Mathcraft! Roll twice, take the higher

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Alrighty, bored at work so I just worked out a probability table of "roll twice, take the higher result". For a d4, I got that the average increases by .75, bringing it to a nice round 3.25 average. For a d20, I got an average increase by ~3.25, giving us an average of about 13.75.

Then I remembered I've never taken a statistics class and probably got this entirely wrong.

Anyone have any real math on "roll twice, take the higher" on a d20?
Well, it seems to me your roll options on a d4 would look like:
(roll, reroll) = result
1,1 =1
1,2 =2
1,3 =3
1,4 =4
2,1 =2
2,2 =2
2,3 =3
2,4 =4
3,1 =3
3,2 =3
3,3 =3
3,4 =4
4,1 =4
4,2 =4
4,3 =4
4,4 =4

Total of results = 50.
Number of results = 16.

50/16 = 3.125

Not sure if I did that right myself or not, but it looks right.

A d20 would have something like 400 roll,reroll sets, right? gotta be a better algorithm.
I'm sleepy, but..

20*{1+2+3+..+20} + for(int i=0, i>20, i++) { SUM(N=0, I, N) }

should give the total of all the roll/reroll results for a d20.
that is:
1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20 (possible results if 'reroll' is 1)
2,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20 (possible results if 'reroll' is 2)
3,3,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20 (possible results if 'reroll' is 3)
etc...
Hrm, new math gives me 15.25 as the value of 2d20 take the higher. (Mix of your math and my original math)

Can anyone confirm?
It involves value frequency.
For example, to get a value of 1, there's one combination. To get a value of 2, there's three combinations (1,2; 2,1; 2,2), to get 3, there's five sets (1,3; 2,3; 3,3; 3,2; 3,1), to get 4 there's 7 sets, and so on.

There are 2 more ways to get any progressively higher number.
1 way to get 1.
3 ways to get 2.
5 ways to get 3.
7 ways to get 4.
9 ways to get 5.

There are 2X-1 ways to get X.

There's a simple and elegant way to turn this into a formula, but I'm at work and can't be bothered. Excel ahoy!

Summation of all values is 5530.
5530/400 = 13.825

Further details: 400 lines. First column is a repeated series from 1 to 20. Second column is a series of twenty 1s, twenty 2s, and so on, until we get twenty 20s. Third column is Max of column 1 and 2. Summation of column three is 5530. 5530/400 = 13.825.
Compared to the average of a single d20 (10.5), it's +3.325.

(This, btw, is why Danger Sense provides less of a benefit than Improved Initiative in general, but makes a difference depending on the "DC" you're trying to hit.)
Roll 1 D20:
Mean: 10.5
Median: 10.5

Roll 2 D20s and take the higher:
Mean:13.825
Median: 15

Normal D20 roll has a mean of 10.5. This may sound like a general increase of 3.325, but that is not a good way of looking at it.

  • About 50% of all rolls on a single d20 will range from 5 to 16.
  • About 50% of all rolls taking the higher of 2 d20s will range 10 to 18.
  • -
  • About 75% of all rolls on a single d20 will range on average 3.5 to 17.5
  • About 75% of all rolls taking the higher of 2 d20s will range from 8 to 19
  • -
  • About 90% of all rolls on a single d20 will range from 2 to 19
  • About 90% of all rolls taking the higher of 2 d20s will range from 5 to 20.
  • -
  • About 95% of all rolls on a single d20 will range on average 1.5 to 19.5
  • About 95% of all rolls taking the higher of 2 d20s will range from 4 to 20.
  • -
  • 50% of the time with a single D20 roll, you will roll under average (10.5).
  • 42% of the time with taking the higher of 2 d20s, you will roll under average(13.825).
  • -
  • 50% of the time with a single D20 roll, you will roll 10 and under.
  • 25% of the time with taking the higher of 2 d20s, you will roll 10 and under.
  • -
  • Needing a 20 for a Critical Hit
  • A single d20 roll has a 5% chance to score a critical hit
  • Taking the higher of 2 d20s has a 9.75% chance to score a critical hit.
  • -
  • Needing a 19 or 20 for a Critical Hit. (Weapon Mastery or Jagged Weapon)
  • A single d20 roll has a 10% chance to score a critical hit.
  • Taking the higher of 2 d20s has a 19% chance to score a critical hit.
  • -
  • Needing an 18 or higher for a Critical Hit. (Daggermaster or Student of Caiphon(with Radiant keyword) PPs.)
  • A single d20 roll has a 15% chance to score a critical hit.
  • Taking the higher of 2 d20s has a 27.75% chance of scoring a critical hit.
  • -
  • Needing a 17 or higher for a Critical Hit. (Raise the Stakes - Rogue Daily Utility Level 16)
  • A single d20 roll has a 20% chance to score a critical hit.
  • Taking the higher of 2 d20s has a 36% chance to score a critical hit.


