Timeline for How do I find the probability of getting specific, distinct, dice rolls on an unfair die in the same group of trials?
Current License: CC BY-SA 4.0
6 events
| when toggle format | what | by | license | comment | |
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| Jan 22, 2021 at 1:38 | vote | accept | ConfusedGachaGamer | ||
| Jan 21, 2021 at 18:33 | comment | added | whuber♦ | Applying the definition 49 times accounts for 1+49 = 50 rolls. | |
| Jan 21, 2021 at 17:30 | comment | added | ConfusedGachaGamer | Why would $n-1$ be used to calculate the probability to the individual steps, and not 50? From my understanding the probability of not $A_i$ would be the probability of failing to get $A_i$ across all 50 rolls. | |
| Jan 20, 2021 at 18:41 | history | edited | whuber♦ | CC BY-SA 4.0 |
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| Jan 20, 2021 at 18:31 | history | edited | whuber♦ | CC BY-SA 4.0 |
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| Jan 20, 2021 at 18:24 | history | answered | whuber♦ | CC BY-SA 4.0 |