Wargaming - Home
Thoughts on wargame design
Sci-fi wargaming ramble
Crossfire WW2 wargaming
The Tank Duel
Here are presented various aspects of Lloyd's work on Crossfire. Click and be comparatively happy.Watch my Crossfire videos on YouTube Advice on play Basing and modelling tips Crossfire at 1:1 scale Description and review of the rules Die roll probability calculations "Hit the Dirt" (scenario book) description and errata Organisations - Orders of Battle Reinforcement rules Scouting rules Suggestions for alternative/extra rules World Crossfire Day 2009 Scenarios
This page may be of use to players of games other than Crossfire. To those who wish to know more about probability, but who know little about the mechanisms used in Crossfire, click HERE for an explanation of the terms used on this page.
I wanted to study the effects of different modifications I might make to the published Crossfire rules, on the way the game plays. For example, I wanted a way to make hitting certain sorts of target harder than it is as the rules are published, but not too hard, or it wouldn't have my desired effect.
Below is a chart showing the percentage chances of causing a Pin or worse, a Suppress or worse, and a mid-way figure half way between these two. For example, 4d6 has an 80.25% chance listed in the "Pin" column. This means that 80.25% of the time, when rolling 4d6, a player will get a Pin, or a Suppress or a Kill result. There is a 40.74% chance listed under the "Suppress" column. This is the chance of a Suppress or a Kill. Note that if you add these two figures together, you get well over 100%. If you want to know what the chance of a Pin and a Pin only is, then you must consult the worked example.
Explanation of terms
- Three six sided dice rolled together.
- Three six sided dice rolled together with a "penalty die", making four six-sided dice in total. The highest or joint-highest result is then ignored. This mechanism does not appear in the published Crossfire rules.
- Roll 2d6 as normal, counting fives and sixes as hits, then roll a third die, but on this third die only a 6 scores a hit. A good way to do this is to roll one die of a different colour, or an average die, which has one 5 on it, but no 6.
- Roll five six-sided dice, and ignore the two highest or joint-highest results.
- Roll three six-sided dice and count rolls of 6 as hits. When firing with a machine gun at troops in a bunker in Crossfire such a roll might be made.
- Make three rolls, each of 3d6, such as when three rifle stands of the same platoon fire at enemy in the open, as a "firegroup".
This table does not take into account the chance of a Kill result (three hits or more). Therefore, the bigger die rolls, like 6d6 do not appear to as great as they really are. I wanted to compare like with like, and there are many situations in Crossfire when 2d6 is rolled, such as rifles firing at targets in cover, and these never have a chance to score a kill.
Above, we can see the effects of modifying 3d6 (the dice used for rifles firing at targets in the open) in five ways: -1 die (rolling 2d6 instead, such as at targets in cover); a penalty die (3d6+1P); -1 pip per die (3d6(6)); knocking off half a die (2.5d6); two penalty dice (3d6+2P). These can also be combined, such as knocking off a die and -1 pip per die (sixes needed), such as when firing at targets in a bunker.
Notice that the effects of these modifications are not the same on the chances of a pin as they are on the chances of a suppress. For example, the chance of a pin or worse with 3d6+1P is lower than the chance with 3d6(6), but the chance of a suppression or worse with 3d6+1P is higher than with 3d6(6). Because of these quirks, I decided that a mid-way figure between the Pin-or-worse figure and the Suppress-or-worse figure, would be a fair measure of the overall effectiveness of the various die rolls, in the context of a Crossfire game, since Pins and Suppressions are the commonest results of firing.
One question I wanted to sort out, was the effect of my rules modifications on the fire-power of typical firegroups. The Crossfire rules state that a typical firegroup is made up of three rifle stands, which each use 3d6. In my version of the game, instead of three stands, each representing a section or squad, making a platoon, I have one stand representing a light-machine gun team, plus one stand representing a rifle team, making up a section. Machine guns roll 4d6. I wanted to make sure that the firepower of 3d6 three times was about the same as of 4d6 plus 3d6. To my relief, they come out as identical (60.5%).
