Let's analyze the scaling behavior of the average number of hits it takes for a 4E Essentials fighter (both knight and slayer) to kill a monster, with some very egregious extrapolations. (We will ignore "Power Strike" in this analysis).
Recall that the average number of hits to kill a monster (N) is the ratio:
N = (monster hit points)/(average amount of damage per successful attack).
The damage done by at-will powers is typically:
[W] + stat mod + magic enhancement + misc.
Stat increases happen at levels 4 and 8, where +1 is added to two stats of choice. We assume one of the stats goes into the primary stat (STR), and the other goes into the secondary stat (DEX).
At the paragon and epic tiers, the stat mod increases happen at levels 11, 14, 18 (paragon) and levels 21, 24, 28 (epic). At levels 14, 18, 24, 28, the stat mod increases are similar to the ones at levels 4 and 8. At levels 11 and 21, the stat mod increases are +1 to every stat.
The magic enhancement for different levels assumes the table:
1 -5 -> +1
6 - 10 -> +2
11 - 15 -> +3
16 - 20 -> +4
21 - 25 -> +5
26 - 30 -> +6
(ie. Magic enhancement increases by +1 every five levels).
To make things simple, we will examine the levels 11-30 of paragon and epic tiers as one entity. Over levels 11-30, the total additional damage contributed by the stat increases and magic enhancement is +7, due to +3 from stat increases and +4 from magic enhancement, by the time one reaches level 30. (Heroic tier by level 10, typically already has a +6 to +8 contributed to the damage, where: +3 to +5 is from the stat mod, +1 from the two stat increases, and +2 from magic enhancement).
One egregious assumption we will make, is that this pattern of stat increases and magic enhancement remains the same every 20 levels as one goes to higher levels beyond level 30. For example, stat increases at levels 31, 34, 38, 41, 44, 48, etc ... and magic enhancement increases of +1 every five levels.
So above level 10, the increase to damage from stat increases and magic enhancement scales approximately as: 7*(level-10)/20
Another egregious assumption we will make, is that the damage of at-will weapon attacks increase as follows.
Slayer
(extrapolating the heroic, paragon, epic "mighty slayer" class ability)
level 5 --> 2 + dex mod damage
level 15 --> 5 + dex mod damage
level 25 --> 8 + dex mod damage
level 35 ---> 11 + dex mod damage
level 45 ---> 14 + dex mod damage
etc ...
(extrapolating the "battle wrath" stance)
level 1 --> +2 damage
level 11 --> +3 damage
level 21 --> +4 damage
level 31 --> +5 damage
level 41 --> +6 damage
etc ...
So for a Slayer always using the "battle wrath" stance, the average damage scales approximately as (above level 10):
avg[W] + level*14/27 + 1 + D + 7*(level-10)/20
where D is the dex mod at level 1.
Knight
(extrapolating the heroic, paragon, epic "weapon mastery" class ability)
level 5 --> +1 damage
level 15 --> +2 damage
level 25 --> +3 damage
level 35 ---> +4 damage
level 45 ---> +5 damage
etc ...
(extrapolating the "battle wrath" stance)
level 1 --> +2 damage
level 11 --> +3 damage
level 21 --> +4 damage
level 31 --> +5 damage
level 41 --> +6 damage
etc ...
So for a Knight always using the "battle wrath" stance, the average damage scales approximately as (above level 10):
avg[W] + level*5/24 + 2 + 7*(level-10)/20
On the monster side, the hit points of various monsters from page 184 of the 4E DMG1 are (assuming this is the same in the 4E Essentials DM Kit box set):
ROLE*(level +1) + CON
where ROLE is:
Artillery, Lurker --> ROLE = 6
Skirmisher, Soldier, Controller --> ROLE = 8
Brute --> ROLE = 10
(Elites double the hit points, while Solos quadruple the hit points).
Now the average number of hits to kill a monster being attacked by a player (of the same level as the monster) is approximately the ratio (for a high level):
Slayer
N = [ROLE*(level+1) + CON]/[avg[W] + level*14/27 + 1 + D + 7*(level-10)/20]
Knight
N = [ROLE*(level+1) + CON]/[avg[W] + level*5/24 + 2 + 7*(level-10)/20]
Taking the level to infinity, the ratios approach the limits of:
Slayer
N -> ROLE/[14/27 + 7/20] = 1.15*ROLE
Knight
N -> ROLE/[5/24 + 7/20] = 1.79*ROLE
For a skirmisher monster (ROLE=8), on average it will take a Slayer around 9 hits and a Knight around 14 hits (without "power strike") to kill the monster as the level goes to infinity.