Starting to see the boon? An Avenger/Rogue wielding going Daggermaster and wielding daggers or an Avenger/Warlock wielding a Radiant(keyword) Weapon going Student of Caiphon will both have 27.75% chance to Crit, and will never roll under 10 75% of the time.

Now excuse me while I start setting up a Half-elf Wis/Str Avenger, with Dilettante ability being Twin Strike, taking the Pact Initiate feat(Star), and going for the Student of Caiphon paragon path with the plan on wielding two Radiant Katars. And the Punisher of the Gods Epic Destiny too. Talk about broken, 48% chance to roll at least one critical hit against a foe using Twin Strike against. ~7.7% chance for two critical hits in that 48% chance.
Seems to me that the same half-elf (twin strike) Avenger would be better served (damage-wise) by going Sneak of Shadows -> Daggermaster with a pair of bloodiron daggers. Sure, the katar is high crit, but the dagger gets the 18-20 crit chance without the enchantment, and the bloodiron is 1d10 (and again at the beginning of the next turn, so effectively 2d10 per plus) rather than the 1d6 from radiant.

On a crit at level 14 (first level you can get a Radiant weapon off creation, and that's only one), the Student would do 6 (base)+ 2d6 (high crit)+ 3d6 (crit effect) +3 (weapon plus) + 3 (property) = 29.5 expected damage from the weapon, with a minimum of 17 and a maximum of 42.

At the same level (14, and you can have TWO bloodiron daggers) the Daggermaster would do 4 (base) + 3d10(crit effect) + 3(weapon plus) = 23.5 expected damage on that turn, with another 16.5 next turn for 40 expected weapon damage; minimum damage on that is 13 and a maximum of 67.

So I guess it depends on whether you're risk-averse or not. The Student/Radiant Katar option has a higher average damage per crit, but bloodiron offers more upside (and potential risk, as well). Unless I've botched the calculations somewhere.

(Disclaimer: calculations are for expected damage on a crit, without to-hit numbers factored in because they use the same 1[w] attack power using the same stat and weapons with the same proficiency bonus. DPR calculations will return lower numbers, and may push the advantage back over to the Student/Radiant Katar route. I may work those out when I get some more time.).
Now excuse me while I start setting up a Half-elf Wis/Str Avenger, with Dilettante ability being Twin Strike, taking the Pact Initiate feat(Star), and going for the Student of Caiphon paragon path with the plan on wielding two Radiant Katars. And the Punisher of the Gods Epic Destiny too. Talk about broken, 48% chance to roll at least one critical hit against a foe using Twin Strike against. ~7.7% chance for two critical hits in that 48% chance.

Unveiled Visage/Punisher of the Gods creates a nasty/ridiculous combo. Unveiled Visage, an Avenger PP, gives you back a Channel Divinity, RRoT, when you spend an AP. So as long as you hit with a RRoT attack, you crit and generate AP until you run out of actions for the round. Hitting with an Armor Splinter/Triumphant Attack (a very good chance) imposes a penalty that means you're gonna hit anything from then on unless you roll double 1's, even with Power Attack. Armor Splinter has two attacks, which gives you 4d20. Really just one of those attacks needs to hit to impose a hefty penalty. With Martial Mastery (reloading and using Armor Splinter every round) this a continual cycle until you miss with Armor Splinter. Toss in Bloodiron/Rending, Two Weapon Opening, and other stuff that's good, and it quickly starts to be stupid.

Punisher of the Gods is absurd.
Math-fu allows us to do this without increasingly large spreadsheets.

