HyperMissingno |

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If you have an ability that lets you roll something twice and select the best option, and are hit with a condition that makes you roll the smae thing twice but take the worse result, what happens?

This came up when my oracle used borrow fortune. She has an ability that lets her roll twice while beating spell resistance or while dispelling. We ruled that the ability and condition cancel each other out, but is there an official rule on this?

skizzerz |

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I'd personally rule it as either they cancel each other out (as in 5e) in order to keep things simple.

If you want to be pedantic about RAW though, the debuff part of *borrow fortune* states "For the next two rounds following your casting of the spell, you must roll two dice each time a d20 roll is called for, keeping the less favorable result." Your ability to roll twice and take the better result calls for 2 d20 rolls, so for each of those rolls you roll twice and take the worse result, and then take the better result of those two results.

(If you want to be *really* pedantic about RAW, you could say that *borrow fortune*'s text to have you roll twice calls for 2 d20 rolls, so each of those rolls call for another 2 d20 rolls, each of which calls for another 2 d20 rolls, and so on ad infinitum. Given you'd be rolling infinite dice, you are guaranteed to get a 1 eventually so your end result is that you always get a nat 1.)

Claxon |

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No official rules answer to my knowledge, considering that the two methods proposed here don't have the same mathematical average according to Ozy (did not double check) says to me that neither is valid. The order you do it in shouldn't matter, but since it does....

I think the only reasonable way to run it is to say the effects cancel one another out.

Steve Geddes |

Mathematical intuition means that, if the average isn't 10.5 for each, the two answers should be equidistant from 10.5. So 9.8 and 11.2 is plausible!

I'd go for just roll once, personally - we're talking trivial differences and less dice rolling equals a quicker game (with fewer arguments about which "should" take precedence).

Mathmuse |

Hmm, well you could:

1) roll twice, take the best, roll two more times and take the best, then take the worst of those two.

or

2) roll twice take the worst, roll two more times and take the worst of those two, then take the best out of those results.

Not sure which is better or worse for you.

Just checked with a little program.

Rolling two sets, taking the worst of each set, then taking the best of those two gets you an average of 9.8 on a d20.

Rolling two sets, taking the best of each set, then taking the worst of those two gets you an average of 11.2 on a d20.

(Correction from Ozy's next posting was applied.)

Statistics should be pretty good on that, 10 million rolls per set for a total of 40M rolls. Unless I screwed up something else that is. ;)

We can analyze all cases rather than relying on a Monte Carlo simulation.

Of the four d20 rolls, label them Worst, Second Worst, Second Best, and Best. It is possible that an adjacent pair or triple could be the same number, but we can still apply distinct labels to identical numbers.

We have 24 possible orders of the four labels based on the actual rolls. But switching the order of the first two has no effect, switching the order of the last two has no effect, and switching the first two with the last two has no effect. We are reduced to three equally likely cases:

Case a: (Worst, 2nd Worst) and (Best, 2nd Best)

1) Picking best of each group gives (2nd Worst, Best) and worst of that is 2nd Worst.

2) Picking worst of each group gives (Worst, 2nd Best) and best of that is 2nd Best.

Case b: (Worst, 2nd Best) and (2nd Worst, Best)

1) Picking best of each group gives (2nd Best, Best) and worst of that is 2nd Best.

2) Picking worst of each group gives (Worst, 2nd Worst) and best of that is 2nd Worst.

Case c: (Worst, Best) and (2nd Worst, 2nd Best)

1) Picking best of each group gives (Best, 2nd Best) and worst of that is 2nd Best.

2) Picking worst of each group gives (Worst, 2nd Worst) and best of that is 2nd Worst.

A quick statistical approximation via beta distributions is that the average Worst is 4.5, the average 2nd Worst is 8.5, the average 2nd Best is 12.5, and the average Best is 16.5. Therefore, the average of Method 1 is 11.2 and the average of Method 2 is 9.8. Ozy's simulation is accurate.

The difference between the two methods is only 1.4. Rather than worry about which goes first, follow Texas Snyper 's, skizerz's, Claxon's, and Steve Geddes' advice and simply roll 1d20.

Or if you care about variance, roll three d20 and take the middle one. If you don't know what "variance" means, in this case it means that highest and lowest rolls are cut out by the combined method, so we cut them out by the three-dice method, too.

Vrischika111 |

...which means the question still exists on which takes precedence, one way or the other.

I'd really rather personally rule a cancel out. Or maybe roll three, take the middle result?

as house rule/interpretation, I'd roll 3:

I'd interpret

* "Roll 2 die, take highest" by "add one dice drop lowest"

* "Roll 2 die, take lowest" by "add one dice drop highest"

this way both can be effective at the same time, resulting in a 3-die roll, dropping highest and lowest

Khudzlin |

Rolling 3 dice and taking the middle result has significantly less variance than rolling a single die (4.48 vs 5.77), so the choice is not neutral (especially if the player gets to choose each time). In the interest of simplicity (less rolling, less thinking after), I'd have the effects cancel each other out.