Lord Johan
Kabalite Warrior
Posts : 169
Join date : 2016-07-21
Location : Coming to a realspace near you
 |  Subject: Mathhammer: What to reroll on your Reapers  Sat Aug 05 2017, 08:54 | |
| In the other thread the issue of how to model rerolls came up and how they are beneficial for Reapers in particular. I figured I would make a separate thread to analyze it because it's interesting in itself and not just in context of spam reapers. I too thought that Reapers would gain extra benefit from rerolls. When planning a turn in advance it's useful to know what you should save a command point for. If you should wait for a better opportunity, etc. Let's examine how much benefit you can expect from saving a command point for some reroll algorithms for bands of 1, 3 and 6 Reapers. We assume you are firing at a T7 target with infinite wounds and 3+ armor, so 1 shot is worth 2/3*2/3*3.5 = 14/9 = approx 1.56 damage. Algorithm 1: "I will re-roll number of shots of X or lower." - Math:
If you re-roll 1 or lower, your avg. shots for a reaper will be (2+3+4+5+6)/5*5/6+1/6*3.5 = 47/12 = approx 3.92, gaining you 0.42 shots and 0.648 damage over a reaper which you didn't save a command point for. If you re-roll 2 or lower, you get (3+4+5+6)/6+2/6*3.5 = 50/12 = approx 4.17 shots, gaining you 0.67 shots and 1.04 damage. If you re-roll 3 or lower, you get (4+5+6)/6+3/6*3.5 = 4.25 shots for 0.75 shots and 1.17 damage gained. However, if you select algorithm 3, you are less likely to be able to use a command point on successive reapers of necessary. You will have used your command point with THRESHOLD/6 probability for every successive reaper you fire. So for re-rolling 1 or lower, after firing n successive reapers the no. of shots is 47/12*(5/6)^n + 3.5*(1-(5/6)^n). 2 or lower, it's 50/12*(4/6)^n + 3.5*(1-(4/6)^n). 3 or lower, it's 4.25*(3/6)^n + 3.5*(1-(3/6)^n). So if we are firing 1 reaper, for this algorithm it's optimal to prepare to reroll a 3 of shots. But if we are firing 3 or 6 or 10 reapers? Do a sum of the function above for n=0...2 and you find for thresholds as follows - Code:
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reapers 3 6 10
reroll 1 11.55 22.66 37.10
reroll 2 11.91 22.82 36.97
reroll 3 11.81 22.48 36.50
So if you are firing 10 reapers reroll 1 is superior and this will gain you 2 shots on avg. Incidentally the infinite sums difference (e.g SUM((47/12*(5/6)^n + 3.5*(1-(5/6)^n))-3.5),n=0 to infinity) corresponds to 2.5. But if you are firing fewer reapers you should reroll 2s and if you are firing just the 1 or 2 you should prepare to reroll 3s. You should switch to reroll 2 if you have more than 2 reapers (but the difference is marginal, 8.125 vs 8.11...) and to reroll 1s if you have 7 or more reapers to go, from solving ((SUM(47/12*(5/6)^n + 3.5*(1-(5/6)^n)),n=0 to x) > (SUM(50/12*(4/6)^n + 3.5*(1-(4/6)^n)),n=0 to x). Summary: For firing 1 reaper, saving your command points for this will get you 1.17 damage on average if you use the reroll 3 shots strategy. For firing 3 reapers, saving your command points for this will get you 1.41 shots and 1.77 damage average. For firing 6 reapers, 1.82 shots and 2.83 damage average. For firing 10 reapers, 2.10 shots and 3.26 damage average if you reroll no. of shots of 1.
Algorithm 2: "I will re-roll damage of X or lower." - Math:
This will increase your damage by 3.5 - X average. You will on average roll 3.5 * 2/3 * 2/3 = 14/9 dice for damage. Your probability of not rolling any X is ((6-X)/6)^dice so your probability of using this stratagem for a given reaper is 1-((6-X)/6)^(14/9), approx 25% for X=1. So for firing N successive reapers the chance that you will have used this stratagem is 1-(1-(1-((6-X)/6)^(14/9)))^n. You will only be able to use it once so it adds (1-(1-(1-((6-X)/6)^(14/9)))^n) * (3.5-X) to your damage. Let's solve. - Code:
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N 1 3 6 10
X
1 0.617 1.432 2.044 2.352
2 0.702 1.274 1.466 1.497
3 0.330 worse
So, this gives you respectable extra damage, but will result in less extra damage than rerolling no. shots on average, but it is very comparable.
