Timeline for Explain the Birthday Paradox
Current License: CC BY-SA 3.0
22 events
| when toggle format | what | by | license | comment | |
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| Apr 28, 2020 at 1:51 | answer | added | Cotton Headed Ninnymuggins | timeline score: 0 | |
| Jan 22, 2019 at 4:48 | answer | added | user53259 | timeline score: 0 | |
| Nov 17, 2017 at 20:46 | history | edited | Henry |
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| Nov 23, 2015 at 23:18 | history | edited | user147263 | CC BY-SA 3.0 |
deleted 218 characters in body
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| Jun 18, 2014 at 13:51 | answer | added | johannesvalks | timeline score: 2 | |
| Jun 18, 2014 at 13:37 | vote | accept | Nib | ||
| Jun 18, 2014 at 13:34 | comment | added | Nib | I'm so sorry @Raskolnikov, your'e right | |
| Jun 18, 2014 at 13:32 | comment | added | Raskolnikov | It's not so much a mistake as it is nonsense. You just divided the amount of ways you can select two different persons from 23 and then divided that by 365^2. Why does this have anything to do with the birthday problem? What happens with the other 21 persons? | |
| S Jun 18, 2014 at 13:30 | review | First posts | |||
| Jun 18, 2014 at 13:31 | |||||
| S Jun 18, 2014 at 13:30 | review | Close votes | |||
| Jun 18, 2014 at 14:19 | |||||
| Jun 18, 2014 at 13:29 | comment | added | Nib | @Raskolnikov, pardon me if I've made any mistake in that math, but wold you please explain exactly what mistake I've made ? | |
| Jun 18, 2014 at 13:27 | comment | added | Raskolnikov | What you computed is the probability of selecting two different persons out of 23 in a draw with replacement. This has nothing to do with the birthday paradox. What you should compute is the complement of the probability of drawing 23 birthdays out of 365 days with all birthdays different. | |
| Jun 18, 2014 at 13:24 | answer | added | puru | timeline score: 6 | |
| Jun 18, 2014 at 13:22 | comment | added | Nib | And also, @Henry, the wikipedia article is saying the same thing as I do till a certain point, after which, I can't understand at all what they're trying to tell me | |
| Jun 18, 2014 at 13:20 | history | edited | Nib | CC BY-SA 3.0 |
Pointed out I'm 13
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| Jun 18, 2014 at 13:19 | comment | added | Nib | Alo, doesn't the birthday paradox state that 2 ppl have the same birthday out of group of 23 ppl, whereas some of the commentators have pointed that the question's a duplicate of one which says 3 ppl have common birthdays, also, in that question, the number of ppl taken in room are 30, not 23, so, that means that my question and that question are quite different. | |
| Jun 18, 2014 at 13:17 | comment | added | Nib | @henry, Ummm, I'm not sure if I can understand those solutions, but here are my calculations: We know that there are 253 possible combinations in the 23 ppl group, now, we also know that, each of them has birthday date A is 1/365, so, probability that two people share their birthday is 1/365^2, and so, any one pair sharing their birthday, the probability will be, 253/365^2, which of course, is very less, hence I believe this paradox is false. | |
| S Jun 18, 2014 at 13:17 | history | suggested | Mario Krenn | CC BY-SA 3.0 |
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| Jun 18, 2014 at 13:15 | review | Suggested edits | |||
| S Jun 18, 2014 at 13:17 | |||||
| Jun 18, 2014 at 13:13 | comment | added | Mario Krenn | It has already been answered here: math.stackexchange.com/questions/25876/… | |
| Jun 18, 2014 at 13:12 | comment | added | Henry | You could read the Wikipedia article en.wikipedia.org/wiki/Birthday_problem or show us your calculations | |
| Jun 18, 2014 at 13:10 | history | asked | Nib | CC BY-SA 3.0 |