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Wij Amsterdam6shooters zijn een vriendelijke club van volwassen wargamers die twee keer per maand wargames (en soms boardgames) met miniaturen organiseren in Amsterdam-Oost. Met passie, met lol, met plezier en met stijl. En heel breed: Alexander de Grote, Napoleon, WOII, 40K, Age of Sigmar, X-Wing: het komt allemaal op tafel. Met ons meedoen kan door gratis te registreren en daarna via de agenda op de site een tafel te boeken. Een eigen leger hebben is niet eens nodig: we hebben alles. Dus wat let je? Angst voor miniatuurtjes?

1 hit for 10 damage, or 10 hits for 1 damage each?

Many wargames have a mechanic that requires a number of damage points to be inflicted for a target to be eliminated. Damage is usually caused by a number of shots, each shot having a certain probability of hitting the target, and each shot causing a number of damage points. Thus, we already have various variables with which we can shape our mechanic: the number of total damage points needed to kill the target; the probability of each shot causing a hit; and the number of damage points per successful hit. Given these variables, it is quite naturally to ask how long it will take before the target is eliminated. And is it better to have a mechanic in which we need 10 hits, causing 1 damage each, or needing one mega-hit causing 10 damage by iself?

2 cruisers closing in on a battleship …
Do we prefer many shots, many of them on target, but each doing a tiny amount of damage?
… or do we rather prefer many shots missing, but the one shot that hits, causing the ship to sink?

 Mathematics

To get a good insight in the mechanic, let’s take a look at the math. First, let us define some of the parameters:

  • p_success: the probability (a number between 0 and 1) with which a shot hits the target. Let’s assume p_success is the same for all shots.Translated into a die roll, a D6 that requires 5+ to hit has p_success = 0.33, a D10 that requires 4+ to hit sets p_success equal to 0.6, and so on.
  • k: the number of successful shots needed to kill the target. Again, to keep things simple, let’s assume that all hits cause the same amount of damage d. Thus, if the target has D damage points, k = D/d (rounded up). E.g., if a successful shot causes 3 damage, and we shoot at a 10-damage target, we need 4 successful shots (10/3, rounded up) to kill the target.
  • n: the total number of shots (trials), included misses. Not every shot will hit the target, as is obvious from p_success. n will always greater or equal than k. If we are lucky, we can kill the target with k number of shots, but the lower the value of p_success, the more shots we will need to reach k successful shots.

So, we can phrase the mathematical problem as follows: what’s the probability of k hits occurring in a sequence of n shots, with the last shot being successful (and thus killing the target)? Or to put it differently: what is the probability we will need n shots, given k successful shots with the n-th shot being the last successful k-th shot?

This problem might look familiar to another problem we have analyzed before, the Buckets of Dice procedure. When rolling  n dice, we are often interested in how many dice will score a success given a certain probability of success for each dice. The binomial distribution expresses the probability that we score k successes, given n dice. But now, we want something slightly different. Instead of fixing the number of trials n, and keeping the number of successes as a changing variable, we now want to fix the required number of successes k, and consider the total number of trials n as a variable. It is a minor change of viewpoint, but we can still use the same mathematical framework.

Before we can write down the equations, we need some additional insights:

  • The first insight we need is that it doesn’t matter which of our shots are successful, except for the last one. If we want to reach 4 successful hits in 10 trials, we know that the 10th shot needs to be a success (the 4th hit), but the previous 3 hits can happen anywhere in the previous sequence of 9 shots.
  • The second insight is that if 3 shots can happen anywhere in a sequence of 9 shots, the mathematics don’t care whether those shots are taken in sequence or all together, as long as all shots are independent from each other. Thus, we can use the binomial distribution as described by the Buckets of Dice method to describe the probability distribution of k-1 shots being succcessful, out of n-1 total number of shots. Using the same notation as in our Buckets of Dice blogpost, we can write this distribution as Bin(k-1, n-1, p_success).

Now we need to combine both observations. The first k-1 succesful hits can happen anywhere within the first n-1 shots. The n-th shot must be the k-th succesful hit, and this shot will be succesful with a probability equal to p_success. Thus, the complete probability distribution for k succesful hits, using n shots, with the last being a success, can be written as:

NBin(k, n , p_success) = Bin(k-1, n-1, p_success) * p_success 

Such a distribution is known in mathematics as a Negative Binomial Distribution, hence the notation Nbin (see also the appendix for some more information).

