Timeline for What is the chance of rolling at least 10 before a 1 on a d20?
Current License: CC BY-SA 4.0
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| Feb 5 at 21:21 | comment | added | Eddymage | @GentlePurpleRain Thanks for this, I added to the answer. Indeed, the difference of just 1 between the terms in the ratio ringed a bell, but I did not pay enough attention. | |
| Feb 5 at 21:20 | history | edited | Eddymage | CC BY-SA 4.0 |
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| Feb 5 at 20:38 | comment | added | GentlePurpleRain | The number of faces on the die is actually irrelevant. Only the number of possible "success" results is important. (If you're trying to get 90 or more on a d100 vs. 10 or more on a d20, the probabilities are exactly the same.) | |
| Feb 5 at 20:37 | comment | added | GentlePurpleRain | You can simplify that final formula even more. Without loss of generalization, you can say that instead of wanting to get a number that is "at least \$t\$", you want to get any number in the set R where the size of R is \$r = d - t + 1\$. (This could be the numbers 4, 17, 23, etc., but the probability is the same.) In the example above, \$r = 20 - 10 + 1= 11\$ (the number of possible rolls that are 10 or higher on a d20). Then \$t = d - r +1\$, and we can substitue that into your formula as \$\frac{d+1-(d-r+1)}{d+2-(d-r+1)}\$, which simplifies to just \$\frac{r}{r+1}\$. | |
| Feb 3 at 9:04 | history | edited | Eddymage | CC BY-SA 4.0 |
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| Feb 2 at 16:46 | history | edited | Eddymage | CC BY-SA 4.0 |
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| Feb 2 at 15:41 | history | edited | Eddymage | CC BY-SA 4.0 |
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| Feb 2 at 15:21 | history | edited | Eddymage | CC BY-SA 4.0 |
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| Feb 2 at 14:21 | vote | accept | András | ||
| Feb 2 at 14:05 | history | edited | Eddymage | CC BY-SA 4.0 |
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| Feb 2 at 13:56 | history | edited | Eddymage | CC BY-SA 4.0 |
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| Feb 2 at 13:40 | history | edited | Eddymage | CC BY-SA 4.0 |
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| Feb 2 at 13:32 | history | answered | Eddymage | CC BY-SA 4.0 |