In older editions of D&D and AD&D, the monsters' hit points was typically the number of d8's rolled and a constant number added sometimes. (We'll ignore the constant number added for simplicity).

For X hit dice, the average number of hit points is:

X avg(d8) = 4.5X (*)

Recall the monster hit point formula from a previous post:

N avg(D)

where N is the number of hits it takes to kill a monster, and avg(D) is the average amount of damage a player does per attack.

In an easy example where a player's weapon does 1d8 damage per attack, the monster hit point formula is:

N avg(1d8) = 4.5N (**)

Equating the average number of hit points from (*) and (**), we get N = X. This means that without any improvements in weapons, it takes on average X hits to kill a monster with X hit dice.

For low level monsters with 1 hit die or less, they basically function like minions which die after one hit. Monsters with 2 hit dice, on average have to be hit twice to kill. (Etc ...)

Let's examine how magic bonuses to damage changes things. The formula N for the number of hits to kill a monster, in general is:

N = 4.5X/avg(D)

For different magic weapon bonuses to damage, for a weapon which normally does d8 damage, we have:

D = d8, avg (d8) = 4.5 -> N = X

D = d8 + 1, avg (d8+1) = 5.5 -> N= 0.82 X

D = d8 + 2, avg (d8+2) = 6.5 -> N= 0.69 X

D = d8 + 3, avg (d8+3) = 7.5 -> N= 0.60 X

D = d8 + 4, avg (d8+4) = 8.5 -> N = 0.53 X

D = d8 + 5, avg (d8+5) = 9.5 -> N = 0.47 X

It takes at least around a +4 or +5 magic weapon bonus to damage, to halve the average number of hits to kill a monster.

## Thursday, February 25, 2010

### Hit points of generic 4E monsters. (part 1)

Continuing from the previous post, let's examine the monster hit point formula:

N avg (D)

where N is the number of hits it takes to kill a monster, and avg (D) is the average damage done by a player per attack.

In 4E, a player can max out their primary combat stat such that it has a stat adjustment of +4 or +5 to attacks and damage. The damage dice is anywhere from a d4 (ie. dagger) to a d12 (ie. greataxe) or 2d6 (ie. maul, heavy flail). Many weapons and magic attacks typically have damage dice of d6 or d8 for at-will powers.

Let's take an example of a player with a primary stat adjustment of +4 for combat attacks and damage, along with d6 damage dice. For such a level 1 player attacking a monster, the average damage per attack would be avg(d6 + 4) = 7.5.

If we want a monster to die after being hit a number of times, the hit points should be:

- 6 hits -> 6(7.5) = 45 hit points

- 5 hits -> 5(7.5) = 37 hit points

- 4 hits -> 4(7.5) = 30 hit points

- 3 hits -> 3(7.5) = 22 hit points

In the case of where the player's damage dice is d8, the average damage per attack would be avg(d8 + 4) = 8.5.

If we want a monster to die after being hit a number of times, the hit points should be:

- 6 hits -> 6(8.5) = 51 hit points

- 5 hits -> 5(8.5) = 42 hit points

- 4 hits -> 4(8.5) = 34 hit points

- 3 hits -> 3(8.5) = 25 hit points

Looking through the 4E Monster Manual, various low level generic monsters have hit points:

Kobold: 24, 27, 36, 42 (level 1-4)

Goblin: 25, 29, 31 (level 1-2)

Hobgoblin: 39, 47 (level 3)

Orc: 46, 64 (level 3-5)

Halfling: 22, 34 (level 1-2)

Human: 37, 42, 47 (level 2-4)

Wolf: 38, 67 (level 2-5)

Wraith: 37 (level 5)

Skeleton: 45, 53 (level 3-5)

Lizardfolk: 50, 54 (level 4-5)

Lycanthrope: 48 (level 3)

Rat: 36, 38 (level 1-2)

Zombie: 40, 46, 54 (level 3-4)

Elf: 32, 39 (level 2)

Gnoll: 50 (level 5)

Gnome: 34, 46 (level 2-3)

Doppelganger: 45 (level 3)

Drake: 29, 38, 40, 48, 77 (level 1-5)

A number of monsters at level 1 or 2 fit into the range of 3 or 4 hit kills, by players with weapons which do d6+4 or d8+4 damage for their at-will powers.

