Let's examine the scenario where one keeps on getting a practical in-game probability Y (from rolling a d20 die "n" times) which appears to consistently deviate from a presumed theoretical probability "p".
We will assume the underlying distribution for rolling a die "n" times is binomial with a theoretical probability p. For n large enough (ie. np(1-p)>10), the normal approximation for the binomial distribution can be used.
The test statistic "z" used to test whether the underlying distribution indeed has a theoretical probability of p, will follow a standard normal distribution with:
z = (Y-p)/sqrt[p(1-p)/n]
For a 95% certainty that the theoretical probability is something other than p, one requires |z| > 1.96 (from a table of areas under the normal distribution). Similarly for a 99% certainty, |z| > 2.575.
For some concrete numbers, we'll look at the case where the presumed theoretical probability is 50% (p = 0.5). From this, we can calculate what the minimum number "n" of d20 die rolls it takes to determine with 95% certainty that the underlying theoretical probability is not 50% (p != 0.5).
Doing some algebra, we get for "n":
n > [1.96/(Y-p)]^2 *[p(1-p)]
for 95% certainty that the theoretical probability is not p. (Similarly for the case of 99% certainty that the theoretical probability is not p, we get n > [2.575/(Y-p)]^2 *[p(1-p)] ).
For the case where |Y-p| = 0.05 which corresponds to a constant +1 bonus or -1 penalty to a d20 roll, one needs to do n > 384 rolls of a d20 in order to determine with 95% certainty that the underlying theoretical probability is not 50% (ie. p != 0.5).
Similarly for different |Y-p| values:
|Y-p| = 0.05 --> n > 384
|Y-p| = 0.10 --> n > 96
|Y-p| = 0.15 --> n > 43
(For |Y-p| = 0.20, the normal approximation to binomial is no longer valid).
For the case where one wants 99% certainty that the underlying theoretical probability is not 50%, we get for different |Y-p| values:
|Y-p| = 0.05 --> n > 663
|Y-p| = 0.10 --> n > 166
|Y-p| = 0.15 --> n > 74
|Y-p| = 0.20 --> n > 41
(For |Y-p| = 0.25, the normal approximation to binomial is no longer valid).
These results suggest that in a generic four-five hour DnD session, there may not be enough d20 die rolls (of a few dozen) with bonuses/penalties of 1 or 2, to determine whether the underlying theoretical probability "p" is not 50% (p != 0.5) with a 95% (or 99%) certainty. One needs more than a hundred or so d20 die rolls to make this determination.
When one is dealing with bonuses/penalties of 4 (or greater), there may be enough d20 die rolls (of a few dozen) in a generic four-five hour DnD game session, such that one can determine with 95% (or 99%) certainty that the underlying theoretical probability "p" is not 50% (p != 0.5).
Bonuses/penalties of 3, are the borderline cases where there may be enough d20 die rolls to determine if the underlying theoretical probability "p" is not 50% (p != 0.5) with 95% or 99% certainty. (One requires at least 43 d20 die rolls to determine this with 95% certainty).
A +5 or +6 magic weapon being used at heroic tier in 4E DnD, or the players are subjected to an "aura" which causes a -4 or -5 penalty to hit, would be noticeable and conclude with 95% (or better) certainty in a four/five hour DnD session that the underlying theoretical probability "p" is not 50% (p != 0.5).
Possibly this explains why so many bonuses/penalties in 4E DnD are +/-1 or +/-2 to the d20 roll, along with all kinds of kludges to prevent the stacking of too many bonuses/penalties on top of one another. Essentially they're attempting to maintain an illusion of "always fighting orcs" with the underlying theoretical probability "p" appearing to be 50% (p=0.5) over a four-five hour DnD game session. (One would need over a hundred or so d20 die rolls with bonuses/penalties of 2, in order to determine with a 95% or better certainty, that the underlying theoretical probability "p" is not 50%).
Sunday, September 5, 2010
In-game probability in practice - part 2.
