Once per turn, when you hit a creature with an attack that deals piercing damage, you can re-roll one of the attack’s damage dice, and you must use the new roll.
It's been like 25 years since I last took probability, so I don't remember how to calculate this.
For my longbow wielding Gloom Stalker, I'm trying to determine the mathematically best time to re-roll a d8 from the bow's damage to have the maximal increase to average damage.
I feel like it's not best to re-roll a result of four, because you then have the exact same odds of rolling a 1-4 again, which doesn't really increase your average roll (since it isn't roll two and take the highest). But, I could be wrong. Intuitively, I feel like re-rolling anything 1-3 would result in the largest increase in average result, but again, I could be completely wrong.
Anyone out there familiar enough with this kind of probability to determine what results on the initial D8 roll are most likely to result in an average increase in damage done, and what would be the net average roll in this situation?
Once per turn, when you hit a creature with an attack that deals piercing damage, you can re-roll one of the attack’s damage dice, and you must use the new roll.
It's been like 25 years since I last took probability, so I don't remember how to calculate this.
For my longbow wielding Gloom Stalker, I'm trying to determine the mathematically best time to re-roll a d8 from the bow's damage to have the maximal increase to average damage.
I feel like it's not best to re-roll a result of four, because you then have the exact same odds of rolling a 1-4 again, which doesn't really increase your average roll (since it isn't roll two and take the highest). But, I could be wrong. Intuitively, I feel like re-rolling anything 1-3 would result in the largest increase in average result, but again, I could be completely wrong.
Anyone out there familiar enough with this kind of probability to determine what results on the initial D8 roll are most likely to result in an average increase in damage done, and what would be the net average roll in this situation?
Thanks for your assistance.
I can calculate this for you if you want, but it would depend on your likelihood of hit percentage, so factors like disadvantage/advantage, number of attacks, not getting your second attack, etc will influence the roll
If I can re-roll a single d8 where I must take the results of the re-roll, what is the optimum number of the original d8 to trigger the re-roll to maximize the average result. It's easy for roll two and take the highest to determine the average, but that isn't this case.
All other variables will be immaterial. Once I have hit, I just want to know when I should re-roll the damage roll. I feel like it should be on rolls of 1-3, but I'm not certain.
Technically, the increase exists for rerolling a 4, but the potential gain there isn't high enough for me personally. I'm of the opinion that rerolling on a 1-3 hits the sweet spot for risk-reward. I went through the math a while back with Savage Attacker. It's a little different in that you can keep either roll, but I was also trying to maximize it's value for multiple attacks (on a fighter).
Brewsky is right though that anything less than average damage will provide an increase on average. 1 less than the highest number that is less than average will be the best for more conservative players who are more willing to keep a moderate amount of damage than lose out by rolling even lower.
If you think about it in terms of rolling at least as well as you did before, then your odds improve. Instead of thinking as rolling better than a 4 as only being a 50/50 proposition, you think of it as rolling at least a 4, which is 5/8 or 62.5% of the time. Rolling at least a 3 is 6/8 or 75%. That may be the easiest way for you to determine what you are comfortable with when it comes to rerolls.
For simple statistics you can just look at the possible results based on what your reroll threshhold is.
No Rerolls
Possible results on a d8 are 1, 2, 3, 4, 5, 6, 7, 8, sum of all possible results is 36, 36/4 = 4.5
Reroll 1s
Possible results on a d8 are 1-1, 1-2, 1-3, 1-4, 1-5, 1-6, 1-7, 1-8, 2, 3, 4, 5, 6, 7, 8, here the sum of all possible results is 71 and there are 15 possible results. 71/15 = 4.7333{r}
I will save the listing out for the others...but here are the average results based on where you put the threshhold:
So the maximum value on the reroll is to reroll any 1s or 2s. It is counterintuitive, I would think like others posed that anything below the average die value you would want to reroll. I think this is because you keep the second result rather than taking the higher or two like advantage.
I said I wouldn't go through the trouble of listing all the possible results, but just in case someone wants to check my math here it is: For clarity the blue numbers are the results that were "rerolled".
As mentioned, if you have multiple attack then your accuracy also comes into play on whether or not you want to reroll the two on your first shot, because what if you hit on your second attack and roll a 1. I could crunch those numbers too if you are interested, but it gets complicated...this is the quick and simple straight forward answer. On a d8 reroll 1s and 2s.