Now to see this information in a more visual form:
This shows the final (mid-way between Pin and Suppress) scores for 3d6, and the various ways of manipulating it. The colours correspond to the colours on the table above. The top row shows a surprisingly even distribution. Note that the difference between 3d6 and 2d6 is twice the difference between 2d6 and 3d6(6). The effect of a penalty die (3d6+1P) is about the same on 3d6 as -1 pip per die (3d6(6)). So, if you want some kind of cover to be better than a wooden building or some bushes (-1 die) but not as good as a bunker (-1 die and -1 pip per die), then either a penalty die or -1 pip per die is the way to do it, at least for 3d6 it is.
This shows all the final (mid) scores for all the dice calculated. You'd have to plot Pin and Suppress chances separately to see all the quirks. The Red numbers are the normal dice, which get closer together as they get higher. Between these are the half-dice. The effects of a penalty die are very similar as the effects of -1 pip per die for 2d6 and 3d6, but -1 pip is a harsher penalty than a penalty die for 4d6 and 5d6. An important difference between 3d6+2P and 2d6+1P, though they look similar on this chart, is that it is possible to get a kill result (three hits) with 3d6+2P, whereas it is impossible with 2d6+1P.
You may be wondering how the above calculations were done. The formulae may surprise you with their complexity. Click HERE to see the method.
CROSSFIRE: Probabilities: explaining Crossfire dice nomenclature
In Crossfire, shooting is resolved by rolling different numbers of six-sided dice, and counting the number that come up five or six. Each die, therefore, has a one-in-three chance of scoring a "hit". If no die scores a hit, then the enemy is unaffected by the shooting. One hit is a "Pin" which stops the enemy advancing, two hits is a "Suppress" which makes the enemy dive for cover, and three hits or more is a "Kill".
Different numbers of dice are rolled depending on what weapon is being used, and what cover the target has. For example a machine gun rolls 4d6, and rifles roll 3d6. Each rolls one die less (3d6 and 2d6 respectively) when the target is in cover. Some types of cover, such as bunkers and pillboxes, do not just take off a die from the number rolled to hit, but also require sixes for hits, not fives and sixes. This is called "minus one pip per die".
Troops can form "firegroups" which all shoot at once at the same target. Three rifle stands might all roll 3d6 against single target in the open. If the first rifle stand rolls a Pin (one hit), as does the second, and the third, then this is neither a Kill nor a Suppress, because the effects are not cumulative.
I think that's all you need to know.
CROSSFIRE: Probabilities: worked example
I will show you a worked example of a probability calculation, which will give the answer to a question which isn't answered on the page which led to this one: "What is the chance of getting a kill result (three hits) on 5d6?" which to non-Crossfire players is, "What is the chance of rolling a five or a six three times or more, when rolling 5d6?"
The magic formula used is this:
In this formula, n is the number of opportunities to score a hit (number of dice rolled), x is the number of hits/successful rolls we are asking about, p is the probability of a success on a single trial (1/3 in the case of a 5 or a 6 on 1d6), and f(x) is the probability of getting x successes in n trials. The curious expression n! means "n factorial". A factorial is the result of multiplying a number by the number below it, and the result by the number below that, and keeping doing this until one gets to multiplying by one. For example, 5!=5x4x3x2x1=120. Some useful factorials are: 0!=1, 1!=1, 2!=2, 3!=6, 4!=24, 5!=120, 6!=720, and 7!=5040.
So, in plain English, the formula is: Take the number of dice being rolled, find the factorial of it, and then divide the result by the result of n minus x factorial, times x factorial. Multiply the result of the first part of the equation by the second part, which is the chance of one success on one die, raised to the power of x. Multiply what you get as a result for multiplying part one by part two, by the result of p subtracted from one, raised to the power of n minus x.
Confused? Let's try plugging in some numbers:
So, for the question we asked above, p=1/3 (chance of a success on one die), n=5 (number of dice), and x=3 (number of successes needed for a kill).
We get: 5!/3!(5-3)!, times 1/3 to the power of three (1/3 cubed), times (1-1/3) to the power of (5-3).
This simplifies to: 120/6(2), times 1/27, times 2/3 squared.
Which simplifies to 10 times 1/27 times 4/9.
Which gives us the answer 40/243.