Let's examine the practical in-game probability "Y" in a typical 4-5 hour DnD game. ("Y" is the number of successful die rolls divided by the total number of die rolls including the misses/failures).
Recall the 95% confidence interval for the in-game probability "Y" (in the normal approximation to binomial distribution):
p - d < Y < p + d
where d = 1.96 * sqrt[p(1-p)/n] and p = theoretical probability.
(For a 99% confidence interval, d = 2.575 * sqrt[p(1-p)/n]).
For this normal approximation to be viable, np(1-p) > 10.
In a typical DnD game, the theoretical probability of success "p" for a d20 roll whether for attacks, skill checks, etc ... is around 50%. (For this normal approximation to binomial distribution to be valid, one needs n > 40 for p=0.5).
If in a typical DnD game a player is doing several dozen or so d20 rolls in the session, let's look at the deviation "d" when n = 41 (ie. borderline case).
So for n = 41 rolls and theoretical probability of success p = 0.5, the deviation "d" of the in-game probability "Y" from "p" for a 95% confidence interval is:
d = 1.96* sqrt[p(1-p)/n] = 0.15
This means that when one is doing 41 rolls of a d20 through a generic 4-5 hour DnD game session, there is a 95% probability that the practical in-game probability "Y" is within the interval 0.35 < Y < 0.65. So 95% of the time, the in-game probability "Y" sees when one is rolling a d20 41 times in a game, can vary from 35% to 65% for a theoretical probability of p=50% (ie. rolling greater than or equal to an 11 for a success on a d20).
If we do this calculation for a 99% confidence interval, the deviation "d" of the in-game probability "Y" from "p" for a 99% confidence interval is:
d = 2.575* sqrt[p(1-p)/n] = 0.20
This means that when one is doing 41 rolls of a d20 through a generic 4-5 hour DnD game, there is a 99% probability that the practical in-game probability "Y" is within the interval 0.30 < Y < 0.70. So 99% of the time, the in-game probability "Y" sees when one is rolling a d20 41 times in a game, can vary from 30% to 70% for a theoretical probability of p=50% (ie. rolling greater than or equal to an 11 for a success on a d20).
Recall the 95% confidence interval for the in-game probability "Y" (in the normal approximation to binomial distribution):
p - d < Y < p + d
where d = 1.96 * sqrt[p(1-p)/n] and p = theoretical probability.
(For a 99% confidence interval, d = 2.575 * sqrt[p(1-p)/n]).
For this normal approximation to be viable, np(1-p) > 10.
In a typical DnD game, the theoretical probability of success "p" for a d20 roll whether for attacks, skill checks, etc ... is around 50%. (For this normal approximation to binomial distribution to be valid, one needs n > 40 for p=0.5).
If in a typical DnD game a player is doing several dozen or so d20 rolls in the session, let's look at the deviation "d" when n = 41 (ie. borderline case).
So for n = 41 rolls and theoretical probability of success p = 0.5, the deviation "d" of the in-game probability "Y" from "p" for a 95% confidence interval is:
d = 1.96* sqrt[p(1-p)/n] = 0.15
This means that when one is doing 41 rolls of a d20 through a generic 4-5 hour DnD game session, there is a 95% probability that the practical in-game probability "Y" is within the interval 0.35 < Y < 0.65. So 95% of the time, the in-game probability "Y" sees when one is rolling a d20 41 times in a game, can vary from 35% to 65% for a theoretical probability of p=50% (ie. rolling greater than or equal to an 11 for a success on a d20).
If we do this calculation for a 99% confidence interval, the deviation "d" of the in-game probability "Y" from "p" for a 99% confidence interval is:
d = 2.575* sqrt[p(1-p)/n] = 0.20
This means that when one is doing 41 rolls of a d20 through a generic 4-5 hour DnD game, there is a 99% probability that the practical in-game probability "Y" is within the interval 0.30 < Y < 0.70. So 99% of the time, the in-game probability "Y" sees when one is rolling a d20 41 times in a game, can vary from 30% to 70% for a theoretical probability of p=50% (ie. rolling greater than or equal to an 11 for a success on a d20).