The relationship with the more well-known Binomial Distribution is clear, but as mentioned before, the viewpoint is slightly different. The Binomial Distribution is mostly concerned with the probability of scoring k successes out of n trials; the Negative Binomial distribition is interested in the probability of needing n trials to score k successes.

Analysis

So, what does this distribution look like? Many spreadsheet programs have the Negative Binomial Distribution built in, but you can also use the above formulation, expressing it as the product of the Binomial Distribution (which also often is a pre-defined function in many spreadsheet programs) multiplied by the probability p_success.

Let’s look at some simple cases first, just to get a feeling of how this function evolves with various parameter settings.

Let’s first set p_success = 2/6, which means rolling 5 or 6 on a D6 in order to score a hit. Let us further assume we want to score 2 succesful hits. The graph below shows the probability of number of total shots n needed.

As you can see, there is zero probability n equals 1 (obviously, since we need 2 successes), and n = 3 or n = 4 is the most likely outcome, each with roughly 15% chance of occuring. The probability we will need more than 4 trials decreases gradually.

Let us now assume we set p_success = 1/6, in other words, a successful hit will on average only happen once every 6 shots. The resulting distribution for the total number of shots n needed is shown below.

You can immediately see the most likely outcomes are n = 5 to 8, and again there is a decreasing probability we will need ever more shots, although the decrease is not as sudden as with p_success = 2/6.

Now let’s very the number of required succesful shots, while keeping p_success at 2/6. The graph below shows the probability distribution.

Since we need a higher number of successes, the graph shifts to right, in this case with a most likely outcome for n = 9 or 10, again with a gentle decrease for higher numbers of n.

I guess you are curious for other values of p_success and s as well, so here are the complete graphs. First, the probabilty disctributions for different values of p_success, fixing the number of successes on 2.

And here a variable number of successes, with fixing the probability for scoring a succesful hit:

Expected value for n

The above graphs might give you some insight in the mathematics, but how can we translate this into useful gaming mechanics? The first thing a game designer might be interested in, is the expected value of the total number of shots n (successes and misses), in function of p_success and s.

We will not go through the mathematical derivation, but the expected value for n (the expected value E(n) is the average value for n  if we would conduct or procedure an infinite number of times), equals s / p_success.

This is not so surprising, if we fill out some numbers:

  • If p_success = 1/6, and s = 1 (we need one successful shot, we a chance of 1/6 of scoring one), we can expect to need 6 shots.
  • If p_success = 3/6, and s = 2, then E(n) = 2*6/3 = 4, which means we can expect to need 4 shots in order to score to 2 successful hits with a success ratio of 50%.

So, what does this formula teach us?

  • doubling the value of s, while keeping p_success a constant, will double the value of E(n). Or more generally, increasing or decreasing the value of s by a certain ratio, will also multiply E(n) by that same ratio.
  • doubling the value of p_success, while keeping s a constant, will halve the value of E(n). Or more generally, increasing or decreasing p_success by a certain ratio, will inversely proportional change the value of E(n).

If we know put some more gaming terms into our equation, and if we want to keep the initial example of having to score a number of hits D on a target, with each hit inflicting d damage points, we can say that s = D/d. But since E(n) = s / p_success, we can now say that:

E(n) = D / (p_success * d)

Thus, we have the expected number of shots (which can translate in a number of turns in the game), the total amount of damage needed, the chance for scoring a hit, and the damage per hit, in one nice formula. You can play around with the values, keeping some fixed while changing others, and see what the outcome is.

Standard deviation
The standard deviation of a stochastic process is a measure for how far any given experiment can deviate from the expected value. After all, the expected value is only an average number, but any single experiment can produce a number lower or higher than the expected value.

The standard deviation for n, is given by the square root of (1 – p_success) * s / (p_success * p_success). This seems like a rather convoluted formula, and you might want to plug in some numbers to see how the standard deviation will change with various parameters, but you can see that the standard deviation:

  • … increases proportional to the square root of s  if p_success is kept constant. Thus, quadrupling s, will only double the standard deviation.
  • … decreases with higher values of p_success, due to p_success in the denominator. Roughly we can say that the standard deviation changes inversely proportional to p_success.