There is also something to be said about the gameplay aspect here that goes beyond average values. If the enemy is very close to dead and you act right before them...or if you rolled a crit and your table does double the value rather than rolling twice...
There are times you would want to take the counterintuitive route of rolling again when it is suboptimal incase you hit the jackpot (8).
Rollback Post to RevisionRollBack
Founding Member of the High Roller Society.(Currently trying to roll max on 4d6)
For simple statistics you can just look at the possible results based on what your reroll threshhold is.
No Rerolls
Possible results on a d8 are 1, 2, 3, 4, 5, 6, 7, 8, sum of all possible results is 36, 36/4 = 4.5
Reroll 1s
Possible results on a d8 are 1-1, 1-2, 1-3, 1-4, 1-5, 1-6, 1-7, 1-8, 2, 3, 4, 5, 6, 7, 8, here the sum of all possible results is 71 and there are 15 possible results. 71/15 = 4.7333{r}
I will save the listing out for the others...but here are the average results based on where you put the threshhold:
So the maximum value on the reroll is to reroll any 1s or 2s. It is counterintuitive, I would think like others posed that anything below the average die value you would want to reroll. I think this is because you keep the second result rather than taking the higher or two like advantage.
I said I wouldn't go through the trouble of listing all the possible results, but just in case someone wants to check my math here it is: For clarity the blue numbers are the results that were "rerolled".
As mentioned, if you have multiple attack then your accuracy also comes into play on whether or not you want to reroll the two on your first shot, because what if you hit on your second attack and roll a 1. I could crunch those numbers too if you are interested, but it gets complicated...this is the quick and simple straight forward answer. On a d8 reroll 1s and 2s.
Notice that the drop off from 1-2 to 1-3 is .0141, whereas the drop off from 1-3 to 1-4 .0364 bringing the 4 or lower down .0505 from the max gain. The variation isn't that great overall, so doing any of them can be useful.
By the way, you can shorthand the rerolled numbers by putting 4.5 in their place and just dividing by 8.
I tend to think of it in simpler terms, if you have a 4, there’s 3 chances you’ll do worse, and 5 chances you’ll do the same or better, so re-roll. If you have a 5, there 3 chances you’ll do better, and 5 you’ll do the same or worse, so don’t.
And if you do it that way overall your DPR is worse. Since the OP was looking for maximum value going to the probability maths is how we get the answer they are looking for.
Rollback Post to RevisionRollBack
Founding Member of the High Roller Society.(Currently trying to roll max on 4d6)
For simple statistics you can just look at the possible results based on what your reroll threshhold is.
No Rerolls
Possible results on a d8 are 1, 2, 3, 4, 5, 6, 7, 8, sum of all possible results is 36, 36/4 = 4.5
Reroll 1s
Possible results on a d8 are 1-1, 1-2, 1-3, 1-4, 1-5, 1-6, 1-7, 1-8, 2, 3, 4, 5, 6, 7, 8, here the sum of all possible results is 71 and there are 15 possible results. 71/15 = 4.7333{r}
I will save the listing out for the others...but here are the average results based on where you put the threshhold:
So the maximum value on the reroll is to reroll any 1s or 2s. It is counterintuitive, I would think like others posed that anything below the average die value you would want to reroll. I think this is because you keep the second result rather than taking the higher or two like advantage.
I said I wouldn't go through the trouble of listing all the possible results, but just in case someone wants to check my math here it is: For clarity the blue numbers are the results that were "rerolled".
As mentioned, if you have multiple attack then your accuracy also comes into play on whether or not you want to reroll the two on your first shot, because what if you hit on your second attack and roll a 1. I could crunch those numbers too if you are interested, but it gets complicated...this is the quick and simple straight forward answer. On a d8 reroll 1s and 2s.
This math seems wrong to me. You are totaling up all the possible roll results and dividing them by the total number of options, but that only gets you an accurate number if each result is equally likely, which in this case, they aren't. You are much more likely to roll an 8 then to roll a 1 and then reroll it into an 8. (1 in 8 chance vs 1 in 64 chance)
Like Jhfffan mentioned, you can replace each rerolled number with 4.5 and then divide by 8. Doing the math this way gets you the following:
Reroll nothing: 4.5
Reroll up to 1s: 4.9375
Reroll up to 2s: 5.25
Reroll up to 3s: 5.4375
Reroll up to 4s: 5.5
Reroll up to 5s: 5.4375
Reroll up to 6s: 5.25
Reroll up to 7s: 4.9375
Reroll everything: 4.5
Based on this, you should reroll on a 4 or less to get the highest damage, but if you have multiple attacks it might be smart to only reroll 2s or 3s in case you get a 1 on your second attack, due to the increase from 3 to 4 being so small.
