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) = p^{k}*(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+ |
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+ |
3+ |
4+ |
5+ |
6+ |
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+ |
4+ |
5+ |
6+ |
7+ |
8+ |
9+ |
10+ |
11+ |
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*p^{k}*(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+ |
3+ |
4+ |
5+ |
6+ |
Expected |
5 |
2 |
1 |
0.5 |
0.2 |
2D6 |
3+ |
4+ |
5+ |
6+ |
7+ |
8+ |
9+ |
10+ |
11+ |
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 |
3+ |
4+ |
5+ |
6+ |
7+ |
8+ |
9+ |
10+ |
11+ |
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.