A full analysis would lead us to far, but again you can try out some numbers yourself. One last thing to note is that you can also compute the relative value of the standard deviation vs E(n). This turns out to sqrt ((1-p_success)/s). Thus, increasing the number or s, while keeping p_success a constant, will make the relative spread narrower around E(n).

Gaming mechanics

What does this all mean for gaming mechanics?

We know that E(n) = D / (p_success * d). Now suppose we want to find values for the different parameters, but keeping the expected value E(n), which can act as a proxy for the number of turns needed to sink a target, the same. Also suppose we keep D a constant (after all, D the total number of damage points, and is in some sense an arbitrary number). For 2 different gaming mechanics, each with different values for d and p_success, and forcing E(n) to remain a constant, we can then say:

p_success_1 * d_1 = p_success_2 * d_2

Thus, the damage points per successful shot should scale inversely proportional to the probability of a successful shot. If we set p_success_1 = 2/6, and d_1 = 6 points, then this is equivalent to setting p_success = 4/6, and d_2 = 3 points.

When we say both procedures are equivalent, they are both equivalent in E(n), the expected number of shots needed to sink the target. But there will be a difference in the standard deviation. However, to compute the standard deviation, we need to set a value for s. s is determined by D and d, since s = D/d. So, the standard deviation, equalling sqrt((1-p_success)*s/p_success*p_success) is now rewritten as sqrt((1-p_success)*D/p_success*p_success*d).

So, let’s plug in some numbers, and let’s set D at 12:

  • p_success = 1/6, d = 12 => s = 1, E(n) = 6, stdev = 5.47
  • p_success = 2/6, d = 6 => s = 2, E(n) = 6, stdev = 3.46
  • p_success = 4/6, d = 3 => s = 4, E(n) = 6, stdev = 1.73

The graph for these 3 settings is shown below.

Playing around with the numbers is of course fun, but there are others things to consider. E.g. setting d = D (i.e. a single shot kills the target), means avoiding bookkeeping (tracking the number of hits, damage points left, …). Setting d different from D implies keeping track of the amount of damage inflicted. Whether or not that’s a good thing, depends on other mechanisms in the rules.

Conclusion

Whether you want one big shoot that kills in an instant, or a sequence of low intensity shots that require many turns to kill, keep in mind that what really matters is the expected number of shots (misses and hits) needed, as well as the standard deviation on that number.

Appendix:

  1. The Negative Binomial Distribution is described in various ways in different textbooks. Often, it is defined as a fixed number of failures in a sequence of trials, thereby reversing the definition of success and failure as we have used in this blogpost. Sometimes, instead of the total number trials n, the number of failures and number of successes are used as parameters, wit n being the sum of the two. But it all ends up describing the same type of distribution. For more information, see https://en.wikipedia.org/wiki/Negative_binomial_distribution

Wargaming Mechanics Blog

Battle Report – Genestealer Cult vs Space Marines

Hello!

It’s another battle report against Stig’s Luna Templars, this time, as promised, a prequel to the arrival of Hive Fleet Goliath, as MarshaKurt Rhinehart leads his men against the Genestealer Cultists.

This photo (and several others) looted from Stig
Playing at 1750 points, Stig used the same list as before, again with Black Templar chapter tactics, a battalion and two vanguard detachments:
HQ: Gravis Captain (Warlord: Oathkeeper, The Crusader’s Helm), Chaplain, 2 Lieutenants
Troops: 2×5 Intercessors, 6 Intercessors with auto bolt rifles
Elites: 2x Redemptors (Heavy onslaught cannons), Company Champion, Emperor’s Champion, Ancient, 5 Reivers 
Fast Attack: 3 Inceptors (Assault bolters)
Heavy Support: 3 Suppressors, 5 Hellblasters, 3 Aggressors
(Apologies if I’ve got some units in the wrong slot)

Here’s my list. It’s a double battalion, both using the Cult of the Four Armed Emperor. This was the first game with the ‘new’ codex, so I didn’t get too funky with extra relics or the bonus warlord traits GSC can get. I went with a trait/relic combo for the Patriarch that would make him hard to kill (-1 to hit, +1 save vs ranged weapons, no overwatch) rather than anything too clever. As for psychic powers, I wasn’t sure which to take, so I semi-randomly dished out 5 of the 6 psychic power cards between my psykers.