In-game probability in practice - part 1.
As one may (or may not) have noticed, rolling a die multiple times does not always follow the theoretical probabilities that closely. For example, it's possible to roll a lot of low numbers on a d20 in a streak. It's also possible to roll a lot of high numbers on a d20 in a row (ie. a hot hand).
So how does the relative frequency of successful die rolls, relate to the theoretical probability for rolling a success?
It turns out, the act of rolling a die multiple times follows a binomial distribution.
To introduce some nomenclature, let's use:
X = number of successes
n = number of trials
p = theoretical probability of success
1-p = theoretical probability of failure
In the case of rolling a die "n" number of times, one will have "X" number of success. (For example, rolling a d20).
The theoretical probability "p" is what one would expect, from examining the DC's for a success. For example, rolling an 11 or over on a d20 is a 50% theoretical probability.
For a binomial distribution, the average value is m = np and the standard deviation is stdev = sqrt[np(1-p)]. For a large enough number of trials "n", one can approximate the binomial distribution with a normal distribution with average value m = np and standard deviation stdev = sqrt[np(1-p)].
(From a probability textbook, such as Sheldon Ross' textbooks on probability, the criteria for the normal approximation being reliable is np(1-p) > 10).
So for rolling a die "n" times and getting "X" successes (where "n" is large), it will follow a standard normal distribution (ie. with mean 0 and variance 1), with:
Z = (X - np)/sqrt[np(1-p)]
Doing some algebra, Z = [(X/n) - p]/sqrt[p(1-p)/n]. In this form, let Y = X/n represents the relative frequency of successes of rolling a die. This would be an estimate "Y" of the theoretical probability "p", from rolling a die "n" times and getting "X" number of successes. We would like to know how "Y" is related to the theoretical "p".
Since Z = (Y-p)/sqrt[p(1-p)/n] approximately follows a standard normal distribution, this means the confidence intervals can be determined for a range of Z's.
For a 95% confidence interval, Z will fall between -1.96 and 1.96 (from a table of areas of a normal distribution). (For a 99% confidence interval, Z will fall between -2.575 an 2.575).
In the case of a 95% confidence interval |Z| < 1.96, this means:
Probability ( |Y-p| < 1.96 * sqrt[p(1-p)/n] ) = 0.95
(Similarly for a 99% confidence interval, Probability ( |Y-p| < 2.575 * sqrt[p(1-p)/n] ) = 0.99 ).
Let the deviation d = 1.96 * sqrt[p(1-p)/n], which represents the deviation of the estimate "Y" from the theoretical probability "p".
From this we can determine how many number of trials "n" (ie. rolls of a die) it takes, such that the deviation between the estimate "Y" and theoretical probability "p" will stay within "d" with a 95% probability.
Doing some algebra, we get "n" in terms of "d":
n = (1.96)^2 * p(1-p)/(d^2)
For various theoretical probabilities "p" and deviations "d", we get:
(For a d20, a 5% deviation d = 0.05 starts to overlap between different values on the d20. If one wants more precision, one can use a 1% deviation d = 0.01).
d = 0.05
p = 0.50 ---> n = 384
p = 0.40 ---> n = 369
p = 0.30 ---> n = 323
p = 0.20 ---> n = 246
p = 0.10 ---> n = 138
d = 0.01
p = 0.50 ---> n = 9604
p = 0.40 ---> n = 9220
p = 0.30 ---> n = 8067
p = 0.20 ---> n = 6147
p = 0.10 ---> n = 3457
These results mean that for 95% of the time when rolling a die to hit a particular DC, the practical in-game probability (ie. number of successes divided by the total number of die rolls including misses/failures) deviating less than 5% from the "theoretical probability", would require at least a few hundred die rolls. (For a 1% deviation between the practical in-game and theoretical probabilities, it would require several thousand die rolls).
Not too surprising that in a generic four-five hour D&D game session where one only does a few dozen or so d20 rolls, the practical in-game probability "Y" does not always converge to the theoretical probability "p". One needs at least several hundred d20 rolls to see this convergence within a 5% tolerance band, 95% of the time.
