In a classic wargaming ruleset, it is often the case that you simply move
all of your units when it's your turn. In more recent rulesets, it has become
quite standard that some form of an activation system is in place. When it's
your turn, you can only act with a subset of units. Which units can be
activated exactly is implemented by random draws, managing a hand of cards, or
activation die rolls. It is the latter we will study more closely in this
article.
Typically, each unit is activated by rolling some target number or higher on
a die. In a rulesystem such as Black Powder, this is done through commander
figures. The commander issues an "order" to a unit, and the commander
has to roll 2D6 vs a target number. If the roll succeeds, the unit can move or
charge or fire or do something else. If the commander fails his command roll,
that commander can no longer activate units for the turn, and the player can
switch to the next commander. Such a commander-centric approach often comes
with a clear organization of units, each cluster of units linked to a specific
commander. The idea is to force some form of military organization on the
decision process of the player.
Other rulesets such as Dragon Rampant or Oathmark use a unit-centric
approach. Each unit has an activation number and can activate by itself. Again,
when an activation roll fails, the turn is over. The main difference with a
commander-centric approach is that you only have one chain of die rolls. The
first failed die roll indicates the end of the turn.
The question now is, how many activations can you expect given specific
probabilities for activating units?
Mathematics
It's always useful to strip a rule to its mathematical essentials. What we
have here is that we roll a die (or 2D6, or something else ...) against a
target number, and we need to roll equal to or higher than the target number.
Thus, in common die notation, we might need a 8+ on a 2D6, or 5+ on a D10. Note
that some rulesets require you to roll lower or equal than the target number,
but that's irrelevant for a statistical analysis. The important notion is that
we have some probability p with which the die roll succeeds, after which
we can continue to the next die roll. When we fail our roll, this ends the
chain of successive die rolls.
For simplicity, we assume that the probability p for success is
always the same. In an actual ruleset, there might be modifiers to the value of
p, due to some special abilities of the commander or the unit, but we
will not consider these situations here.
It is obvious that the first die roll will succeed with probability p,
and will fail with probability 1-p. When the roll succeeds, we roll
again, and again will score a success with probability p and fail with probability
1-p. Etc. Thus, to score exactly 1 success, we need the first roll the
succeed, and the second roll to fail. To score 2 successes, we need the first
two rolls to succeed, and the third to fail, and so on. In general, to score
exactly k successes, we need the first k rolls to succeed, and roll
k+1 to fail. The total combined probability for scoring exactly k
consecutive successes is therefore:
prob(p, k) = pk*(1-p)
Let's fill in some numbers. Suppose we need a 5+ on a simple D6. Hence p
= 1/3. We get the following table listing the probability for scoring exactly
that number of successes:
|
D6
|
5+
(p = 0.33)
|
|
0
|
0.67
|
|
1
|
0.22
|
|
2
|
0.07
|
|
3
|
0.02
|
|
4
|
0.01
|
|
5
|
0.00
|
|
6
|
0.00
|
|
7
|
0.00
|
|
9
|
0.00
|
|
10
|
0.00
|
You can see that we have
a 67% chance that we score exactly 0 successes, we have 22% chance we
score exactly 1 success. The probabilities for an increasing number of
successes drops fairly rapidly.
Increasing or decreasing p produces other combined probabilities, so
here we have expanded the table for p values ranging from 5/6 to 1/6.
Note that we could also compute probabilities for more than 10 successes, but
these numbers become very small, although you can see from the table we still
have a 3% chance that we will score exactly 10 successes when having to roll anything
but a 1 (2+).
|
D6
|
2+
(p = 0.93)
|
3+
(p = 0.67)
|
4+
(p = 0.5)
|
5+
(p = 0.33)
|
6+
(p = 0.17)
|
|
0
|
0.17
|
0.33
|
0.50
|
0.67
|
0.83
|
|
1
|
0.14
|
0.22
|
0.25
|
0.22
|
0.14
|
|
2
|
0.12
|
0.15
|
0.13
|
0.07
|
0.02
|
|
3
|
0.10
|
0.10
|
0.06
|
0.02
|
0.00
|
|
4
|
0.08
|
0.07
|
0.03
|
0.01
|
0.00
|
|
5
|
0.07
|
0.04
|
0.02
|
0.00
|
0.00
|
|
6
|
0.06
|
0.03
|
0.01
|
0.00
|
0.00
|
|
7
|
0.05
|
0.02
|
0.00
|
0.00
|
0.00
|
|
9
|
0.03
|
0.01
|
0.00
|
0.00
|
0.00
|
|
10
|
0.03
|
0.01
|
0.00
|
0.00
|
0.00
|
And the table below gives the probabilities when rolling 2D6 vs the
corresponding target numbers. Note we have rounded the results to two digits.
