Timeline for Calculating the Birthday Problem The Other Way Around
Current License: CC BY-SA 3.0
12 events
| when toggle format | what | by | license | comment | |
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| Nov 17, 2017 at 21:00 | history | edited | Henry |
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| Oct 9, 2013 at 18:44 | vote | accept | Daniel Scocco | ||
| Oct 9, 2013 at 18:42 | history | edited | Daniel Scocco | CC BY-SA 3.0 |
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| Oct 9, 2013 at 18:41 | answer | added | Daniel Scocco | timeline score: 1 | |
| Oct 9, 2013 at 18:06 | answer | added | Thomas Andrews | timeline score: 2 | |
| Oct 9, 2013 at 17:58 | history | edited | Daniel Scocco | CC BY-SA 3.0 |
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| Oct 9, 2013 at 17:54 | comment | added | Daniel Scocco | @Thomas Andrews, see my comment above. Once the individual probabilities get calculated correctly it should never go above 1. It's going above one cause the value of each probability is wrong. Once you get this right the correct way to do this is to add the individual probabilities, since it's an OR situation (or the second person share a birthday, or the third, or the fourth, etc ) | |
| Oct 9, 2013 at 17:52 | comment | added | Thomas Andrews | So is the probability zero? No. Is the probability ever greater than $1$? It will be if you add with $184$ people in the room. So, no, it doesn't have to be one or the other. The point is, your quest to find a "direct" way of solving this is doomed. | |
| Oct 9, 2013 at 17:51 | comment | added | Brendan W. Sullivan | It's step 3. It's simply not linear like that. P(person 3 shares bday with neither person 1 nor person 2) $\neq$ P(person 3 doesn't share bday with person 1) + P(person 3 doesn't share bday with person 2). | |
| Oct 9, 2013 at 17:50 | comment | added | Daniel Scocco | @Thomas, I agree with counting some cases twice. But you said I can't add the probabilities, but I don't agree. I might not be calculated each individual probability correctly, but once I do it to get the right answer I'll certainly need to add them, right? | |
| Oct 9, 2013 at 17:46 | comment | added | Thomas Andrews | What are you doing with those indidual probabilities? You can't add them, or you'll get a total greater than one when you have 183 people. You can't multiply them, because you'll get $0$. The problem is that you are counting some cases twice - if person $1$ and $2$ share a birthday, and $3$ and $4$ share a birthday, you've counted that case twice. | |
| Oct 9, 2013 at 17:41 | history | asked | Daniel Scocco | CC BY-SA 3.0 |