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At least how many persons are required in a group so that the probability of two persons were born on the same day of the week is $0.5$?

I simplified it so there's only 365 days. Next find the probability that no one is born on the same day of the week then subtract by $1$ to get the solution.

But the answer is wrong, so I'm kind of out of ideas.

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  • $\begingroup$ Don't you mean at least $0.5$? $\endgroup$ – bof Mar 28 '15 at 2:54
  • $\begingroup$ I'm not completely sure what you mean by simplifying it, but note that your question is concerned with days of the week, that is, there are 7 different possibilities. There's probably a way to tackle the problem considering all $365$ days of the year, but the issue is that $365$ is not a multiple of $7$, so there are more of one day than there are of the other, which must be taken into account when computing probabilities. The way you've phrased your question its probably easier to assume all $7$ days are uniformly distributed with individual probabilities $1/7$. $\endgroup$ – Reveillark Mar 28 '15 at 2:57
  • $\begingroup$ @Reveillark thank you, I think my mistake is I used the 365 days birthday problem way to solve but the question asks about only for the 7 days. $\endgroup$ – Hong Yuan Mar 28 '15 at 3:00
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    $\begingroup$ @HyhY Try next time to add what you tried to your post, otherwise people sometimes downvote your question. I upvoted your question, because I see you did effort. But best is to add that to your post immediatly to not get downvoted. ;) $\endgroup$ – Pedro Mar 28 '15 at 3:03
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    $\begingroup$ @Pedro ok, thank you for advice. $\endgroup$ – Hong Yuan Mar 28 '15 at 3:06
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Given a group of $n$ people, the sample space of days of the week in which they were born is $\Omega = \left \{ 1, ..., 7 \right \}^n$, which give $\left | \Omega \right |=7^n$.

Let $A$ be the event that at least two of them were born on the same day. Then $P(A)=1-P(A^c)\ $. Now $A^c$ is the set that none of them share a birthday, so you need to pick $n$ different days out of $7$. This can be done in ${7 \choose n}$ ways. Assuming the people are ordered, given a set of $n$ days we have to consider all permutations, hence:

$$P(A^c)={n!{7 \choose n} \over 7^n}$$

Notice that if $n≥8$ then obviously at least two share a birthday, which agrees with our formula since $P(A^c)=0$ for $n>7$. Now use the formula above to compute the probabilities for $2≤n≤7$.

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This is a variant on a very classic question known as the "Birthday Paradox." Instead of trying to calculate the probability that two people have the same birthday, consider the probability that two people have different birthdays for a given size group. If you can get this probability under .5, then you know the probability of two people having the same birthday is greater than .5

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  • $\begingroup$ The Birthday Paradox is concerned with all $365$ days of the year, while this question asks about days of the week. I'm not sure it applies here. $\endgroup$ – Reveillark Mar 28 '15 at 2:58
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    $\begingroup$ Same concept applied to a different domain no? 7 possible birthdays instead of 365 (6 if leap year considered) $\endgroup$ – Mike Andrejkovics Mar 28 '15 at 2:59
  • $\begingroup$ Hmmm, maybe I was too quick to dismiss it, its probably (no pun intended) the same concept scaled down. The paradox factor is kind of reduced since the numbers are smaller and easier to work with. You need $8$ people rather than $366$ (not counting leap years) to get to pigeonhole status. $\endgroup$ – Reveillark Mar 28 '15 at 3:05
  • $\begingroup$ That's a good point- with only 7 possible days the result is hardly counterintuitive, but the math is essentially the same $\endgroup$ – Mike Andrejkovics Mar 28 '15 at 3:14

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