Timeline for When do I have 80% chance of 6 rolls of either 1, 2 or 3?
Current License: CC BY-SA 4.0
16 events
| when toggle format | what | by | license | comment | |
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| Nov 3, 2021 at 16:55 | vote | accept | RobbingDaHood | ||
| Oct 29, 2021 at 6:11 | history | edited | RobbingDaHood | CC BY-SA 4.0 |
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| Oct 28, 2021 at 23:02 | comment | added | Sextus Empiricus | For such a small problem you can always use a Markov Chain. You have 6x6x6x2 states for the number of rolls of 1, 2, 3, and whether the last roll was a 4 or not. You compute the transitions each step and monitor the probability to be in a final state. | |
| Oct 28, 2021 at 22:28 | answer | added | Henry | timeline score: 3 | |
| Oct 28, 2021 at 18:30 | history | edited | RobbingDaHood | CC BY-SA 4.0 |
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| Oct 28, 2021 at 18:20 | history | edited | RobbingDaHood | CC BY-SA 4.0 |
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| Oct 28, 2021 at 17:55 | comment | added | RobbingDaHood | Edited: Deleted | |
| Oct 28, 2021 at 15:31 | comment | added | carlo | this is very difficult. doing simulations seems the fastest way out of it. | |
| Oct 28, 2021 at 13:36 | history | edited | RobbingDaHood | CC BY-SA 4.0 |
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| Oct 28, 2021 at 13:27 | comment | added | RobbingDaHood | @fblundun: ahh true. Will edit the text. | |
| Oct 28, 2021 at 13:26 | comment | added | RobbingDaHood | @bernhard correct, edited the description. | |
| Oct 28, 2021 at 13:25 | history | edited | RobbingDaHood | CC BY-SA 4.0 |
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| Oct 28, 2021 at 13:20 | comment | added | Bernhard | Are 1, 2 and 3 of equal probability? | |
| Oct 28, 2021 at 13:13 | comment | added | fblundun | If I understand the game correctly, I think you only need 32 rolls to ensure a 100% chance of success. | |
| S Oct 28, 2021 at 12:53 | review | First questions | |||
| Oct 28, 2021 at 12:53 | |||||
| S Oct 28, 2021 at 12:53 | history | asked | RobbingDaHood | CC BY-SA 4.0 |