That's not how the math works. If you replace the die you are rerolling by 4.5 instead of actually rerolling it you are oversimplfying and will end up throwing off the average result.
It is the difference between 1, 2, 2, 3, 3, 4, 4, 5, 5, 6, 6, 7, 7, 8, 8 and 4.5, 2, 3, 4, 5, 6, 7, 8.
(a+b)/c does not equal a/c + b/c.
If you want to look at the odds, assuming you are rerolling your 1 then you have a 1 in 15 chance of getting a 1 in the final result and a 2 in 15 chance of getting any of the other values on the d8. 1/15 = .0666 repeating, so by rerolling 1s you have about a 13.5% chance of getting any of the other results and a 6.5% chance of getting a 1. You can use these probabilities to get the average result as well.
That's not how the math works. If you replace the die you are rerolling by 4.5 instead of actually rerolling it you are oversimplfying and will end up throwing off the average result.
It is the difference between 1, 2, 2, 3, 3, 4, 4, 5, 5, 6, 6, 7, 7, 8, 8 and 4.5, 2, 3, 4, 5, 6, 7, 8.
(a+b)/c does not equal a/c + b/c.
If you want to look at the odds, assuming you are rerolling your 1 then you have a 1 in 15 chance of getting a 1 in the final result and a 2 in 15 chance of getting any of the other values on the d8. 1/15 = .0666 repeating, so by rerolling 1s you have about a 13.5% chance of getting any of the other results and a 6.5% chance of getting a 1. You can use these probabilities to get the average result as well.
Bit of a rounding error, but this method bears out the same result as I got before.
This isn’t correct. You’re assigning equal values of rolling the new dice as the old dice, and this is not correct.
When re-rolling 1s on a 1d8, the chance of you rolling a 1 is 1-in-8, and the chance of rolling a 1 on the reroll again is 1-in-8. So the chance of rolling a 1 is 1/64 which is 1.5625%.
The chance of rolling a 2 in the first roll is 1/8, and the chance of rolling a 1-then-a-2 is 1/64, so the odds are 1/8 + 1/64 = 9/64 which is 14.0625. And that’s the same for all other numbers 2-8.
7 numbers with a probability of 14.0625% plus 1 number with a probability of 1.5625% = 100% probability.
For simple statistics you can just look at the possible results based on what your reroll threshhold is.
No Rerolls
Possible results on a d8 are 1, 2, 3, 4, 5, 6, 7, 8, sum of all possible results is 36, 36/4 = 4.5
Reroll 1s
Possible results on a d8 are 1-1, 1-2, 1-3, 1-4, 1-5, 1-6, 1-7, 1-8, 2, 3, 4, 5, 6, 7, 8, here the sum of all possible results is 71 and there are 15 possible results. 71/15 = 4.7333{r}
I will save the listing out for the others...but here are the average results based on where you put the threshhold:
So the maximum value on the reroll is to reroll any 1s or 2s. It is counterintuitive, I would think like others posed that anything below the average die value you would want to reroll. I think this is because you keep the second result rather than taking the higher or two like advantage.
I said I wouldn't go through the trouble of listing all the possible results, but just in case someone wants to check my math here it is: For clarity the blue numbers are the results that were "rerolled".
As mentioned, if you have multiple attack then your accuracy also comes into play on whether or not you want to reroll the two on your first shot, because what if you hit on your second attack and roll a 1. I could crunch those numbers too if you are interested, but it gets complicated...this is the quick and simple straight forward answer. On a d8 reroll 1s and 2s.
