Timeline for How many rolls of sum of 6d6 for an expectation of 90% chance of hitting three consecutive 6d6 totals each less than or equal to each other?
Current License: CC BY-SA 4.0
4 events
| when toggle format | what | by | license | comment | |
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| Jul 7, 2020 at 13:21 | comment | added | Stephan Kolassa | @whuber: thanks, good points. I edited the +2 in. | |
| Jul 7, 2020 at 13:21 | history | edited | Stephan Kolassa | CC BY-SA 4.0 |
added 241 characters in body
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| Jul 7, 2020 at 13:18 | comment | added | whuber♦ | (+1) You need to add $2$ to your answers because you aren't counting the first two rolls. For confirmation of your answer, note that a slight overestimate can be obtained by assuming each roll is a uniform random value in $[0,1],$ for which an exact solution can be found (giving $15$ as the upper bound). | |
| Jul 7, 2020 at 12:36 | history | answered | Stephan Kolassa | CC BY-SA 4.0 |