One shouldn't be shocked at seeing somebody with a "hot hand" streak of rolling lots of high numbers on a d20, or somebody else on a losing streak of rolling lots of low numbers on a d20.
So how does the relative frequency of successful die rolls, relate to the theoretical probability for rolling a success?
It turns out, the act of rolling a die multiple times follows a binomial distribution.
To introduce some nomenclature, let's use:
X = number of successes
n = number of trials
p = theoretical probability of success
1-p = theoretical probability of failure
In the case of rolling a die "n" number of times, one will have "X" number of success. (For example, rolling a d20).
The theoretical probability "p" is what one would expect, from examining the DC's for a success. For example, rolling an 11 or over on a d20 is a 50% theoretical probability.
For a binomial distribution, the average value is m = np and the standard deviation is stdev = sqrt[np(1-p)]. For a large enough number of trials "n", one can approximate the binomial distribution with a normal distribution with average value m = np and standard deviation stdev = sqrt[np(1-p)].
(From a probability textbook, such as Sheldon Ross' textbooks on probability, the criteria for the normal approximation being reliable is np(1-p) > 10).
So for rolling a die "n" times and getting "X" successes (where "n" is large), it will follow a standard normal distribution (ie. with mean 0 and variance 1), with:
Z = (X - np)/sqrt[np(1-p)]
Doing some algebra, Z = [(X/n) - p]/sqrt[p(1-p)/n]. In this form, let Y = X/n represents the relative frequency of successes of rolling a die. This would be an estimate "Y" of the theoretical probability "p", from rolling a die "n" times and getting "X" number of successes. We would like to know how "Y" is related to the theoretical "p".
Since Z = (Y-p)/sqrt[p(1-p)/n] approximately follows a standard normal distribution, this means the confidence intervals can be determined for a range of Z's.
For a 95% confidence interval, Z will fall between -1.96 and 1.96 (from a table of areas of a normal distribution). (For a 99% confidence interval, Z will fall between -2.575 an 2.575).
In the case of a 95% confidence interval |Z| < 1.96, this means:
Probability ( |Y-p| < 1.96 * sqrt[p(1-p)/n] ) = 0.95
(Similarly for a 99% confidence interval, Probability ( |Y-p| < 2.575 * sqrt[p(1-p)/n] ) = 0.99 ).
Let the deviation d = 1.96 * sqrt[p(1-p)/n], which represents the deviation of the estimate "Y" from the theoretical probability "p".
From this we can determine how many number of trials "n" (ie. rolls of a die) it takes, such that the deviation between the estimate "Y" and theoretical probability "p" will stay within "d" with a 95% probability.
Doing some algebra, we get "n" in terms of "d":
n = (1.96)^2 * p(1-p)/(d^2)
For various theoretical probabilities "p" and deviations "d", we get:
(For a d20, a 5% deviation d = 0.05 starts to overlap between different values on the d20. If one wants more precision, one can use a 1% deviation d = 0.01).
d = 0.05
p = 0.50 ---> n = 384
p = 0.40 ---> n = 369
p = 0.30 ---> n = 323
p = 0.20 ---> n = 246
p = 0.10 ---> n = 138
d = 0.01
p = 0.50 ---> n = 9604
p = 0.40 ---> n = 9220
p = 0.30 ---> n = 8067
p = 0.20 ---> n = 6147
p = 0.10 ---> n = 3457
These results mean that for 95% of the time when rolling a die to hit a particular DC, the practical in-game probability (ie. number of successes divided by the total number of die rolls including misses/failures) deviating less than 5% from the "theoretical probability", would require at least a few hundred die rolls. (For a 1% deviation between the practical in-game and theoretical probabilities, it would require several thousand die rolls).
Not too surprising that in a generic four-five hour D&D game session where one only does a few dozen or so d20 rolls, the practical in-game probability "Y" does not always converge to the theoretical probability "p". One needs at least several hundred d20 rolls to see this convergence within a 5% tolerance band, 95% of the time.