N avg (D)

where N is the number of hits it takes to kill a monster, and avg (D) is the average damage done by a player per attack.

In 4E, a player can max out their primary combat stat such that it has a stat adjustment of +4 or +5 to attacks and damage. The damage dice is anywhere from a d4 (ie. dagger) to a d12 (ie. greataxe) or 2d6 (ie. maul, heavy flail). Many weapons and magic attacks typically have damage dice of d6 or d8 for at-will powers.

Let's take an example of a player with a primary stat adjustment of +4 for combat attacks and damage, along with d6 damage dice. For such a level 1 player attacking a monster, the average damage per attack would be avg(d6 + 4) = 7.5.

If we want a monster to die after being hit a number of times, the hit points should be:

- 6 hits -> 6(7.5) = 45 hit points

- 5 hits -> 5(7.5) = 37 hit points

- 4 hits -> 4(7.5) = 30 hit points

- 3 hits -> 3(7.5) = 22 hit points

In the case of where the player's damage dice is d8, the average damage per attack would be avg(d8 + 4) = 8.5.

If we want a monster to die after being hit a number of times, the hit points should be:

- 6 hits -> 6(8.5) = 51 hit points

- 5 hits -> 5(8.5) = 42 hit points

- 4 hits -> 4(8.5) = 34 hit points

- 3 hits -> 3(8.5) = 25 hit points

Looking through the 4E Monster Manual, various low level generic monsters have hit points:

Kobold: 24, 27, 36, 42 (level 1-4)

Goblin: 25, 29, 31 (level 1-2)

Hobgoblin: 39, 47 (level 3)

Orc: 46, 64 (level 3-5)

Halfling: 22, 34 (level 1-2)

Human: 37, 42, 47 (level 2-4)

Wolf: 38, 67 (level 2-5)

Wraith: 37 (level 5)

Skeleton: 45, 53 (level 3-5)

Lizardfolk: 50, 54 (level 4-5)

Lycanthrope: 48 (level 3)

Rat: 36, 38 (level 1-2)

Zombie: 40, 46, 54 (level 3-4)

Elf: 32, 39 (level 2)

Gnoll: 50 (level 5)

Gnome: 34, 46 (level 2-3)

Doppelganger: 45 (level 3)

Drake: 29, 38, 40, 48, 77 (level 1-5)

A number of monsters at level 1 or 2 fit into the range of 3 or 4 hit kills, by players with weapons which do d6+4 or d8+4 damage for their at-will powers.

### How many hit points should a monster have?

In an older WotC article by David Noonan, it is suggestive that monsters last around 5 rounds. How exactly this is implemented, the article does not state precisely.

Perhaps this "5 rounds" means the number of times it takes to hit a monster, before it is killed by the players?

To investigate this question, we need to recall a few things.

Recall that the average number of attacks it takes to hit a monster N times is N/p, where p is the probability of hitting a monster.

Also recall that the average amount of damage done by one attack is p avg(D), where D is the damage dice (ie. such as 1d6+2) and p is the probability of hitting a monster. (avg(D) is the expectation value of D).

If a monster dies after being hit N times, it will take on average N/p attacks to kill it. So on average the number of hit points it should have, is the average number of attacks multiplied by the average amount of damage done by each attack. Assuming each attack is done by the same player with the same weapon on the same monster, the number of hit points the monster should have is:

(N/p)(p avg(D)) = N avg(D)

Interesting that the number of monster hit points in this scenario, is independent of the probability to hit the monster. N avg(D) is only dependent on the number of hits a monster will take before dying, and the average damage done by the player.

Perhaps this "5 rounds" means the number of times it takes to hit a monster, before it is killed by the players?

To investigate this question, we need to recall a few things.

Recall that the average number of attacks it takes to hit a monster N times is N/p, where p is the probability of hitting a monster.

Also recall that the average amount of damage done by one attack is p avg(D), where D is the damage dice (ie. such as 1d6+2) and p is the probability of hitting a monster. (avg(D) is the expectation value of D).

If a monster dies after being hit N times, it will take on average N/p attacks to kill it. So on average the number of hit points it should have, is the average number of attacks multiplied by the average amount of damage done by each attack. Assuming each attack is done by the same player with the same weapon on the same monster, the number of hit points the monster should have is:

(N/p)(p avg(D)) = N avg(D)

Interesting that the number of monster hit points in this scenario, is independent of the probability to hit the monster. N avg(D) is only dependent on the number of hits a monster will take before dying, and the average damage done by the player.