The average result of the best of n rerolls of a die of size M:

Number of microstates where 'b' is the best of the rolls = b^n - (b-1)^n

=> Probability that 'b' is the best of the rolls = (b^n - (b-1)^n)/M^n

=> Average is Sum_{b=1,M} (b Prob)

=> Average is M^-n (HarmonicNumber[M, -1 - n] - HarmonicNumber[-1 + M, -1 - n] - HarmonicNumber[-1 + M, -n])


For the best of 2 d4s, this is indeed 3.125
For the best of 2 d20s, this is 13.825
Since it came up above, the average roll for best of 4d20 is 16.4833
I think it should be pointed out that the average value for rolling d20 twice is almost never useful information. The value that shows on a d20 isn't a scaling factor relative to its worth. Example: you are trying to hit something with a defense of 20, and you have +10 to your attack roll:

Your "average attack roll" is going to be 10+13.825, or 23.8, compared to 20.5 rolling once. So you might be tempted to think that rolling twice improves your chance to hit by 15% (i.e. by 3 numbers on the die).

In actuality though, averaging doesn't provide meaningful information, because if you need a 10 to hit, a 1 is exactly as valuable as a 9, and a 10 is exactly as valuable as a 19. The number of times you can roll a d20 twice and not get at least one number higher than a 9 is .45^2 or 20.25% of the time. So you will hit the target's defense 79.75% of the time, compared to 55% of the time with only one die.

That means 24.75% of the time you will hit because of rolling a second die. Compare that to the 15% value you'd expect if you averaged them.
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.
...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
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.
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?
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
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.)
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.
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.
...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.
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.
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?
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 3.0
How much damage should I shoot for?
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
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?
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.
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
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.
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)
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
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.
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.
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...
True. Maybe it's because you didn't express as a table of percentages. ;)

Never mind. All the various versions of the answer to the OPs question have now surfaced, including your answer, generalisations, a table of the actual numbers, and the answer to the question that they actually were interested in (regard OoE interacting with elven reroll). As Molecule pointed out, only the 'best of 4 rerolls' version is relevant, so that's easy enough. Although WOLead may prefer to turn it into a table.
...
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...

I saw it. I just wanted it in a form that can be seen and used quickly in a gaming session. So I created a table that gives the specific percentages. Formulas may be compact, but I'd rather have a quick-reference table than a formula where I have to pull out a calculator.
I can't imagine a situation where you'd use the first column at the gaming table. You'd probably be more interested in knowing your chances of hitting a certain value (your second column) rather than your chances of hitting a certain value exactly. For the OP, it might be worth tabulating the formula for the best of 4 dice as well.
Has anyone also run the MathCraft for Holy Ardor from the Ardent Champ PP as well?  Im just curious what percentage this ups your critical chances depending on your target "to hit" numbers.
Instead of averages, it is indeed easier to think in terms of "odds to hit". How does rolling twice increase your chances of hitting. This table shows it pretty well.



























































































































































 required roll to hitchance to hitimproved chance to hit by rolling twiceincrease in odds
ACwith a +5 1-(chance to miss)*(chance to miss) 
25205.00%9.75%4.75%
241910.00%19.00%9.00%
231815.00%27.75%12.75%
221720.00%36.00%16.00%
211625.00%43.75%18.75%
201530.00%51.00%21.00%
191435.00%57.75%22.75%
181340.00%64.00%24.00%
171245.00%69.75%24.75%
161150.00%75.00%25.00%
151055.00%79.75%24.75%
14960.00%84.00%24.00%
13865.00%87.75%22.75%
12770.00%91.00%21.00%
11675.00%93.75%18.75%
10580.00%96.00%16.00%
9485.00%97.75%12.75%
8390.00%99.00%9.00%
7295.00%99.75%4.75%


note, you always miss on a 1, and you have a 0.25% chance of rolling double 1's (snake eyes)
Has anyone also run the MathCraft for Holy Ardor from the Ardent Champ PP as well?  Im just curious what percentage this ups your critical chances depending on your target "to hit" numbers.



It depends on how you read it.  I believe the "true" (and basically worthless as a feature) reading is that you crit only if the numbers are the same and that number would have hit.  In that case, if you hit on a die roll of N or higher, you have 21 - N combinations that will give you a crit.  However, at least one of these (20,20) is trivial because it already crits, so your true number of additional rolls is 20-N, or 19-N if you crit on nineteens.  In addition, each double combination only appears once, which means each one has a 1/400 chance of happening.