One shouldn't be shocked at seeing somebody with a "hot hand" streak of rolling lots of high numbers on a d20, or somebody else on a losing streak of rolling lots of low numbers on a d20.
In-game probability in practice.
We'll be reposting the "In-game probability in practice" posts from therpgsite.
Wednesday, August 25, 2010
4E powers with half-damage on miss.
Let's examine the case for new encounter powers which produce half-damage on a miss. (These are purportedly being introduced in 4E Essentials). Let's assume the new encounter powers still follow the same damage of the Heinsoo 4E D&D encounter powers, but with the "half-damage on miss" tacked on.
For non-striker encounter powers with half-damage on a miss and a probability p of a hitting a monster of the same level, the average damage per round scales approximately as (after level 10):
0.5(p+1)*[level*(average[W])/7 + 7*(level-10)/20]
So as the level goes to infinity, "R" approaches:
R -> 2*ROLE/{(p+1)*[(average[W])/7 + 7/20]}
For different [W] weapons attacking a skirmisher monster (ROLE=8) with the player having a p=50% of hitting the monster, we have average number of rounds "R" as the level goes to infinity:
average[d12] = 6.5 --> R = 8.34
average[d10] = 5.5 --> R = 9.39
average[d8] = 4.5 --> R = 10.74
average[d6] = 3.5 --> R = 12.55
average[d4] = 2.5 --> R = 15.08
On average, for a player hypothetically repeatedly spamming a new encounter power (with half-damage on a miss) against a skirmisher monster of the same level, the average number of rounds "R" to kill the monster is approximately shorter by 45% compared to at-will powers, as the level goes to infinity.
For completeness, there's at-will powers with half-damage on a miss. (IIRC, there's a few monsters with such an at-will power from the 4E MM2 and/or MM3).
With half-damage on a miss, the at-will with half-damage on a miss reduces the number of rounds R to kill a monster by a factor = p/(1+p) in comparison to ordinary at-will powers. (p is the probability of hitting a monster).
For different p:
p = 100% --> factor = 50%
p = 75% --> factor = 42.6%.
p = 50% --> factor = 33.3%
p = 25% --> factor = 20%
So with a probability 50% of hitting a monster, an at-will power with half-damage on a miss will reduce the number of rounds to kill the monster by 33.3%, in comparison to the same at-will power without the half-damage on a miss part.
For non-striker encounter powers with half-damage on a miss and a probability p of a hitting a monster of the same level, the average damage per round scales approximately as (after level 10):
0.5(p+1)*[level*(average[W])/7 + 7*(level-10)/20]
So as the level goes to infinity, "R" approaches:
R -> 2*ROLE/{(p+1)*[(average[W])/7 + 7/20]}
For different [W] weapons attacking a skirmisher monster (ROLE=8) with the player having a p=50% of hitting the monster, we have average number of rounds "R" as the level goes to infinity:
average[d12] = 6.5 --> R = 8.34
average[d10] = 5.5 --> R = 9.39
average[d8] = 4.5 --> R = 10.74
average[d6] = 3.5 --> R = 12.55
average[d4] = 2.5 --> R = 15.08
On average, for a player hypothetically repeatedly spamming a new encounter power (with half-damage on a miss) against a skirmisher monster of the same level, the average number of rounds "R" to kill the monster is approximately shorter by 45% compared to at-will powers, as the level goes to infinity.
For completeness, there's at-will powers with half-damage on a miss. (IIRC, there's a few monsters with such an at-will power from the 4E MM2 and/or MM3).
With half-damage on a miss, the at-will with half-damage on a miss reduces the number of rounds R to kill a monster by a factor = p/(1+p) in comparison to ordinary at-will powers. (p is the probability of hitting a monster).
For different p:
p = 100% --> factor = 50%
p = 75% --> factor = 42.6%.
p = 50% --> factor = 33.3%
p = 25% --> factor = 20%
So with a probability 50% of hitting a monster, an at-will power with half-damage on a miss will reduce the number of rounds to kill the monster by 33.3%, in comparison to the same at-will power without the half-damage on a miss part.
