Timeline for Birthday Problem for 3 people
Current License: CC BY-SA 3.0
14 events
| when toggle format | what | by | license | comment | |
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| Apr 13, 2017 at 12:20 | history | edited | CommunityBot |
replaced http://math.stackexchange.com/ with https://math.stackexchange.com/
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| Sep 7, 2013 at 4:55 | comment | added | Thomas | Interesting, I wonder how many people you need for four people to share a birthday with 50% probability. If we made a graph of the function f(n) of the number of people needed to be 50% sure that n people have the same birthday, what would it look like? | |
| Sep 7, 2013 at 4:52 | vote | accept | Thomas | ||
| Sep 6, 2013 at 9:27 | history | edited | Caleb Stanford | CC BY-SA 3.0 |
added update
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| Sep 6, 2013 at 9:12 | comment | added | Caleb Stanford | @Henry, Using this answer, I calculate the probably for 84 people to be $.464549...$. You're right; the approximation is further off than I thought. But it's still reasonably accurate. | |
| Sep 6, 2013 at 7:38 | comment | added | Henry | I think that with $84$ people in a room, the probability of three people sharing a birthday seems to be lower than $0.47$ (and for $83$ people still lower) | |
| Sep 6, 2013 at 6:21 | comment | added | Thomas | Ok, that makes more sense. I was thinking 46 was a bit small. Especially given that the probability for 30 people is less than 0.05. | |
| Sep 6, 2013 at 6:11 | comment | added | Caleb Stanford | @Thomas There was an error in my algebra. I've fixed it now, but it's 83 people in stead of 46. | |
| Sep 6, 2013 at 6:10 | history | edited | Caleb Stanford | CC BY-SA 3.0 |
Fixed algebra mistake
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| Sep 6, 2013 at 6:07 | comment | added | Thomas | Interesting that it is about double of the amount needed to ensure 50-50 for two people having the same birthday. | |
| Sep 6, 2013 at 6:03 | comment | added | Caleb Stanford | This comment in particular: math.stackexchange.com/questions/25876/… | |
| Sep 6, 2013 at 6:02 | comment | added | Caleb Stanford | @vantonio1992 check out Byron's answer in the linked question. $1/365^2$ is the chance of a given three people having the same birthday. | |
| Sep 6, 2013 at 6:00 | comment | added | vantonio1992 | Shoudn't the probability be over $365^3$ instead of $365^2?$ | |
| Sep 6, 2013 at 5:58 | history | answered | Caleb Stanford | CC BY-SA 3.0 |