As such, Holy Ardor will increase your chance to crit by (20-N)/400, or (19-N)/400 if you crit on nineteens.  If N is 8 (which seems like a pretty reasonable number), that means your chance to crit is increased by 3%.  Compare this to the feat which increases your crit range to 19-20: that will increase your chance to crit from 9.75% to 19%, for an increase of 9.25%.  So that version is quite weak as a PP feature.

There is a different reading though (and again, I don't think this is the way it actually works, but it wouldn't be overpowered to read it this way by any means), which would be that any doubles that aren't ones result in a crit (so 2,2 would crit even if you need an 18 to hit).  In that case, regardless of what you need to hit the target, you will get all of the cases from (2,2) to (19,19) to count as crits when they didn't before.  That's 4.5% extra chance if you normally crit on twenties, or 4.25% chance if you normally hit on nineteens.  Still quite a bit weaker than the crit range feat, but a little better than the other reading.

There is a different reading though (and again, I don't think this is the way it actually works, but it wouldn't be overpowered to read it this way by any means), which would be that any doubles that aren't ones result in a crit (so 2,2 would crit even if you need an 18 to hit).  In that case, regardless of what you need to hit the target, you will get all of the cases from (2,2) to (19,19) to count as crits when they didn't before.  That's 4.5% extra chance if you normally crit on twenties, or 4.25% chance if you normally hit on nineteens.  Still quite a bit weaker than the crit range feat, but a little better than the other reading.


To be fair, this reading also adds to your hit chance, as even if you get a double that normally misses, the crit overrides the miss.

There is a different reading though (and again, I don't think this is the way it actually works, but it wouldn't be overpowered to read it this way by any means), which would be that any doubles that aren't ones result in a crit (so 2,2 would crit even if you need an 18 to hit).  In that case, regardless of what you need to hit the target, you will get all of the cases from (2,2) to (19,19) to count as crits when they didn't before.  That's 4.5% extra chance if you normally crit on twenties, or 4.25% chance if you normally hit on nineteens.  Still quite a bit weaker than the crit range feat, but a little better than the other reading.


To be fair, this reading also adds to your hit chance, as even if you get a double that normally misses, the crit overrides the miss.




Actually it doesn't increase your to-hit either.  This is an oft-missed rule:

Precision: Some class features and powers allow you to score a critical hit when you roll numbers other than 20 (only a natural 20 is an automatic hit).  (Compendium/PHB1)
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Actually it doesn't increase your to-hit either.  This is an oft-missed rule:

Precision: Some class features and powers allow you to score a critical hit when you roll numbers other than 20 (only a natural 20 is an automatic hit).  (Compendium/PHB1)


So....the only value of getting a crit on a (2,2) if you don't hit on a 2 is if you have a weapon/implement that has a nonstandard effect on crit?
If you ascribe to that reading of that PP feature, I would say so.  Personally, I have always interpreted that feature as actually scoring the crit if you roll the same number, regardless of if it would have hit.

The entry on critical hit says: You score a critical hit if you roll a natural 20 and it would also have normally hit.  Some features allow an expanded range, but you still need to hit. 

Basically: If you roll a number that would result in a crit, and it hits, then you crit.  The feature says "If you roll the same number twice, you crit."  To me, that means that it is bypassing the qualifiers that would prevent a critical hit (i.e. the number isn't enough), because it says you score a critical, not that you can score a critical, meaning (IMO) that specific beats general, and you actually score the critical hit.  Regardless, that debate has come and gone, and I don't think anyone is going to change their mind (and I reaaaally don't want to restart a semantic argument), but if you agree that a double 2 does not crit because it does not hit, then I can't see how you wouldn't agree that it just plain doesn't hit given the rest of the critical hit definition about only a natural 20 hitting.  Which makes that PP terribad, and is another reason I disagree with that interpretation Yell
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To be fair, this reading also adds to your hit chance, as even if you get a double that normally misses, the crit overrides the miss.



Oh yeah, good point.  So this reading makes it a decently strong PP feature.