Scaling of 4E daily powers.
Let's do the same for daily powers, where the the daily powers have half-damage on a miss.
Recall for half-damage on a miss: R = 2N/(1+p).
For a daily power, the average damage per hit scales approximately as (after level 10):
level*(average[W])/5 + 7*(level-10)/20.
For the easy case where the daily powers of non-strikers always miss, "R" approaches
R -> 2*ROLE/{p*[(average[W])/5 + 7/20]} = ROLE/{p*[(average[W])/10 + 7/40]}
as the level goes to infinity.
Comparing this to the expression "R" for at-wills, it means that daily powers always missing and producing half-damage, is slightly worse than generic daily powers.
For non-striker daily powers with half-damage on a miss and a probability p of a hitting a monster of the same level, the average damage per round scales approximately as (after level 10):
0.5(p+1)*[level*(average[W])/5 + 7*(level-10)/20]
So as the level goes to infinity, "R" approaches:
R -> 2*ROLE/{(p+1)*[(average[W])/5 + 7/20]}
For different [W] weapons attacking this skirmisher monster (ROLE=8) with the player having a p=50% of hitting the monster, we have average number of rounds "R" as the level goes to infinity:
average[d12] = 6.5 --> R = 6.46
average[d10] = 5.5 --> R = 7.36
average[d8] = 4.5 --> R = 8.53
average[d6] = 3.5 --> R = 10.16
average[d4] = 2.5 --> R = 12.55
On average, for a player hypothetically repeatedly spamming a daily power (with half-damage on a miss) against a skirmisher monster of the same level, the average number of rounds "R" to kill the monster is approximately shorter by a half compared to at-will powers, as the level goes to infinity.
This means a repeatedly "spammed" daily power (with half damage on a miss) attacking a skirmisher (of the same level), is on average damage-wise approximately equal to two at-wills in general. If the daily power always misses and always produces half damage, it is damage-wise approximately equal to one at-will power.
Recall for half-damage on a miss: R = 2N/(1+p).
For a daily power, the average damage per hit scales approximately as (after level 10):
level*(average[W])/5 + 7*(level-10)/20.
For the easy case where the daily powers of non-strikers always miss, "R" approaches
R -> 2*ROLE/{p*[(average[W])/5 + 7/20]} = ROLE/{p*[(average[W])/10 + 7/40]}
as the level goes to infinity.
Comparing this to the expression "R" for at-wills, it means that daily powers always missing and producing half-damage, is slightly worse than generic daily powers.
For non-striker daily powers with half-damage on a miss and a probability p of a hitting a monster of the same level, the average damage per round scales approximately as (after level 10):
0.5(p+1)*[level*(average[W])/5 + 7*(level-10)/20]
So as the level goes to infinity, "R" approaches:
R -> 2*ROLE/{(p+1)*[(average[W])/5 + 7/20]}
For different [W] weapons attacking this skirmisher monster (ROLE=8) with the player having a p=50% of hitting the monster, we have average number of rounds "R" as the level goes to infinity:
average[d12] = 6.5 --> R = 6.46
average[d10] = 5.5 --> R = 7.36
average[d8] = 4.5 --> R = 8.53
average[d6] = 3.5 --> R = 10.16
average[d4] = 2.5 --> R = 12.55
On average, for a player hypothetically repeatedly spamming a daily power (with half-damage on a miss) against a skirmisher monster of the same level, the average number of rounds "R" to kill the monster is approximately shorter by a half compared to at-will powers, as the level goes to infinity.
This means a repeatedly "spammed" daily power (with half damage on a miss) attacking a skirmisher (of the same level), is on average damage-wise approximately equal to two at-wills in general. If the daily power always misses and always produces half damage, it is damage-wise approximately equal to one at-will power.
Scaling of 4E encounter powers.
Let's examine the scaling behavior of the encounter and daily powers, in the unrealistic scenario where they can be repeatedly spammed every round. We will calculate the average number of rounds "R" to kill a monster by a player of the same level.