### Average value of a fair die.

Just for completeness, we will show why the average value of a fair dX die is (X+1)/2.

A fair dX die will have X sides with the numbers 1,2, ... to X, which have an equal likelihood of being rolled. Hence the probability of rolling any number is 1/X.

The average value is the sum of the die side values weighted by the probabilities of being rolled. Mathematically, for example the average value for d4 is:

1(1/4) + 2(1/4) + 3(1/4) + 4(1/4) = (1+2+3+4)/4

while for a d6, the average value is:

1(1/6) + 2(1/6) + 3(/16) + 4(1/6) + 5(1/6) + 6(1/6) = (1+2+3+4+5+6)/6

In general, the average value for a fair dX die is:

1(1/X) + 2(1/X) + ... + (X-1)/X + X/X = [1+2+...+(X-1)+X]/X

Recall from high school algebra, the sum of the first n numbers 1, 2, ..., n-1, n is: n(n+1)/2. (This can be shown to be true for any n by mathematical induction).

So for the average value of dX, we get: [X(X+1)/2]/X = (X+1)/2

A fair dX die will have X sides with the numbers 1,2, ... to X, which have an equal likelihood of being rolled. Hence the probability of rolling any number is 1/X.

The average value is the sum of the die side values weighted by the probabilities of being rolled. Mathematically, for example the average value for d4 is:

1(1/4) + 2(1/4) + 3(1/4) + 4(1/4) = (1+2+3+4)/4

while for a d6, the average value is:

1(1/6) + 2(1/6) + 3(/16) + 4(1/6) + 5(1/6) + 6(1/6) = (1+2+3+4+5+6)/6

In general, the average value for a fair dX die is:

1(1/X) + 2(1/X) + ... + (X-1)/X + X/X = [1+2+...+(X-1)+X]/X

Recall from high school algebra, the sum of the first n numbers 1, 2, ..., n-1, n is: n(n+1)/2. (This can be shown to be true for any n by mathematical induction).

So for the average value of dX, we get: [X(X+1)/2]/X = (X+1)/2

### Average damage.

We will examine the average amount of damage when a player attacks a target (ie. monster).

The easiest way to look at average damage, is to examine the expectation value of the damage rolled.

We start off with a probability p of hitting a target, and rolling D amount of damage for a hit. (D is just a placeholder to represent dice damage, such as 1d6+2).

As a concrete example, let's look at an example where the damage is 1d4 and the probability of hitting is p. (We will ignore critical success and failures).

With probability p of hitting, the possible damage from rolling 1d4 is: 1, 2, 3, and 4 for which each of these damage values have a probability of 1/4. So mathematically, the expectation value of the damage is:

p[1(1/4) + 2 (1/4) + 3(1/4) + 4(1/4)] + (1-p)(0)

which after simplifying is: p[avg(1d4)] = p 2.5

The first term is when the attack hits, while the second term (1-p)(0) is when the attack misses.

Recall that the average value of a single d4 die is avg(1d4) = [1(1/4)+2(1/4)+3(1/4)+4(1/4)] = 2.5. In general, the average value of a single fair die dX is avg(dX) = (X+1)/2. (ie. A single fair die dX has X sides, with numbers 1, 2, ..., X on them, with a probability of 1/X of rolling any number from 1 to X).

To bring this example one step up, let's examine the previous concrete example for two attacks. The probability of hitting a target is still p, and the damage is still 1d4.

The expectation value of the damage of these two attacks is:

p^2 [avg(1d4 +1 d4)] + p(1-p)[avg(1d4)] + (1-p)p[avg(1d4)] + (1-p)^2 (0)

The first term is when both attacks hit, while the second and third terms is when one attacks hits while the other misses. The fourth term is when both attacks miss.

Doing some algebra and noticing that since the 1d4's are independently rolled, avg(1d4+1d4) = avg(1d4) + avg(1d4), we get: 2p avg(1d4) for the average damage done by two attacks.