Let's compare these results to the older Heinsoo 4E classes.
It turns out the 4E Essentials Slayer using "battle wrath" without using "Power Strike", is approximately equivalent damage-wise to a non-striker Heinsoo 4E class using a weapon with damage dice [W]=d10. The 4E Essentials Knight using "battle wrath" without using "Power Strike", is approximately equivalent damage-wise to a non-striker Heinsoo 4E class using a weapon with damage dice [W]=d4.
Tuesday, September 14, 2010
4E Essentials D&D scaling analysis articles.
We'll be reposting the series 4E Essentials D&D scaling analysis articles from therpgsite.
Sunday, September 12, 2010
In-game probability in practice - part 5.
Here are the 95% and 99% confidence intervals for the practical in-game probabilities "Y" after rolling a die n times, with an underlying theoretical probability of 50% (p = 0.5).
(They were calculated using a program which implemented the Clopper-Pearson method, when the normal approximation is no longer valid for n < 40).
95%
n = 2 --> (0.0126, 0.9874)
n = 4 --> (0.0676, 0.9324)
n = 6 --> (0.1181, 0.8819)
n = 8 --> (0.1507, 0.8430)
n = 10 --> (0.1871, 0.8129)
n = 12 --> (0.2109, 0.7891)
n = 14 --> (0.2304, 0.7696)
n = 16 --> (0.2465, 0.7535)
n = 18 --> (0.2602, 0.7381)
n = 20 --> (0.2720, 0.7280)
n = 22 --> (0.2822, 0.7178)
n = 24 --> (0.2912, 0.7088)
n = 26 --> (0.2993, 0.7007)
n = 28 --> (0.3065, 0.6935)
n = 30 --> (0.3130, 0.6870)
n = 32 --> (0.3189, 0.6811)
n = 34 --> (0.3243, 0.6757)
n = 36 --> (0.3292, 0.6708)
n = 38 --> (0.3338, 0.6662)
n = 40 --> (0.3380, 0.6620)
99%
n = 2 --> (0.0025, 0.9975)
n = 4 --> (0.0294, 0.9706)
n = 6 --> (0.0663, 0.9337)
n = 8 --> (0.0999, 0.9001)
n = 10 --> (0.1283, 0.8717)
n = 12 --> (0.1522, 0.8478)
n = 14 --> (0.1724, 0.8276)
n = 16 --> (0.1897, 0.8103)
n = 18 --> (0.2047, 0.7953)
n = 20 --> (0.2177, 0.7823)
n = 22 --> (0.2293, 0.7707)
n = 24 --> (0.2396, 0.7604)
n = 26 --> (0.2489, 0.7511)
n = 28 --> (0.2572, 0.7428)
n = 30 --> (0.2649, 0.7351)
n = 32 --> (0.2718, 0.7282)
n = 34 --> (0.2782, 0.7281)
n = 36 --> (0.2841, 0.7152)
n = 38 --> (0.2895, 0.7105)
n = 40 --> (0.2946, 0.7054)
As one can see for less than 4 or 5 rolls of a d20, with a 95% to 99% certainty, the practical in-game probabilities "Y" include the cases of 5% (ie. only hitting on a d20 roll of 20) and 95% (ie. hit on any roll of a d20, except on a 1).
Also in a generic DnD combat encounter which lasts 8 rounds or less, with 95% to 99% certainty, it should not be surprising at all to see a player which only hits once on all of their d20 rolls during the encounter (ie. Pinky's cousin), or a player which only misses once on all their d20 rolls during the encounter (ie. hot hand).
(Pinky is a "cursed" d20 which never rolls a success).
For completion, the number of rounds "n" where it is not surprising to see one with "Pinky's cousin" or a "hot hand", with 95% to 99% certainty.(They were calculated using a program which implemented the Clopper-Pearson method, when the normal approximation is no longer valid for n < 40).