|
2D6
|
3+
(p=0.97)
|
4+
(p=0.92)
|
5+
(p=0.83)
|
6+
(p=0.72)
|
7+
(p=0.58)
|
8+
(p=0.42)
|
9+
(p=0.28)
|
10+
(p=0.17)
|
11+
(p=0.08)
|
|
0
|
0.03
|
0.08
|
0.17
|
0.28
|
0.42
|
0.58
|
0.72
|
0.83
|
0.92
|
|
1
|
0.03
|
0.08
|
0.14
|
0.20
|
0.24
|
0.24
|
0.20
|
0.14
|
0.08
|
|
2
|
0.03
|
0.07
|
0.12
|
0.14
|
0.14
|
0.10
|
0.06
|
0.02
|
0.01
|
|
3
|
0.03
|
0.06
|
0.10
|
0.10
|
0.08
|
0.04
|
0.02
|
0.00
|
0.00
|
|
4
|
0.02
|
0.06
|
0.08
|
0.08
|
0.05
|
0.02
|
0.00
|
0.00
|
0.00
|
|
5
|
0.02
|
0.05
|
0.07
|
0.05
|
0.03
|
0.01
|
0.00
|
0.00
|
0.00
|
|
6
|
0.02
|
0.05
|
0.06
|
0.04
|
0.02
|
0.00
|
0.00
|
0.00
|
0.00
|
|
7
|
0.02
|
0.05
|
0.05
|
0.03
|
0.01
|
0.00
|
0.00
|
0.00
|
0.00
|
|
8
|
0.02
|
0.04
|
0.04
|
0.02
|
0.01
|
0.00
|
0.00
|
0.00
|
0.00
|
|
9
|
0.02
|
0.04
|
0.03
|
0.01
|
0.00
|
0.00
|
0.00
|
0.00
|
0.00
|
|
10
|
0.02
|
0.03
|
0.03
|
0.01
|
0.00
|
0.00
|
0.00
|
0.00
|
0.00
|
But how many successes can we expect on average? For that,
we need to compute the expected value of the number of successes. We need to
multiply each possible outcome of the number of successes by its probability. Thus:
Expected successes = 0*prob_0_successes +
1*prob_1_success + 2*prob_2_successes + …
We could compute this by hand, but there is a closed
formula. Indeeed, when we use the above formula for scoring k successes, we get
the following sum:
Expected successes = sum(k*pk*(1-p)) for all
values of k = 0 … infinity
Lucklily, there are closed form solutions for this sum (I won’t
bother you with the details, but it results in:
Expected successes = p/(1-p)
This produces the following results for a single D6, and for
rolling 2D6:
|
D6
|
2+
(p = 0.93)
|
3+
(p = 0.67)
|
4+
(p = 0.5)
|
5+
(p = 0.33)
|
6+
(p = 0.17)
|
|
Expected
|
5
|
2
|
1
|
0.5
|
0.2
|
|
2D6
|
3+
(p=0.97)
|
4+
(p=0.92)
|
5+
(p=0.83)
|
6+
(p=0.72)
|
7+
(p=0.58)
|
8+
(p=0.42)
|
9+
(p=0.28)
|
10+
(p=0.17)
|
11+
(p=0.08)
|
|
Expected
|
35.00
|
11.00
|
5.00
|
2.60
|
1.40
|
0.71
|
0.38
|
0.20
|
0.09
|
Thus, when we roll 2D6 vs 7+, we can expect on average to
score 1.4 successes. If all units would activate on a 7+, we can expect to activate
1.4 units in a given turn.
Increasing the number of chains
The above analysis holds for a single chain – we break off
the die rolls when we roll our first failure. However, rules such as Black
Powder use commanders. When a commander has failed its order, the following
commander can still give orders. Since this is a completely new chain of die
rolls, we can simply add the expected values together to get the total number
of expected order in a given turn.
|
2D6
Commanders
|
3+
(p=0.97)
|
4+
(p=0.92)
|
5+
(p=0.83)
|
6+
(p=0.72)
|
7+
(p=0.58)
|
8+
(p=0.42)
|
9+
(p=0.28)
|
10+
(p=0.17)
|
11+
(p=0.08)
|
|
1
|
35.00
|
11.00
|
5.00
|
2.60
|
1.40
|
0.71
|
0.38
|
0.20
|
0.09
|
|
2
|
70.00
|
22.00
|
10.00
|
5.20
|
2.80
|
1.43
|
0.77
|
0.40
|
0.18
|
|
3
|
105.00
|
33.00
|
15.00
|
7.80
|
4.20
|
2.14
|
1.15
|
0.60
|
0.27
|
|
4
|
140.00
|
44.00
|
20.00
|
10.40
|
5.60
|
2.86
|
1.54
|
0.80
|
0.36
|
|
5
|
175.00
|
55.00
|
25.00
|
13.00
|
7.00
|
3.57
|
1.92
|
1.00
|
0.45
|
|
6
|
210.00
|
66.00
|
30.00
|
15.60
|
8.40
|
4.29
|
2.31
|
1.20
|
0.55
|
You can see that if we would have 3 commanders, each giving successful
orders on a 8+, we can expect a total 2.14 units to activate during the turn.
If we would have a 6+ commanders, a 7+ commander, and a 8+ commander, we can
expect 2.60 + 1.40 + 0.71 = 4.71 successful orders during the turn.
Game mechanics
It is important to consider these numbers when designing
scenarios. If both sides can activate the same numbers of units on average
during a turn, this makes it difficult for e.g. the more numerous side in an
attack-defence scenario. If the attacker and defender can both activate the
same number of units, the numerical superiority of the attacker can only
translate in successive attack waves, not in creating a “strong point” or “focal
point” where he can obtain a numerical advantage.
We would expect that a given force can activate a number of
units proportional to its total number of units. Giving a higher number of
commanders, or better commanders, or providing a bonus for unit activation, can
maintain this balance.