This math seems wrong to me. You are totaling up all the possible roll results and dividing them by the total number of options, but that only gets you an accurate number if each result is equally likely, which in this case, they aren't. You are much more likely to roll an 8 then to roll a 1 and then reroll it into an 8. (1 in 8 chance vs 1 in 64 chance)
Like Jhfffan mentioned, you can replace each rerolled number with 4.5 and then divide by 8. Doing the math this way gets you the following:
Reroll nothing: 4.5
Reroll up to 1s: 4.9375
Reroll up to 2s: 5.25
Reroll up to 3s: 5.4375
Reroll up to 4s: 5.5
Reroll up to 5s: 5.4375
Reroll up to 6s: 5.25
Reroll up to 7s: 4.9375
Reroll everything: 4.5
Based on this, you should reroll on a 4 or less to get the highest damage, but if you have multiple attacks it might be smart to only reroll 2s or 3s in case you get a 1 on your second attack, due to the increase from 3 to 4 being so small.
This also means if you have Savage Attacker or some other reroll method, you should also reroll 5s as well if you have Piercer (again, considering only one attack). Math checks out 🙂
Where are you getting 1/15 chance of getting a 1? There is a 1/8 chance you get a 1 to begin with, and then another 1/8 chance to reroll into another 1. Multiply those together and you get a 1/64 chance of ending up with a 1. That is a 1.5% chance to get a 1 with a reroll, not 6.5%. Just because there are 15 different outcomes, does not mean that all 15 have the same likelihood of happening.
If you reroll all of your 1s, a 1 essentially turns into 1d8. Rerolling a die will on average result in the reroll being 4.5. This means that a 1 is actually on average a 4.5 when taking the average values after a reroll, so the math would be taking the average of 4.5, 2, 3, 4, 5, 6, 7, 8.
You are treating the two rolls as unrelated when they are not. The rolls are linked because you only roll a second time if the first roll meets the criteria for a reroll. (If you roll a 3 the first time then you don't roll a second time, so the possibility of 3-1 is non-existent).
Because of this there are not 64 possible results. There are 7 possible acceptable results on the first roll (2 - 8) and if you end up with the one unacceptable result then there are 8 possible results on the second roll (1 - 8). 7+8 is 15. That is where the 15 comes from.
And again...You can't replace the first term with 4.5...it doesn't work that way. You are wrong. If it did work that way then the average value of 4.5, 2, 3, 4, 5, 6, 7, 8. and 1, 2, 2, 3, 3, 4, 4, 5, 5, 6, 6, 7, 7, 8, 8 would be the same, and they aren't. By averaging the 1 early (into the 4.5) you are giving the other values (2 - 8) more "weight" in the final average the 1 should be 1/15th of the average, but instead it is 1/64th of it. It is an "average of an average" error.
A fact that we teach in our OLAP class is that you can’t take the average of averages and hope it will match the average. This is a common enough mistake for people working with databases and doing number crunching. It is only true if all of the averages are computed over sets having the same cardinality, otherwise it is false. In fancy terms, the average is not distributive though it is algebraic. This phenomenon has a name: the fact that the average of averages is not the average is an instance of Simpson’s Paradox.
Here is an example, consider the following list of numbers:
3
4
6
5
4.5
The average is 4.5. However, we can split the list in two: The average of the first list is 3.5:
3
4
The average of the second list is approximately 5.2:
6
5
4.5
However, the average of the two average is (5.2 +3.5)/2 which is less than 4.5!
This only works if the two sets have a different number of elements.
Just because there are 15 different outcomes, does not mean that all 15 have the same likelihood of happening.
Also also I literally addressed this in the post when I said that a 1 has a ~6.5% chance of happening compared to the ~13.5% chance of the other results. 1/15 vs 2/15.
Rollback Post to RevisionRollBack
Founding Member of the High Roller Society.(Currently trying to roll max on 4d6)
There is a dice calculator out there that could solve this debate easily, but I cannot figure out how to implement rerolls on it. I have a decent head for coding languages, but can't make sense of this.
If anyone wants to take a crack at it go to anydice.com
Rollback Post to RevisionRollBack
Founding Member of the High Roller Society.(Currently trying to roll max on 4d6)
There is a dice calculator out there that could solve this debate easily, but I cannot figure out how to implement rerolls on it. I have a decent head for coding languages, but can't make sense of this.
If anyone wants to take a crack at it go to anydice.com
Relevant text from the Piercer Feat:
It's been like 25 years since I last took probability, so I don't remember how to calculate this.
For my longbow wielding Gloom Stalker, I'm trying to determine the mathematically best time to re-roll a d8 from the bow's damage to have the maximal increase to average damage.
I feel like it's not best to re-roll a result of four, because you then have the exact same odds of rolling a 1-4 again, which doesn't really increase your average roll (since it isn't roll two and take the highest). But, I could be wrong. Intuitively, I feel like re-rolling anything 1-3 would result in the largest increase in average result, but again, I could be completely wrong.
