|
4 years ago ::
Mar 14, 2009 - 5:13PM
#21
|
Date Joined:
Nov 16, 2005
|
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)?.
The first is a special case of one of the results in the posts above:
Probability of result 'b' for best of ndM = ( b^n - (b-1)^n )/M^n
Therefore,
Probability of result 'b' for best of 2d20 = ( 2 b - 1 )/400
Note that above I also gave the result for getting at least result 'T' in the best of n dM. For 2d20 it is Prob(T) = 1 - (T-1)^2 /400
I'm not quite sure what the second question is - as far as I can tell there's only three rerolls there. And elven accuracy doesn't let you choose the best - you have to take the reroll. How would you reroll both?
|
|
|
|
4 years ago ::
Mar 14, 2009 - 5:25PM
#22
|
Date Joined:
Jan 23, 2008
|
Oath of enmity specifically states that if a power lets you reroll, you reroll both die and take better.
The problem here is you'd only want to do elven accuracy if you know that neither roll were sufficent. Thus i'd only use it with a powerful power that happens to miss.
Doing any calculations with elven accuracy to increase your chance of critting should only be taken into account when you really want that crit. Else you have to live with the reroll, and there's a chance that your reroll did worse than your first 2d20.
DPR King Candidates 3.0How much damage should I shoot for?
Show
You're fired : 1 Kills Per 5 Rounds = .2 KPR
Fair Striker : 2 Kills Per 5 Rounds = .4 KPR
Highly Optimized : 3 Kills Per 5 Rounds = .6 KPR
Nerfbat please : 4 Kills Per 5 Rounds = .8 KPR
It's OVER 9000!!!!!: 5 Kills Per 5 Rounds = 1+ KPR
DPR? KPR? KP4R? Bless you
Show
DPR = Damage Per round ~= Chance to hit * damage on a hit
KPR = Kills Per Round. 1 Kill = 8*Level+24 damage
= DPR/(8*level+24)
KPNR = Kills Per N Rounds. How many standards can you kill in N rounds?
|
|
|
|
4 years ago ::
Mar 14, 2009 - 6:36PM
#23
|
Date Joined:
Jan 15, 2009
|
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.
Elven accuracy doesn't work quite the same way as OoE because, as anomalous man said, you don't get to pick the highest number, you just use the second number. Your chance of hitting with the first two rolls is exactly the same as your chance of hitting with the second two rolls though (unless you have the wild elf luck and/or elven precision feats). Since you are interacting with the result halfway through, it doesn't quite make sense to ask what the probability of each number coming up is; it's based on your decision to reroll or not.
<Ioun> they're apparently making a MolIsCool pp
|
|
|
|
4 years ago ::
Mar 14, 2009 - 7:58PM
#24
|
Date Joined:
Nov 16, 2005
|
Right, so you need to ask a more specific question. If you want to hit and you know your target defence, then you take (best of 2 rolls) and only reroll if you haven't made your target. In that case, your probability of hitting it first time is:
Prob_OoE_only = 1 - (T-1)^2/400
So your chance of hitting it with a reroll option is
Prob_OoE_only + (1 - Prob_OoE_only)*Prob_OoE_only
which is also, unsurprisingly, exactly the same as the chance of rolling your target with a "best of 4 rolls", or
P(T) = 1 - (T-1)^4/160000
If, however, you're just trying to maximise your roll, then you'd only reroll if you rolled less than the best of two average. From my formula above, that's 13.825, so you'd reroll on a 13 or less, and keep your roll on a 14 or more. That gives you an average roll of:
(1-Prob_OoE_only(14))*13.825 + Sum_{j=14, 20}(j (2 j -1)/400 )
which equals:
Average result of OoE plus optional reroll: 14.9345
|
|
|
|
4 years ago ::
Mar 15, 2009 - 3:23AM
#25
|
Date Joined:
Jul 16, 2008
|
Guys, you are forgetting... you would probably need dice from this guy: http://www.gamescience.com/for your numbers to be right. and be rolling on that green casino carpet thing. hmmm.
|
|
|
|
4 years ago ::
Mar 15, 2009 - 8:47AM
#26
|
Date Joined:
Mar 15, 2001
|
Elven accuracy doesn't work quite the same way as OoE because, as anomalous man said, you don't get to pick the highest number, you just use the second number. Your chance of hitting with the first two rolls is exactly the same as your chance of hitting with the second two rolls though (unless you have the wild elf luck and/or elven precision feats). Since you are interacting with the result halfway through, it doesn't quite make sense to ask what the probability of each number coming up is; it's based on your decision to reroll or not.