HQ: Patriarch (Warlord: Shadow Stalker, Amulet of the Voidwyrm), 2x Magus, Primus
Troops: 3×10 Neophyte Hybrids (assorted weapons), 2×10 Acolyte Hybrids (one squad with 4 melee weapons, one with 4 demolition charges), 10 Brood Brothers
Elites: Kelermorph, 15 Purestrain Genestealers
Heavy Support: 3x Cult Leman Russ (1 Battle cannon, 1 Vanquisher cannon, 1 Exterminator autocannon)
Transports: Cult Chimera

Since I have the digital codex, it occurred to me to whip up some quick blip tokens on spare 32mm bases:

They seem to have been well received when I posted them on twitter, but ironically they didn’t get used since I was the attacker in the mission and as such got first turn automatically. We played the Hold Your Gains mission from Vigilus Defiant, which mean 1/3 of the Space Marine force started in the centre, getting assailed on either side by the Genestealer Cult, until their reinforcements arrive. The winner would be the force with the highest total power level within 12 inches of the centre point at the end of the game.

The defenders set up in a fortified position, while the Cult infantry and tanks rolled in from either flank. Several of the Genestealer units were not immediately present, instead laying somewhere in ambush.
The first turn was relatively quiet. Some shots from the Neophytes plinked off wounds from Intercessors here and there, while the battle cannon got a good round of shooting at the Hellblasters, though their barricade saved some, leaving 3 left. The Exterminator also did some damage on the other flank to more Intercessors. In return a few of the cultists were gunned down, and the Hellblasters returned fire, stripping 6 wounds off the Leman Russ, however it’s Devoted Crew (stratagem) fought on regardless of the damage.

Turn 2 for the Cult, I moved up and brought on the Hybrids with melee weapons in a Perfect Ambush, and the Primus. The rest of the ambushers waited for their moment to counter attack. Shooting from the tanks finished off the Hellblasters and the last of the Intercessors on the right flank. The Acolytes made a big charge in to the remaining Intercessors on the left, easily cutting them down. At this stage the Luna Crusader’s presence was down to three characters, the Chaplain, Company Champion and Lieutenant, but the reinforcements were about to arrive!
Stig rolled for his reserves to arrive, and got everything apart from the Suppressors. A big block of black and yellow marines arrived, flanked by the Redemptors, while the Aggressors popped in a corner to line up a charge on the damaged Leman Russ, the Reivers dropped in behind the Neophytes in cover. Inceptors dropped in to face off against the Acolyte Hybrids.

A brutal shooting phase ensued, with the Acolytes reduced to a single model, while the two Neophyte units in the open were removed. The charge phase was a disappointment however, even with rerolls nothing made it in. To add insult to injury, the last Acolyte passed his morale on a 1, allowing me to play the Cult Reinforcements stratagem to get 4 of them back (naturally the mining weapons).
The Luna Crusaders had punched back, but now was time for the trap to be sprung!
The Patriarch emerged from a hatch to the sewers, looking down and the Inceptors from the shadows. Elsewhere the Purestrains turned up on the right flank in another Perfect Ambush, lining up the Intercessors and a Redemptor, while the demolition charge Acolytes were Laying in Wait for the other Redemptor, supported by the Kelermorph.
Psychic powers went off all over, using both familiars to attempt 5 this turn. Might from Beyond buffed the Purestrains, while some mortal wounds went on one of the Lieutenants, softening him up for the Kelermorph. I attempted to Mind Control the Redemptor, but the Templars said no and denied it with a stratagem. 


The Shooting phase started well with the Kelermorph killing his target, granting a reroll 1 buff for the Acolytes throwing all their demo charges and destroying the Redemptor. Several other units took casualties, but the rest of the damage was done in the fight phase.

The Patriarch made short work of the Inceptors with a big charge using my last command point to reroll a one. The Purestrains survived the overwatch to reach the Intercessors and the remaining dread, finishing off the Primaris marines but leaving the Redemptor largely intact. The Primus and Magus on the right made it into the Champion and Chaplain, but failing to kill them (and the Primus got battered in response.) Marshall Rhinehart wasn’t waiting around, using his warlord trait to heroically intervene 6 inches into the demolitions acolytes, cutting several of them down.