But we're not there yet...
You see, 40/243 is the chance that three dice will turn up as a five or a six, when five dice are rolled. But if four dice or all five dice were to come up 5s and 6s, then these too would be "kill" results. The question was "What is the chances of rolling a five or a six three times or more, when rolling 5d6?" To find this, we have to add to 40/243 the chances of 4 dice rolling hits, and of five dice rolling hits. Therefore we need to use the formula again for the values of x=4 and x=5.
The probability of getting three or more successes is prob(x>=3)=f(3)+f(4)+f(5).
If we use the magic formula at the top for x=4 (the chance that four dice will score hits) we get 120/4!(1!) times 1/3 to the power of 4, times 2/3 to the power of 1. This simplifies to 5 times 1/81 times 2/3 = 10/243.
Now for x=5: 120/120(0!) times 1/3 to the power of 5, times 2/3 to the power of 0. To work this one out, you have to know that 0!=1, and any number to the power of 0 is 1. This gives us 1 times 1/243, times 2/3 = 2/729.
So f(3)+f(4)+f(5)= 40/243 + 10/243 + 2/729 = 152/729.
If you want to express this as a percentage, then divide 100 by 729 and multiply by 152 = 20.85048%.
You might want to know the answer to "What is the chance of rolling at least one success with five dice?" In Crossfire terms, this would be the chance of a Pin result or better on 5d6. You could use the above method, and calculate the values for f(1), f(2), f(3), f(4), and f(5), and then add them all together. However, we know that all the possibilities added together must add up to 100%. Another way of putting this is: f(0)+f(1)+f(2)+f(3)+f(4)+f(5)=1. "f(0)" is the chance that after rolling five dice, not a single one has rolled a success. To find out the chance of a Pin or better, therefore, we can use the formula 1-f(0).
f(0)=5!/0!(5-0)! times 1/3 to the power of 0, times 2/3 to the power of 5-0. I'll tell you now that when x=0, the first part of the equation always comes out as 1, and the second part of the equation also comes out as 1. So we have 1 times 1 times 2/3 to the power of 5. This gives us 32/243.
32/243 now has to be subtracted from 1, giving us the answer: 211/243, or 86.831268%. You really are very likely to score a Pin result with 5d6.
Now you can work out all manner of your own probability calculations. To work out the chances of rolling sixes, rather than fives and sixes, just make p=1/6 instead of 1/3. To calculate 3d6+1P, use n=4 and then add one to the minimum value of x. Doubtless, though, you will soon want to know how to work out the answer to questions where the value of p is not constant. For example, "What is the chance of a fire group's causing a suppression, when the fire group is a machine gun rolling 4d6 and a rifle stand rolling 3d6?" To tackle this question, you must first calculate the chances of 3d6 and 4d6 causing a suppression or worse. I'll tell you that the answers are 7/27 and 33/81. These, then, are your two values of p. How to combine them? You have to work out all the combinations: one could suppress, the other could suppress, neither could suppress, both could suppress. In this case, there are four combinations. The chance that each combination occurs is one value of p multiplied by the other.
So the chance that only the MG suppresses is 33/81 (chance of two or more success in 4 trials) times 20/27 (chance that 3d6 does NOT score a suppress or worse) = 660/2187.
The chance that only the rifles suppress is 7/27 times 48/81 = 336/2187.
The chance that both suppress is 33/81 times 7/27 = 231/2187.
The chance that neither suppresses is 48/81 times 20/7 (we get these figures by subtracting the chance of a success from 1) = 960/2187.
To check that we haven't made a mistake, we can add all the numbers together: 660/2187 + 336/2187 + 231/2187 + 960/2187 = 2187/2187 = 1 (huzza!)
We are not interested, though, in the chance that neither causes a suppression, so we add up only the first three combinations: 66/2187 + 336/2187 + 231/2187 = 1227/2187 or as a percentage 56.104206%.
Now don't say that I never tell you anything.
I am enormously indebted to Andy Gordon, a professional statistician, for his providing me with the magic formulae, and instructions on how to use them.
So, do you reckon that you understand probability now? If you do, try the Monty Hall Dilemma in the Games section.