Algorithm 3. "I will reroll failed to-wound rolls"- Math:
You will fire 3.5 shots, of which 2/3 hit, and then those shots wound 2/3 of the time. So the chance that you get a roll that does not wound is 1-(2/3)^(3.5*2/3) = approx 0.6117 and the chance of all rolls wounding is (2/3)^(7/3)
It will give you 2/3 * 3.5 = 7/3 = 2.333... damage the one time you use it.
You will not have used this after firing n reapers with probability of ((2/3)^(7/3))^n = (2/3)^(7n/3), so you will have used it with probability of 1-(2/3)^(7n/3) and can expect (1-(2/3)^(7n/3)) * 7/3 damage from this strategy after n reapers.
So for 1 reaper this is 1.427, for 2 reapers 1.981, for 3 reapers 2.197, and for 6 reapers 2.325 damage.
Comparing the top 2 strategies... 2/3*2/3*3.5*SUM(50/12*(4/6)^n + 3.5*(1-(4/6)^n)-3.5), n = 0 to 3 = 2.497, 2/3*2/3*3.5*SUM(50/12*(4/6)^n + 3.5*(1-(4/6)^n)-3.5), n = 0 to 2 = 2.183.
Algorithm 4. "I will reroll failed to-hit and to-wound rolls"- Math:
For 1 reaper you will do this for "to-hit" rolls with probability of
1-(2/3)^(3.5) and this will give you 2/3*2/3*3.5 wounds
and for "to-wound" rolls with probability of
(1-(1-(2/3)^(3.5)))*(1-(2/3)^(7/3)) giving you
(1-(2/3)^(3.5))*2/3*2/3*3.5+(1-(1-(2/3)^(3.5)))*(1-(2/3)^(7/3))*2/3*3.5 = a = 1.524... damage
For reaper 2, the probability you didn't use it on reaper 1 is (2/3)^(3.5) * (2/3)^(7/3) = b. So the general expression is
a + b*a + b^2*a +...
which gives for 2 reapers 1.668 damage
and for 3 reapers it's going to be worse than strict re-roll to wound
Putting it all together, finally:1. By saving your reroll for reapers you can expect about 1.5 wounds if you have 1 reaper, about 2 and 1/5 if you have 3 reapers, about 3 if you have 6 reapers, and about 3 1/4 if you have 10 reapers. If you can take more wounds by not saving your reroll for reapers then you should. 2. If you have 1 to 3 reapers you should prepare to reroll failed to wound rolls. If you only have 1 reaper you should also re-roll misses. If you have more than 3 reapers you should save the command point to reroll no. of shots of 2 or less. If you have more than 7 reapers to go you should save it to reroll no. of shots of 1. The reason why it works this way is because rerolling no. of shots is ultimately better trending towards 3.888... extra damage for rerolling no. of shots if you had infinite extra reapers, but you are less likely to be able to reroll a 1 for shots, so if you save your command point for that with few reapers you will end up with less damage. In real life where you get extra information, of course, if you roll a 1 for shots then you should re-roll after the fact, since this will yield more damage than saving it for re-rolls to wound. Rerolling damage is the worst strategy of all tested vs T7 targets. But if you were against a T9 opponent reroll 1 for damage would actually be the best if you have many reapers, since you would divide the other strategies damage by 2. Thoughts? | |
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Jimsolo
Dracon
Posts : 3212
Join date : 2013-10-31
Location : Illinois
| | Subject: Re: Mathhammer: What to reroll on your Reapers Sat Aug 05 2017, 15:25 | |
| Very interesting, and relevant to my army direction. Lots to digest here. | |
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Dizlen
Hellion
Posts : 42
Join date : 2017-05-09
| | Subject: Re: Mathhammer: What to reroll on your Reapers Mon Aug 07 2017, 11:46 | |
| Nice write up. With the few games that I have been using the reaper, I've found that I am usually using the rerolls to try and get extra shots in. Will need to take this into consideration now  | |
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