Anyone out there familiar enough with this kind of probability to determine what results on the initial D8 roll are most likely to result in an average increase in damage done, and what would be the net average roll in this situation?
Thanks for your assistance.
I can calculate this for you if you want, but it would depend on your likelihood of hit percentage, so factors like disadvantage/advantage, number of attacks, not getting your second attack, etc will influence the roll
I will try and make a spreadsheet for you.
I wasn't looking for anything that complicated.
I was specifically looking for:
If I can re-roll a single d8 where I must take the results of the re-roll, what is the optimum number of the original d8 to trigger the re-roll to maximize the average result. It's easy for roll two and take the highest to determine the average, but that isn't this case.
All other variables will be immaterial. Once I have hit, I just want to know when I should re-roll the damage roll. I feel like it should be on rolls of 1-3, but I'm not certain.
To maximize, it’s anything below average (4.5). So 1-4 reroll.
Theres a chance you hit your second attack and roll 1-3 and it would be better served, but that’s where the complications start adding up.
If you have other abilities like GWF Fighting Style, it may actually be better to reroll 1-5 at times.
Technically, the increase exists for rerolling a 4, but the potential gain there isn't high enough for me personally. I'm of the opinion that rerolling on a 1-3 hits the sweet spot for risk-reward. I went through the math a while back with Savage Attacker. It's a little different in that you can keep either roll, but I was also trying to maximize it's value for multiple attacks (on a fighter).
Brewsky is right though that anything less than average damage will provide an increase on average. 1 less than the highest number that is less than average will be the best for more conservative players who are more willing to keep a moderate amount of damage than lose out by rolling even lower.
If you think about it in terms of rolling at least as well as you did before, then your odds improve. Instead of thinking as rolling better than a 4 as only being a 50/50 proposition, you think of it as rolling at least a 4, which is 5/8 or 62.5% of the time. Rolling at least a 3 is 6/8 or 75%. That may be the easiest way for you to determine what you are comfortable with when it comes to rerolls.
For simple statistics you can just look at the possible results based on what your reroll threshhold is.
No Rerolls
Possible results on a d8 are 1, 2, 3, 4, 5, 6, 7, 8, sum of all possible results is 36, 36/4 = 4.5
Reroll 1s
Possible results on a d8 are 1-1, 1-2, 1-3, 1-4, 1-5, 1-6, 1-7, 1-8, 2, 3, 4, 5, 6, 7, 8, here the sum of all possible results is 71 and there are 15 possible results. 71/15 = 4.7333{r}
I will save the listing out for the others...but here are the average results based on where you put the threshhold:
No Rerolls: 4.5
Reroll 1s: 4.7333
Reroll 1-2: 4.7727
Reroll 1-3: 4.7586
Reroll 1-4: 4.7222
Reroll 1-5: 4.6744
Reroll 1-6: 4.62
Reroll 1-7: 4.5614
Always Reroll: 4.5
So the maximum value on the reroll is to reroll any 1s or 2s. It is counterintuitive, I would think like others posed that anything below the average die value you would want to reroll. I think this is because you keep the second result rather than taking the higher or two like advantage.
I said I wouldn't go through the trouble of listing all the possible results, but just in case someone wants to check my math here it is: For clarity the blue numbers are the results that were "rerolled".