Turn 3 for the Marines, the Suppressors arrived, dropping into a commanding position (and proceeded to have a limited effect on the tough Cult armour.)

The Aggressors moved in, their flamers damaging the Leman Russ further before the charge. The last Reiver tossed a stun grenade at the Neophytes to avoid their overwatch and charge in.

In the shooting phase the two characters on the right flank killed off the Magus with their bolt pistols, allowing them to charge out and tie up the Exterminator and help kill off the Genestealers.

More characters doing work for the Crusaders, the Emperor’s Champion cutting down the other Magus, while the Marshal and the Ancient finished off the Acolytes. Rather embarrassingly, having reduced it to 3 wounds, the Aggressors were unable to finish off the Leman Russ in combat, losing one to overwatch to boot. 
My turn 4, running out of assets, but still with a few tricks up the sleeves, the Cult tried to hold the middle ground. The Russes broke out of combat, while the Vanquisher lined up and blew up the last Redemptor (causing wounds on the nearby characters). The Kelermorph tried to assassinate Marshal Rhinehart, but his armour just about protected him, leaving him on two wounds. The last Acolytes charge the Emperor’s champion to avenge their fallen Magus, but the rock saws were not quite able to cut through, leaving him alive on a single wound, and able to wipe the unit.
The Luna Crusaders were now down to mostly just a few injured characters, but they would not withdraw from this fight!

With little left on the field, the Marshall and the Emperor’s Champion joined forces to take on the deadliest threat – the Patriarch. Wreathed in shadow, the xenos proved a tough challenge…

But the Crusaders proved their worth, taking the beast’s head! Other action: The Ancient punched out the Kelermorph, the Company Champion charged the Exterminator again, and the Aggressors incinerated the last of the hybrids, before charging and once again failing to kill the Leman Russ.

Turn 5, time to get bodies in the middle. The Brood Brothers finally disembarked from the Chimera, and both units opened fire on the remaining characters. 

The Chimera gunned down the Emperor’s Champion, while he Marshall was peppered with lasfire and finally downed with a frag grenade.

One tank down, the Aggressors finally got their target. In the centre many of the Brood Brothers were cut down, but they held. The Ancient and wounded Chaplain holding the objective against them. The game did not end. On to turn 6.

Shooting from the remaining Russes (one of which had amusingly run over the Company Champion) and desperate charges from the Brood Brothers and the Chimera failed to remove the Ancient and Chaplain, who charged in himself to try and cut down the last of the cult infantry.

At the end of turn 6, Stig rolled the dice and got a 2. This was the way the world ended. The Brood Brothers had held on, and the Chimera had snuck into the 12 inch radius, but the closest Leman Russ wasn’t wholly inside, so the combined power level of the Ancient and the Chaplain trumped them, giving the Luna Crusaders a very narrow win!
What a game! A massacre on both sides, and it came down to some heroics from the Crusader’s characters to win the game.In hindsight I shouldn’t have gone for the kill on them with the Russes, and instead should have parked them on the narrow edge of the barricades to get one or both within 12 of the centre, but I really wanted to kill them! In the end though I’m pleased with the narrative of the Ancient standing on the objective, holding the banner high, while more Imperial reinforcements pour into the city to destroy the last of the cult. Just in time for the Tyranids to arrive, of course.

Redtoof’s 40K

Tale of Tuesdays: SpaceShrimp Month 1 – Champions of Death

Introducing the first of my BloodBowl team for Tale of Tuesdays! Champions of Death – Spaceshrimp flavour.

So despite being a really easy going and playful person in general, I actually take this hobby quite "seriously". What I mean by that is, I am not a fan of out of place conversions or paint schemes. If painting your Primaris like Buzz Lightyear is your thing, then power to you! Just totally not my approach at all. However, I think Bloodbowl lends it's self really well to having a bit of fun with the team names and themes and being a little bit silly.
More after the jumpTale of Painters

WE ROCK. WE ROLL SIXES. WE WARGAME. WE NEVER LOSE.

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