$ ways to choose the shared birthday of the other three. So in this case, we have
$$4 \cdot 364 \cdot 363$$ overcounts.
The number of total successes, taking into account the two overcounting cases above, is $$35 \cdot 365 \cdot 364^4 - 4 \cdot 364 \cdot 363 - 364$$
Dividing this by the number of total possibilities, we get our final answer:
$$\frac{35 \cdot 365 \cdot 364^4 - 4 \cdot 364 \cdot 363 - 364}{365^7}=\frac{224267551925508}{863078009304453125} \approx 0.0002598$$
It's not very likely.
answered Nov 18, 2014 at 3:42
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For "exactly 3", select a birthday and 3 people to share it, and for the 4 others, select any remaining birthdays such that 4 don't share, and half of the cases where another 3 share a different birthday (to avoid over counting). Divide by ways to select any birthday for seven people.
$$\mathsf P(C=3) = \frac{365{7\choose 3}(364^4-363-364{4\choose 3}363/2)}{365^7}$$
Can you now do "at least 3"? (Hint: It might be easier to use the Law of Complements.)
answered Nov 18, 2014 at 3:40
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$\begingroup$ Wow, I noticed that my final answer was equivalent to yours (just distributed out), but had a missing $365$ in the first product. Anyway, I fixed it. Turns out I had written the idea corresponding to the $365$ but simply forgot to put it in my answer. How do you keep track of so many cases? +1 $\endgroup$ Commented Nov 18, 2014 at 3:51
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$\begingroup$ It seems a bit messy ... is there a nicer way to do the problem? $\endgroup$ Commented Nov 18, 2014 at 3:56
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$\begingroup$ Not really, @ZubinMukerjee . It's a pretty messy problem. $\endgroup$ Commented Nov 18, 2014 at 3:59
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$\begingroup$ @GrahamKemp , i am not getting select any remaining birthdays such that 4 don't share, and half of the cases where another 3 share a different birthday (to avoid over counting).,please ellaborate :) $\endgroup$– lauraCommented May 25, 2017 at 5:26
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I was in a surgery last week with 5 hospital personnel and one patient. 6 people in room. Me, the anesthetist and the patient all shared September 13 as our birthday. Even greater odds against.
