Timeline for Counting fails and successes in dice pools in Anydice
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
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| Sep 19, 2019 at 10:41 | history | edited | Ilmari Karonen | CC BY-SA 4.0 |
use Carcer's d{} trick
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| May 25, 2019 at 20:58 | comment | added | Macallan Lehane | Wow. Thank you for that. Really helpful explanation and I'm gonna have to wrap my head about it (which is something I always love about probabilities). And the anydice script is going to very helpful too! Thanks! | |
| May 25, 2019 at 20:47 | vote | accept | Macallan Lehane | ||
| May 25, 2019 at 14:01 | comment | added | Carcer | @MacallanLehane conditional probability is complicated and rarely aligns with human intuition. If you rolled 7d6 and randomly removed 3 dice from it, the expected successes on the dice that remained would be the distribution for 4d6; but when you selectively remove dice from the pool based on other criteria, the expectation of what's left is affected in a more complex way. | |
| May 25, 2019 at 13:52 | comment | added | Ilmari Karonen | @MacallanLehane: If you're assuming exactly three successes, then the remaining four dice can only be failures or neither. If you're assuming at least three successes, things get more complicated; see my edit above. | |
| May 25, 2019 at 13:49 | comment | added | Ilmari Karonen | @Carcer: Reading the question more carefully, you do have a point: the OP's logic has some mistakes that I mentally corrected while reading it. I've edited my answer. | |
| May 25, 2019 at 13:47 | history | edited | Ilmari Karonen | CC BY-SA 4.0 |
added 1527 characters in body
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| May 25, 2019 at 8:20 | comment | added | Macallan Lehane | Can you explain a bit about why that logic is wrong? (I do worry its too simple, and that's often misleading in probabilities!). 7d6 has whatever chance of getting 2 fails. But if we posit that 7d6 has been rolled and 3 of them are successes, then those three dice noonfer have any possibility of being fails, so the overall possibility of 2 fails must diminish based on this information. My assumption would be that it would return to whatever possibility 4d6 has, as those four other dice are free to be successes, fails or neither. | |
| May 24, 2019 at 17:13 | comment | added | Carcer | I think your results demonstrate that the OP's simple logic is not correct, not that it is! (the OP suggests that expected failures in 7d6 assuming 3+ successes might be equal to expected failures in 4d6) | |
| May 24, 2019 at 15:59 | history | answered | Ilmari Karonen | CC BY-SA 4.0 |