Reroll 1-2s: 1 2 3 4 5 6 7 8 1 2 3 4 5 6 7 8 3 4 5 6 7 8 Count: 22 Total: 108 108/22 = 4.7722
Reroll 1-3s:1 2 3 4 5 6 7 8 1 2 3 4 5 6 7 8 1 2 3 4 5 6 7 8 4 5 6 7 8 Count: 29 Total: 138 138/29 = 4.7586
Reroll 1-4s:1 2 3 4 5 6 7 8 1 2 3 4 5 6 7 8 1 2 3 4 5 6 7 8 1 2 3 4 5 6 7 8 5 6 7 8 Count: 36 Total: 170 170/36 = 4.7222
Reroll 1-5s: 1 2 3 4 5 6 7 8 1 2 3 4 5 6 7 8 1 2 3 4 5 6 7 8 1 2 3 4 5 6 7 8 1 2 3 4 5 6 7 8 6 7 8 Count: 43 Total: 201 201/43 = 4.6744
Reroll 1-6s: 1 2 3 4 5 6 7 8 1 2 3 4 5 6 7 8 1 2 3 4 5 6 7 8 1 2 3 4 5 6 7 8 1 2 3 4 5 6 7 8 1 2 3 4 5 6 7 8 7 8 Count: 50 Total: 230 230/50 = 4.62
Reroll 1-7s: 1 2 3 4 5 6 7 8 1 2 3 4 5 6 7 8 1 2 3 4 5 6 7 8 1 2 3 4 5 6 7 8 1 2 3 4 5 6 7 8 1 2 3 4 5 6 7 8 1 2 3 4 5 6 7 8 8 Count: 57 Total: 260 260/57 = 4.5614
Reroll all: 1 2 3 4 5 6 7 8 1 2 3 4 5 6 7 8 1 2 3 4 5 6 7 8 1 2 3 4 5 6 7 8 1 2 3 4 5 6 7 8 1 2 3 4 5 6 7 8 1 2 3 4 5 6 7 8 1 2 3 4 5 6 7 8 Count: 64 Total: 288 288/64 = 4.5
As mentioned, if you have multiple attack then your accuracy also comes into play on whether or not you want to reroll the two on your first shot, because what if you hit on your second attack and roll a 1. I could crunch those numbers too if you are interested, but it gets complicated...this is the quick and simple straight forward answer. On a d8 reroll 1s and 2s.
Founding Member of the High Roller Society. (Currently trying to roll max on 4d6)
There is also something to be said about the gameplay aspect here that goes beyond average values. If the enemy is very close to dead and you act right before them...or if you rolled a crit and your table does double the value rather than rolling twice...
There are times you would want to take the counterintuitive route of rolling again when it is suboptimal incase you hit the jackpot (8).
Founding Member of the High Roller Society. (Currently trying to roll max on 4d6)
@Bounces: That was what I was looking for. Thanks for the info.
Notice that the drop off from 1-2 to 1-3 is .0141, whereas the drop off from 1-3 to 1-4 .0364 bringing the 4 or lower down .0505 from the max gain. The variation isn't that great overall, so doing any of them can be useful.
By the way, you can shorthand the rerolled numbers by putting 4.5 in their place and just dividing by 8.
I tend to think of it in simpler terms, if you have a 4, there’s 3 chances you’ll do worse, and 5 chances you’ll do the same or better, so re-roll. If you have a 5, there 3 chances you’ll do better, and 5 you’ll do the same or worse, so don’t.
And if you do it that way overall your DPR is worse. Since the OP was looking for maximum value going to the probability maths is how we get the answer they are looking for.
Founding Member of the High Roller Society. (Currently trying to roll max on 4d6)
This math seems wrong to me. You are totaling up all the possible roll results and dividing them by the total number of options, but that only gets you an accurate number if each result is equally likely, which in this case, they aren't. You are much more likely to roll an 8 then to roll a 1 and then reroll it into an 8. (1 in 8 chance vs 1 in 64 chance)
Like Jhfffan mentioned, you can replace each rerolled number with 4.5 and then divide by 8. Doing the math this way gets you the following:
Based on this, you should reroll on a 4 or less to get the highest damage, but if you have multiple attacks it might be smart to only reroll 2s or 3s in case you get a 1 on your second attack, due to the increase from 3 to 4 being so small.
That's not how the math works. If you replace the die you are rerolling by 4.5 instead of actually rerolling it you are oversimplfying and will end up throwing off the average result.
It is the difference between 1, 2, 2, 3, 3, 4, 4, 5, 5, 6, 6, 7, 7, 8, 8 and 4.5, 2, 3, 4, 5, 6, 7, 8.
(a+b)/c does not equal a/c + b/c.
If you want to look at the odds, assuming you are rerolling your 1 then you have a 1 in 15 chance of getting a 1 in the final result and a 2 in 15 chance of getting any of the other values on the d8. 1/15 = .0666 repeating, so by rerolling 1s you have about a 13.5% chance of getting any of the other results and a 6.5% chance of getting a 1. You can use these probabilities to get the average result as well.
(1 * 1/15) + (2 * 2/15) + (3 * 2/15) + (4 * 2/15) + (5 * 2/15) + (6 * 2/15) + (7 * 2/15) + (8 * 2/15)
.06667 +.26667 + .40000 + .53333 + .66667 + .80000 + .93333 + 1.06667 = 4.73334
Bit of a rounding error, but this method bears out the same result as I got before.
