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4 years ago ::
Mar 13, 2009 - 4:25PM
#11
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Date Joined:
Nov 16, 2005
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I already gave the probability of rolling a target number 'b' with n rerolls of a dM:
b^n - (b-1)^n
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M^n
If you want the probability of hitting at least a target number 'T', then we simply sum that from b=T to b=M:
Prob = M^-n (Zeta[-n, M] - Zeta[-n, 1 + M] - Zeta[-n, -1 + T] + Zeta[-n, T])
You're right that this is usually the key number, but it depends extra inputs (the defense) in a nontrivial way. The average values calculated by the formula in the posts above give you a much faster feel for how good your rerolls are going to be.
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4 years ago ::
Mar 13, 2009 - 4:35PM
#12
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Date Joined:
Aug 22, 2005
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...man, I'm so glad I didn't try to work that out on first principles.  I was actually at work, in the midst of a huge spreadsheet, so I just plugged in numbers. :P
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4 years ago ::
Mar 13, 2009 - 4:36PM
#13
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Date Joined:
Jan 15, 2009
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I already gave the probability of rolling a target number 'b' with n rerolls of a dM:
b^n - (b-1)^n
----------------
M^n
If you want the probability of hitting at least a target number 'T', then we simply sum that from b=T to b=M:
Prob = M^-n (Zeta[-n, M] - Zeta[-n, 1 + M] - Zeta[-n, -1 + T] + Zeta[-n, T])
You're right that this is usually the key number, but it depends extra inputs (the defense) in a nontrivial way. The average values calculated by the formula in the posts above give you a much faster feel for how good your rerolls are going to be.
The extra inputs are necessary to obtain useful information though. For non-boolean entries (e.g. damage rolls) average rolls are fine; in fact they're exactly what you want. But for something where a 1 and a 9 are valued exactly the same, you NEED that extra input to give yourself any truly useful information (beyond "rolling twice improves your chance to hit", which really doesn't require any analysis at all).
Rerolls against something where you already hit on a 2 give you 4.75%; rerolls where you hit on an 11 give you 25%; rerolls where you only hit on a 20 once again give you 4.75%. This is compared to an expected value of about 16.6% across the board if you just look at the average.
<Ioun> they're apparently making a MolIsCool pp
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4 years ago ::
Mar 13, 2009 - 5:27PM
#14
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Forgive me if I'm wrong, but in the context of DPR calculations, aren't the average "to-hit" values important? That is, assuming you hit on an 11 (a common assumption, it seems), then you have a 50% chance of hitting, and you work the DPR from there. However, with the Avenger's "roll twice" ability, then the expected value of 2d20 comes into play in the function rather than the standard "50% chance of hitting." Right? Or am I misunderstanding the DPR calculations?
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4 years ago ::
Mar 13, 2009 - 5:41PM
#15
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Date Joined:
Jan 15, 2009
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Forgive me if I'm wrong, but in the context of DPR calculations, aren't the average "to-hit" values important? That is, assuming you hit on an 11 (a common assumption, it seems), then you have a 50% chance of hitting, and you work the DPR from there. However, with the Avenger's "roll twice" ability, then the expected value of 2d20 comes into play in the function rather than the standard "50% chance of hitting." Right? Or am I misunderstanding the DPR calculations?
This is accurate, although with the advent of weapon expertise hitting on an 11 seems to be a bit of a harsh assumption. Here is how the math works out with OoE though, given being able to hit on a certain number:
7: 91% chance to hit
8: 87.75% chance to hit
9: 84% chance to hit
10: 79.75% chance to hit
11: 75% chance to hit
<Ioun> they're apparently making a MolIsCool pp
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4 years ago ::
Mar 13, 2009 - 5:48PM
#16
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Date Joined:
Nov 16, 2005
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Special cases to the rescue!
Although more than 1 reroll comes up a bit, I suppose that a single reroll is the most common. Let me give vastly simplifed formulae for the common cases.
Best of two rolls of a dM:
Average result = (4M-1)(M+1)/(6M)
Probability of hitting target result 'T' is... P(T,M) = 1 - ( (T-1)/M )^2
Edit: This second formula made me suspicious. Results for Zeta functions limited to the integers, and some retrospectively obvious reasoning, and we have a far simpler result...
Best of N rolls for a dM:
Probability of hitting target result T = 1 - ( (T-1)/M )^N
(There isn't an especially simple version for the average.)
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4 years ago ::
Mar 13, 2009 - 5:56PM
#17
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Date Joined:
Nov 16, 2005
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Forgive me if I'm wrong, but in the context of DPR calculations, aren't the average "to-hit" values important? That is, assuming you hit on an 11 (a common assumption, it seems), then you have a 50% chance of hitting, and you work the DPR from there. However, with the Avenger's "roll twice" ability, then the expected value of 2d20 comes into play in the function rather than the standard "50% chance of hitting." Right? Or am I misunderstanding the DPR calculations?
Yes and no. They're important to DPR, but you can't assume you roll the average all the time, just as you can't assume you roll a 10.5 all the time for a single d20. You have to use the probability formula of hitting your target (either trivial for 1 die, or formula in the previous post for 2 or more dice) in order to calculate DPR. The average roll is strongly indicative of where that probability might be, but it's not quite enough to get it right by itself.
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4 years ago ::
Mar 14, 2009 - 3:15PM
#18
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Date Joined:
Feb 12, 2009
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By brute-forcing the calculations with a Java app I programmed in my spare time, I found that the average difference between the higher and lower results on a 2d20 roll was 7, based on a sample set of 1000000 pairs of randomly-generated numbers, the smaller of each subtracted from the larger and the resulting array of integers averaged.
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4 years ago ::
Mar 14, 2009 - 4:10PM
#19
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Date Joined:
Nov 16, 2005
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...and clearly the lower one has a greater than 50% chance of being below average. It's easy enough to calculate this quantity exactly, but I don't think it's useful for valuing the effectiveness of rerolls.
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4 years ago ::
Mar 14, 2009 - 4:35PM
#20
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Date Joined:
Mar 15, 2001
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So, as a non-math guy, here's what I'd like to know:
* What are the % chances of getting each distinct result with 2d20, take the higher roll? I can figure out that getting a 1 is a 0.25% chances (1 in 400), but the other results are harder to figure out.
* What are the chances of each die result for 2d20, take the higher, then rerolling both (Oath of Enmity + Elven Accuracy)?.
If I had those two things (a 1 to 20 list for each), it should be easy enough to figure out my chances to hit in any particular situation.
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