95%
n = 2 --> (0.0126, 0.9874)
n = 4 --> (0.0676, 0.9324)
n = 6 --> (0.1181, 0.8819)
n = 8 --> (0.1507, 0.8430)
n = 10 --> (0.1871, 0.8129)
n = 12 --> (0.2109, 0.7891)
n = 14 --> (0.2304, 0.7696)
n = 16 --> (0.2465, 0.7535)
n = 18 --> (0.2602, 0.7381)
n = 20 --> (0.2720, 0.7280)
n = 22 --> (0.2822, 0.7178)
n = 24 --> (0.2912, 0.7088)
n = 26 --> (0.2993, 0.7007)
n = 28 --> (0.3065, 0.6935)
n = 30 --> (0.3130, 0.6870)
n = 32 --> (0.3189, 0.6811)
n = 34 --> (0.3243, 0.6757)
n = 36 --> (0.3292, 0.6708)
n = 38 --> (0.3338, 0.6662)
n = 40 --> (0.3380, 0.6620)
99%
n = 2 --> (0.0025, 0.9975)
n = 4 --> (0.0294, 0.9706)
n = 6 --> (0.0663, 0.9337)
n = 8 --> (0.0999, 0.9001)
n = 10 --> (0.1283, 0.8717)
n = 12 --> (0.1522, 0.8478)
n = 14 --> (0.1724, 0.8276)
n = 16 --> (0.1897, 0.8103)
n = 18 --> (0.2047, 0.7953)
n = 20 --> (0.2177, 0.7823)
n = 22 --> (0.2293, 0.7707)
n = 24 --> (0.2396, 0.7604)
n = 26 --> (0.2489, 0.7511)
n = 28 --> (0.2572, 0.7428)
n = 30 --> (0.2649, 0.7351)
n = 32 --> (0.2718, 0.7282)
n = 34 --> (0.2782, 0.7281)
n = 36 --> (0.2841, 0.7152)
n = 38 --> (0.2895, 0.7105)
n = 40 --> (0.2946, 0.7054)
As one can see for less than 4 or 5 rolls of a d20, with a 95% to 99% certainty, the practical in-game probabilities "Y" include the cases of 5% (ie. only hitting on a d20 roll of 20) and 95% (ie. hit on any roll of a d20, except on a 1).
Also in a generic DnD combat encounter which lasts 8 rounds or less, with 95% to 99% certainty, it should not be surprising at all to see a player which only hits once on all of their d20 rolls during the encounter (ie. Pinky's cousin), or a player which only misses once on all their d20 rolls during the encounter (ie. hot hand).
(Pinky is a "cursed" d20 which never rolls a success).
One hit only (or only one miss on a "hot hand")
95% certainty --> n < 8
99% certainty --> n < 9
Two hits only (or only two misses on a "hot hand")
95% certainty --> n < 11
99% certainty --> n < 13
Three hits only (or only three misses on a "hot hand")
95% certainty --> n < 14
99% certainty --> n < 16
Four hits only (or only four misses on a "hot hand")
95% certainty --> n < 17
99% certainty --> n < 19
Five hits only (or only five misses on a "hot hand")
95% certainty --> n < 19
99% certainty --> n < 22
Sunday, September 5, 2010
In-game probability in practice - part 4.
For completion, the remaining |Y-p| cases for determining with 95% certainty that the underlying theoretical probability is not 50% (ie. p != 0.5). This was done by laboriously going through cumulative binomial distribution calculations, where the normal approximation is no longer valid.
|Y-p| = 0.20 --> n > 29
|Y-p| = 0.25 --> n > 19
|Y-p| = 0.30 --> n > 13
|Y-p| = 0.35 --> n > 10
|Y-p| = 0.40 --> n > 8
|Y-p| = 0.45 --> n > 7
Within a generic four-five hour DnD game with a few dozen or so d20 rolls made, one would certainly notice a bonus/penalty of 5 or greater to the d20 rolls.
|Y-p| = 0.20 --> n > 29
|Y-p| = 0.25 --> n > 19
|Y-p| = 0.30 --> n > 13
|Y-p| = 0.35 --> n > 10
|Y-p| = 0.40 --> n > 8
|Y-p| = 0.45 --> n > 7
Within a generic four-five hour DnD game with a few dozen or so d20 rolls made, one would certainly notice a bonus/penalty of 5 or greater to the d20 rolls.
In-game probability in practice - part 3.
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%).
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%).
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.
Subscribe to:
Posts (Atom)