Timeline for How to generate a random number between 1 and 10 with a six-sided die?
Current License: CC BY-SA 3.0
43 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 21, 2017 at 7:41 | comment | added | Sentinel | Use all but 2 of the edges, colouring individual edges to represent one to ten, rerolling when an unmarked edge is closest to the observer | |
| Apr 20, 2017 at 12:54 | comment | added | daniel | Following on from the base 6 idea, discard the first roll unless it is a 1 or a 2, discard the second roll if the first was a 2 and the second was a 5 or a 6. Your answers is 36 times the first roll plus the second roll mod 10. Or discard all rolls that are not a 6, use von Neumanns extractor on the results, when you have 4 binary results convert to decimal, if the result is above 10 repeat from the start. | |
| Jun 18, 2015 at 20:28 | answer | added | celtschk | timeline score: 2 | |
| S Jun 18, 2015 at 14:02 | history | bounty ended | MJD | ||
| S Jun 18, 2015 at 14:02 | history | notice removed | MJD | ||
| Jun 18, 2015 at 3:27 | answer | added | Michael Anderson | timeline score: 1 | |
| Jun 17, 2015 at 13:10 | answer | added | Xoff | timeline score: 1 | |
| Jun 17, 2015 at 12:41 | answer | added | Christian Blatter | timeline score: 1 | |
| Jun 14, 2015 at 9:33 | answer | added | wythagoras | timeline score: 0 | |
| Jun 12, 2015 at 10:14 | answer | added | Shubham Avasthi | timeline score: 0 | |
| Jun 12, 2015 at 1:07 | answer | added | Peter | timeline score: 0 | |
| S Jun 11, 2015 at 13:53 | history | bounty started | MJD | ||
| S Jun 11, 2015 at 13:53 | history | notice added | MJD | Reward existing answer | |
| Jun 11, 2015 at 3:02 | history | tweeted | twitter.com/#!/StackMath/status/608831627217743872 | ||
| Jun 9, 2015 at 1:45 | answer | added | mathreadler | timeline score: 0 | |
| Jun 8, 2015 at 15:17 | comment | added | Julia Hayward | Go to gaming shop. "Hello, I'd like a 1d10, would you take this lovely 1d6 in part exchange?" :) | |
| Jun 8, 2015 at 14:20 | answer | added | mathmandan | timeline score: 1 | |
| Jun 8, 2015 at 14:06 | answer | added | Jay | timeline score: 2 | |
| Jun 8, 2015 at 11:52 | comment | added | CJ Dennis | Cut off one of the faces so you have a 5-sided die. Then it's much easier! | |
| Jun 8, 2015 at 7:21 | answer | added | Fattie | timeline score: 25 | |
| Jun 8, 2015 at 3:42 | answer | added | Mark Joshi | timeline score: 4 | |
| Jun 8, 2015 at 3:35 | answer | added | user985366 | timeline score: 4 | |
| Jun 8, 2015 at 1:22 | comment | added | Vim | I think it's time to create a "dice" tag | |
| Jun 7, 2015 at 22:16 | comment | added | Daniel R Hicks | Simplest is to roll the die twice (or roll two of them). Add the first value and 6 times the second value. If the result is greater than 10, discard the result and try again. A modification to produce fewer discards is to produce your sum, then, if it's < 10 & <= 20, subtract 10, < 20 & <= 30 subtract 20, < 30 discard. | |
| Jun 7, 2015 at 19:58 | history | protected | Community Bot | ||
| Jun 7, 2015 at 11:08 | answer | added | Chronocide | timeline score: 4 | |
| Jun 7, 2015 at 3:57 | answer | added | peterh | timeline score: 12 | |
| Jun 6, 2015 at 22:09 | answer | added | NovaDenizen | timeline score: 42 | |
| Jun 6, 2015 at 20:26 | answer | added | user2836778 | timeline score: 3 | |
| Jun 6, 2015 at 14:28 | answer | added | MJD | timeline score: 26 | |
| Jun 6, 2015 at 13:09 | comment | added | David K | Also see math.stackexchange.com/q/1200627/139123 and math.stackexchange.com/q/1273214/139123 | |
| Jun 6, 2015 at 12:45 | answer | added | MJD | timeline score: 3 | |
| Jun 6, 2015 at 12:41 | history | edited | P.A. | CC BY-SA 3.0 |
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| Jun 6, 2015 at 12:41 | comment | added | Alexey Burdin | I'd write the sequence as 0ドル\le x \le 1,,円 x=0.d_1d_2d_3\dots$ base 6ドル$ and read it as a decimal number (like Arithmetic coding ). @P.A. | |
| Jun 6, 2015 at 12:41 | answer | added | Jack D'Aurizio | timeline score: 103 | |
| Jun 6, 2015 at 12:39 | answer | added | Clement C. | timeline score: 1 | |
| Jun 6, 2015 at 12:39 | comment | added | David Mitra | For 2., roll one die once. If a "6" results, roll again? | |
| Jun 6, 2015 at 12:37 | comment | added | Did | Indeed, throwing the dice $n$ times yields 6ドル^n$ equiprobable results, whose set cannot be partitioned into 10ドル$ subsets of the same size since 10ドル$ does not divide 6ドル^n$. | |
| Jun 6, 2015 at 12:37 | comment | added | Travis Willse | Of course, I'd rather just buy a D10. | |
| Jun 6, 2015 at 12:36 | comment | added | Travis Willse | Since 6ドル$ is not divisible by all the factors of 10ドル,ドル there is no such method, but one can still ask for the method whose expected number of rolls required to produce a number is as small as possible. | |
| Jun 6, 2015 at 12:35 | history | edited | P.A. | CC BY-SA 3.0 |
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| Jun 6, 2015 at 12:33 | review | First posts | |||
| Jun 6, 2015 at 12:40 | |||||
| Jun 6, 2015 at 12:32 | history | asked | P.A. | CC BY-SA 3.0 |