(In more detail, the values of the total damage done by 1d4 in the first attack and another 1d4 in the second attack: are 2, 3, 4, 5, 6, 7, 8 with probabilities

1/16 for 2 damage

2/16 for 3 damage

3/16 for 4 damage

4/16 for 5 damage

3/16 for 6 damage

2/16 for 7 damage

1/16 for 8 damage.

The average total damage value of two d4 rolls is:

2(1/16) + 3(2/16) + 4(3/16) + 5(4/16) + 6(3/16) + 7 (2/16) + 8(1/16)

which gives: 5 = 2.5 + 2.5

One can do a similar exercise and show that for doing three d4 rolls for damage, the average total damage will be 7.5 = 2.5 + 2.5 + 2.5).

If one has the patience, once can show that for three attacks the average damage done by the three attacks is: 3p avg(1d4).

More generally, if individual attacks have different to-hit probabilities of p1, p2, ..., pN and their respective damage dice rolls are D1, D2, ..., DN, one can show using mathematical induction that the average damage done by these N attacks in sequential order from 1 to N is:

p1 avg(D1) + p2 avg(D2) + ... + pN avg(DN)

(To show this, one uses avg(D1 + D2 + ... + DN) = avg(D1) + avg(D2) + ... + avg(DN) since the damage dice rolls are independent of one another).

The easiest way to look at average damage, is to examine the expectation value of the damage rolled.

We start off with a probability p of hitting a target, and rolling D amount of damage for a hit. (D is just a placeholder to represent dice damage, such as 1d6+2).

As a concrete example, let's look at an example where the damage is 1d4 and the probability of hitting is p. (We will ignore critical success and failures).

With probability p of hitting, the possible damage from rolling 1d4 is: 1, 2, 3, and 4 for which each of these damage values have a probability of 1/4. So mathematically, the expectation value of the damage is:

p[1(1/4) + 2 (1/4) + 3(1/4) + 4(1/4)] + (1-p)(0)

which after simplifying is: p[avg(1d4)] = p 2.5

The first term is when the attack hits, while the second term (1-p)(0) is when the attack misses.

Recall that the average value of a single d4 die is avg(1d4) = [1(1/4)+2(1/4)+3(1/4)+4(1/4)] = 2.5. In general, the average value of a single fair die dX is avg(dX) = (X+1)/2. (ie. A single fair die dX has X sides, with numbers 1, 2, ..., X on them, with a probability of 1/X of rolling any number from 1 to X).

To bring this example one step up, let's examine the previous concrete example for two attacks. The probability of hitting a target is still p, and the damage is still 1d4.

The expectation value of the damage of these two attacks is:

p^2 [avg(1d4 +1 d4)] + p(1-p)[avg(1d4)] + (1-p)p[avg(1d4)] + (1-p)^2 (0)

The first term is when both attacks hit, while the second and third terms is when one attacks hits while the other misses. The fourth term is when both attacks miss.

Doing some algebra and noticing that since the 1d4's are independently rolled, avg(1d4+1d4) = avg(1d4) + avg(1d4), we get: 2p avg(1d4) for the average damage done by two attacks.

(In more detail, the values of the total damage done by 1d4 in the first attack and another 1d4 in the second attack: are 2, 3, 4, 5, 6, 7, 8 with probabilities

1/16 for 2 damage

2/16 for 3 damage

3/16 for 4 damage

4/16 for 5 damage

3/16 for 6 damage

2/16 for 7 damage

1/16 for 8 damage.

The average total damage value of two d4 rolls is:

2(1/16) + 3(2/16) + 4(3/16) + 5(4/16) + 6(3/16) + 7 (2/16) + 8(1/16)

which gives: 5 = 2.5 + 2.5

One can do a similar exercise and show that for doing three d4 rolls for damage, the average total damage will be 7.5 = 2.5 + 2.5 + 2.5).

If one has the patience, once can show that for three attacks the average damage done by the three attacks is: 3p avg(1d4).

More generally, if individual attacks have different to-hit probabilities of p1, p2, ..., pN and their respective damage dice rolls are D1, D2, ..., DN, one can show using mathematical induction that the average damage done by these N attacks in sequential order from 1 to N is:

p1 avg(D1) + p2 avg(D2) + ... + pN avg(DN)

(To show this, one uses avg(D1 + D2 + ... + DN) = avg(D1) + avg(D2) + ... + avg(DN) since the damage dice rolls are independent of one another).

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