Of course, if you know your target number, it doesn't matter. Against a tough foe, I can usually guess the target number (at least for AC) within the first two rounds of combat.
Anyway, here's what I've been able to come up with for chances of hitting a specific number with 2d20 take the highest:
01 - 1/400, 0.25% (automatic)
02 - 3/400, 0.75% (99.75% to get or beat)
03 - 5/400, 1.25% (99% to get or beat)
04 - 7/400, 1.75% (97.75% to get or beat)
05 - 9/400, 2.25% (96% to get or beat)
06 - 11/400, 2.75% (93.75% to get or beat)
07 - 13/400, 3.25% (91% to get or beat)
08 - 15/400, 3.75% (87.75% to get or beat)
09 - 17/400, 4.25% (84% to get or beat)
10 - 19/400, 4.75% (79.75% to get or beat)
11 - 21/400, 5.25% (75% to get or beat)
12 - 23/400, 5.75% (69.75% to get or beat)
13 - 25/400, 6.25% (64% to get or beat)
14 - 27/400, 6.75% (57.75% to get or beat)
15 - 29/400, 7.25% (51% to get or beat)
16 - 31/400, 7.75% (43.75% to get or beat)
17 - 33/400, 8.25% (36% to get or beat)
18 - 35/400, 8.75% (27.75% to get or beat)
19 - 37/400, 9.25% (19% to get or beat)
20 - 39/400, 9.75% (9.75% to get a natural 20)
|
|
|
|
4 years ago ::
Mar 15, 2009 - 3:19PM
#27
|
Date Joined:
Nov 16, 2005
|
Anyway, here's what I've been able to come up with for chances of hitting a specific number with 2d20 take the highest
Dude, I've given the compact formula multiple times: (2 b - 1)/400
|
|
|
|
4 years ago ::
Mar 15, 2009 - 6:52PM
#28
|
|
|
Dude, I've given the compact formula multiple times: (2 b - 1)/400
Yes, yes. You gave the formula first. But Fedifensor provided more information in that post then just a formula. Sure anyone could theoretically come about the same conclusions using that formula, but its easier to just see a "table" of the data. I had the same data of table made up when I was making my conclusions in the earlier post, when the OP was just trying to find the average of the "roll 2 d20s and take the higher". I didn't post the table, as it wasn't asked, and just placed in the data for comparison of the d20 to the "roll 2 d20s and take the higher".
Again, Fedifensor gives the data from that formula, along with the % chance of getting such a number, and then puts the % chance of matching or beating that number. Until then, only a tiny portion of it existed in Molecule's post earlier on.
Lastly, formulas help a mathematician or those able to manipulate them know what is going on. Hard data is more useful at telling the masses what happens. Use hard data, unless you know you are serving the info to a mathematician or a programmer.
There is a reason why I hated a portion of statistics, and it wasn't the math. It was always the conclusion where you had to take the results of the math and put them into a paragraph of words, in a certain format, for the bloody English/Business majors to understand.
|
|
|
|
4 years ago ::
Mar 15, 2009 - 7:20PM
#29
|
Date Joined:
Jan 15, 2009
|
Of course, if you know your target number, it doesn't matter. Against a tough foe, I can usually guess the target number (at least for AC) within the first two rounds of combat.
I don't quite understand this statement. Does your DM make you choose to use elven accuracy before you know whether you've hit or not? That's not how it's supposed to work; you don't need to have any idea at all about what your opponent's AC is to use elven accuracy effectively. You can choose to use it after you know you've missed.
<Ioun> they're apparently making a MolIsCool pp
|
|
|
|
4 years ago ::
Mar 15, 2009 - 10:59PM
#30
|
Date Joined:
Aug 22, 2005
|
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.
...
I was totally ignored, apparently.
In the 4th post, right after 2d20 was asked about, I already said you have a 1/400 chance of a 1, a 3/400 chance of a 2, 5/400 chance of a 3, and so on... 
|
|
|
