Timeline for Group of $r$ people at least three people have the same birthday?
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24 events
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| Nov 17, 2017 at 21:10 | history | edited | Henry |
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| Apr 13, 2017 at 12:21 | history | edited | CommunityBot |
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| Mar 28, 2017 at 12:52 | history | edited | Mithlesh Upadhyay | CC BY-SA 3.0 |
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| Dec 20, 2015 at 14:49 | comment | added | robjohn♦ | Note that the first two terms of 3. match with the first two terms of my answer. That is, $1-P(\text{no one has the same birthday})=1-\frac{365!}{(365-n)!}$. However, $P(\text{any $2$ have the same birthday})$ is unclear. First, it sounds as if it might include a case where $3$ people have the same birthday. Secondly, in 3., the third term is added when it should be subtracted (after changing it to be the probability of at least one pair, but no triples). | |
| Dec 20, 2015 at 7:30 | vote | accept | Mithlesh Upadhyay | ||
| Dec 20, 2015 at 7:06 | comment | added | anon | Consider the following scenario: persons A and B have birthday X while persons C and D have birthday Y. This sort of situation is not accounted for if you're only counting ways there can be precisely one pair of persons sharing a birthday. It is possible for multiple pairs of persons to share birthdays without any three sharing a birthday. | |
| S Dec 20, 2015 at 6:12 | history | bounty ended | Mithlesh Upadhyay | ||
| S Dec 20, 2015 at 6:12 | history | notice removed | Mithlesh Upadhyay | ||
| Dec 20, 2015 at 6:05 | history | edited | Mithlesh Upadhyay | CC BY-SA 3.0 |
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| Dec 19, 2015 at 13:41 | history | edited | Mithlesh Upadhyay | CC BY-SA 3.0 |
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| Dec 18, 2015 at 8:45 | history | edited | Mithlesh Upadhyay | CC BY-SA 3.0 |
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| S Dec 14, 2015 at 4:35 | history | suggested | Olorun |
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| Dec 14, 2015 at 4:19 | review | Suggested edits | |||
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| Dec 14, 2015 at 4:18 | answer | added | robjohn♦ | timeline score: 14 | |
| Dec 13, 2015 at 17:14 | answer | added | Markus Scheuer | timeline score: 6 | |
| Dec 13, 2015 at 15:19 | history | tweeted | twitter.com/StackMath/status/676058757764390914 | ||
| Dec 13, 2015 at 13:49 | comment | added | zhoraster | There is no correct answer among 1-4. However, 3 answers a different question. Namely, call a pair $\{i,j\}$ a collision if persons $i$ and $j$ have the same birthday. Then the probability that there are at least two collisions is given by 3. | |
| S Dec 13, 2015 at 6:13 | history | bounty started | Mithlesh Upadhyay | ||
| S Dec 13, 2015 at 6:13 | history | notice added | Mithlesh Upadhyay | Improve details | |
| Nov 24, 2015 at 16:49 | history | edited | Michael Hardy | CC BY-SA 3.0 |
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| Nov 24, 2015 at 16:46 | history | edited | Mithlesh Upadhyay | CC BY-SA 3.0 |
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| Nov 24, 2015 at 16:00 | answer | added | Pieter21 | timeline score: 3 | |
| Nov 24, 2015 at 15:36 | comment | added | Peter | The problem is quite complicated. First of all, we have to remove the case of pairwise different birthdays. But then, we have to remove the case of $1$ pair , $2$ pairs , ... $[r/2]$ pairs with the same birthday (but every pair has another birthday). | |
| Nov 24, 2015 at 15:14 | history | asked | Mithlesh Upadhyay | CC BY-SA 3.0 |