Recall that a non-striker repeatedly using at-will powers against a monster of the same level, "R" approaches the limit
R = N/p -> N -> ROLE/{p*[(average[W])/10 + 7/20]}
as the level goes to infinity.
For different [W] weapons attacking this skirmisher monster (ROLE=8) with the player having a p=50% of hitting the monster, we have average number of rounds "R" as the level goes to infinity:
average[d12] = 6.5 --> N = 8, R = 16
average[d10] = 5.5 --> N = 8.89, R = 17.78
average[d8] = 4.5 --> N = 10, R = 20
average[d6] = 3.5 --> N = 11.43, R = 22.86
average[d4] = 2.5 --> N = 13.33, R = 26.66
For non-striker encounter powers, the average damage per hit scales approximately as (after level 10):
level*(average[W])/7 + 7*(level-10)/20.
So for non-striker encounter powers, the average damage per round scales approximately as (after level 10):
p*[level*(average[W])/7 + 7*(level-10)/20].
So as the level goes to infinity, "R" approaches:
R -> N/p = ROLE/{p*[(average[W])/7 + 7/20]}
For different [W] weapons attacking this skirmisher monster (ROLE=8) with the player having a p=50% of hitting the monster, we have average number of rounds "R" as the level goes to infinity:
average[d12] = 6.5 --> R = 12.51
average[d10] = 5.5 --> R = 14.01
average[d8] = 4.5 --> R = 16.12
average[d6] = 3.5 --> R = 18.82
average[d4] = 2.5 --> R = 22.63
On average, for a player hypothetically repeatedly spamming an encounter power against a skirmisher monster of the same level, the average number of rounds "R" to kill the monster is shorter by approximately 4 rounds (compared to at-will powers) as the level goes to infinity. (For a 50% probability of hitting the monster, this would mean that it takes approximately 2 less hits to kill the skirmisher monster).
More generally, a hypothetical spammed encounter power takes approximately 15% to 20% less rounds to kill a monster (of the same level).
Recall that a non-striker repeatedly using at-will powers against a monster of the same level, "R" approaches the limit
R = N/p -> N -> ROLE/{p*[(average[W])/10 + 7/20]}
as the level goes to infinity.
For different [W] weapons attacking this skirmisher monster (ROLE=8) with the player having a p=50% of hitting the monster, we have average number of rounds "R" as the level goes to infinity:
average[d12] = 6.5 --> N = 8, R = 16
average[d10] = 5.5 --> N = 8.89, R = 17.78
average[d8] = 4.5 --> N = 10, R = 20
average[d6] = 3.5 --> N = 11.43, R = 22.86
average[d4] = 2.5 --> N = 13.33, R = 26.66
For non-striker encounter powers, the average damage per hit scales approximately as (after level 10):
level*(average[W])/7 + 7*(level-10)/20.
So for non-striker encounter powers, the average damage per round scales approximately as (after level 10):
p*[level*(average[W])/7 + 7*(level-10)/20].
So as the level goes to infinity, "R" approaches:
R -> N/p = ROLE/{p*[(average[W])/7 + 7/20]}
For different [W] weapons attacking this skirmisher monster (ROLE=8) with the player having a p=50% of hitting the monster, we have average number of rounds "R" as the level goes to infinity:
average[d12] = 6.5 --> R = 12.51
average[d10] = 5.5 --> R = 14.01
average[d8] = 4.5 --> R = 16.12
average[d6] = 3.5 --> R = 18.82
average[d4] = 2.5 --> R = 22.63
On average, for a player hypothetically repeatedly spamming an encounter power against a skirmisher monster of the same level, the average number of rounds "R" to kill the monster is shorter by approximately 4 rounds (compared to at-will powers) as the level goes to infinity. (For a 50% probability of hitting the monster, this would mean that it takes approximately 2 less hits to kill the skirmisher monster).
More generally, a hypothetical spammed encounter power takes approximately 15% to 20% less rounds to kill a monster (of the same level).
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