Founding Member of the High Roller Society. (Currently trying to roll max on 4d6)
This isn’t correct. You’re assigning equal values of rolling the new dice as the old dice, and this is not correct.
When re-rolling 1s on a 1d8, the chance of you rolling a 1 is 1-in-8, and the chance of rolling a 1 on the reroll again is 1-in-8. So the chance of rolling a 1 is 1/64 which is 1.5625%.
The chance of rolling a 2 in the first roll is 1/8, and the chance of rolling a 1-then-a-2 is 1/64, so the odds are 1/8 + 1/64 = 9/64 which is 14.0625. And that’s the same for all other numbers 2-8.
7 numbers with a probability of 14.0625% plus 1 number with a probability of 1.5625% = 100% probability.
1 * 0.015625 = 0.015625
(2+3+4+5+6+7+8) * 0.140625 = 4.9218
4.9218+0.015625 = 4.9374
Which is close enough to 4.5+2+3+…+8 / 8.
This also means if you have Savage Attacker or some other reroll method, you should also reroll 5s as well if you have Piercer (again, considering only one attack). Math checks out 🙂
Where are you getting 1/15 chance of getting a 1? There is a 1/8 chance you get a 1 to begin with, and then another 1/8 chance to reroll into another 1. Multiply those together and you get a 1/64 chance of ending up with a 1. That is a 1.5% chance to get a 1 with a reroll, not 6.5%. Just because there are 15 different outcomes, does not mean that all 15 have the same likelihood of happening.
If you reroll all of your 1s, a 1 essentially turns into 1d8. Rerolling a die will on average result in the reroll being 4.5. This means that a 1 is actually on average a 4.5 when taking the average values after a reroll, so the math would be taking the average of 4.5, 2, 3, 4, 5, 6, 7, 8.
You are treating the two rolls as unrelated when they are not. The rolls are linked because you only roll a second time if the first roll meets the criteria for a reroll. (If you roll a 3 the first time then you don't roll a second time, so the possibility of 3-1 is non-existent).
Because of this there are not 64 possible results. There are 7 possible acceptable results on the first roll (2 - 8) and if you end up with the one unacceptable result then there are 8 possible results on the second roll (1 - 8). 7+8 is 15. That is where the 15 comes from.
And again...You can't replace the first term with 4.5...it doesn't work that way. You are wrong. If it did work that way then the average value of 4.5, 2, 3, 4, 5, 6, 7, 8. and 1, 2, 2, 3, 3, 4, 4, 5, 5, 6, 6, 7, 7, 8, 8 would be the same, and they aren't. By averaging the 1 early (into the 4.5) you are giving the other values (2 - 8) more "weight" in the final average the 1 should be 1/15th of the average, but instead it is 1/64th of it. It is an "average of an average" error.
https://lemire.me/blog/2005/10/28/average-of-averages-is-not-the-average/
A fact that we teach in our OLAP class is that you can’t take the average of averages and hope it will match the average. This is a common enough mistake for people working with databases and doing number crunching. It is only true if all of the averages are computed over sets having the same cardinality, otherwise it is false. In fancy terms, the average is not distributive though it is algebraic. This phenomenon has a name: the fact that the average of averages is not the average is an instance of Simpson’s Paradox.
Here is an example, consider the following list of numbers:
The average is 4.5. However, we can split the list in two:
The average of the first list is 3.5:
The average of the second list is approximately 5.2:
However, the average of the two average is (5.2 +3.5)/2 which is less than 4.5!
This only works if the two sets have a different number of elements.
Founding Member of the High Roller Society. (Currently trying to roll max on 4d6)
Also also I literally addressed this in the post when I said that a 1 has a ~6.5% chance of happening compared to the ~13.5% chance of the other results. 1/15 vs 2/15.
Founding Member of the High Roller Society. (Currently trying to roll max on 4d6)
There is a dice calculator out there that could solve this debate easily, but I cannot figure out how to implement rerolls on it. I have a decent head for coding languages, but can't make sense of this.
If anyone wants to take a crack at it go to anydice.com
Founding Member of the High Roller Society. (Currently trying to roll max on 4d6)
https://anydice.com/program/260ed
As indicated, the average is 4.94. And yes, Anydice’s functions are kinda messy. Haha