Matt Price
United States San Francisco California
Member of the San Francisco Game Group since 2005
This is a customized Bane Tower from the game Man o' War

The capriciousness of killing the creatures in this game doesn’t jive well with my gaming group. We’d like to add a little more strategy to using creatures, and thought that assigning them hit points would help.
My question to you statisticians out there is: how many hit points should we give the creatures? And my gut instinct says to assign 8. But I’m not a statistician – any way to rationalize 8 (or any other number)?
Assuming you hit the creatures on 1in6 chance (ignore bonus strikes, etc., for the moment), your chance of killing them is 2.78% with each die you roll, right? (ie, the chance of rolling a 2 with two dice)
So, on average, how many dice (not including the second roll to attempt the critical hit) would you have to roll to get about a 50% chance of killing a creature? (This is where my knowledge of statistics fails me… How do you calculate this?)
Then, given that each roll has a 16.67% chance of hitting under noncreature circumstances (for simplicity’s sake, I’m still not considering bonus strikes, etc. here), on average how many hits would we expect to inflict on a creature rolling that many dice? I’m guesstimating this should be half of the creature’s total hit points.
Anyone else out there have any thoughts on this?


John Lopez
United States Tucson Arizona
Get off my lawn!
The explanation: Impossible Triangle + TW (my company initials) = my logo.

I have created a Google spreadsheet that gives the chance that a number of dice will both hit and critical a creature with a 1/6 chance being applied at each step. (This would represent a volley of arrows quite well, for example).
http://spreadsheets.google.com/pub?key=pDvuaRUEyeA7jMwo3TN70...
You can see that I have run this out to 9 dice: the cumulative odds of successfully killing with that many dice applied (say 4 archers and 1 archer that moved) is 22.3%. You will also notice that each additional die is giving (at this point) another 2.2% increase to the odds of success. This increase will continue to occur with each new die, but the increase does continue to diminish.
At 20 dice, the odds are 43%. At 25 dice the odds finally cross the finish line at 50.5%


John Lopez
United States Tucson Arizona
Get off my lawn!
The explanation: Impossible Triangle + TW (my company initials) = my logo.

For those who need to know the inner workings of such a spreadsheet.
The first column (A) is simply the number of dice we start with. The third column (C) is the number of "hits" we are looking for in this instance. The second column (B) computes a Binomial Distribution and returns the probability that exactly that number of hits occurs.
Column four (D) is the odds of a critical happening on the number of dice that hit. This is simply 1(5/6)^H where H is the number of hits that made it through.
Column five (E) is the odds of both events occurring in sequence. As these are independent die rolls, we can multiply the odds of the specific number of hits with the odds of those hits resulting in a critical. Finally we sum the resulting odds of the various hit possibilities to find the overall odds of success.
Column six (F) is the delta that the new die brought to the party.
There is a function that approximates such tedious calculations, but it has a high degree of inaccuracy for a small numbers of trials. Additionally, I think that the "odds of both events" column is interesting to those not familiar with the really lousy chances that independent, low probability trials cause.
As a further aside, these computations are off by a whopping 1% due to rounding errors by the time I reached the 25th iteration. I used Excel and pasted the numbers over as Google's spreadsheet is slow as molasses for this kind of thing.


John Lopez
United States Tucson Arizona
Get off my lawn!
The explanation: Impossible Triangle + TW (my company initials) = my logo.

Last post in this sequence. Your "expected hits" for a given number of dice and a 1:6 chance of hitting is simply D/6 where D is the number of dice rolled. For 25 dice, this is 4.16 hits. This would seem to indicate that 8 hit points might be accurate if your goal is to have twice as many hit points as a 50% chance of killing would emulate. This would require 50 dice to fully kill the creature.
Recomputing this spreadsheet with a 2:6 chance of hitting (typical melee) shows that at 9 dice we are already at a 40.2% chance of killing the monster. (If we simply naively doubled the odds of success, we would assume 44.8%, not all that far off). 12 dice is 49.6%, close enough for government work.


Steve Cates
United States Visalia California

Well if your using a red banner unit to attack it's going to be way easier than a green banner unit. So during gameplay one would make decisions based on how many dice they get to roll and whether they want to attempt a critical hit with that type of unit. With hit points those decisions may be different. Also, critial hits on retreats that are blocked is a very difficult thing to quantify. I'd say just leave it out of the calculation all together. Pursuit on Mounted units is another factor but I think that could be played as normal.
But that aside:
Green 1/6*1/6*2 dice * x = .50, x=9 ... 9x2 = 18hp Blue 1/6*2/6*3 * x = .50, x=3 ... 3x3 = 9hp Red 1/6*2/6*4 * x = .50, x=2.25 ... 4x2.25 = 9hp
1/6 for the banner color on the second roll, 2/6 for banner and sheild on first roll for short sword and long sword, as an assumption I made Blue and Red always short or long sword and Green always Common Bow, of course that is common but not always the case.
x is the amount of tries it would take with said number of dice to reach a 50% probablity.
Now the question is how often does one attack the creature with Green, Blue, or Red? I guess 9 would be the best to go with but it's all based on your assumption.


Blue Jackal
United States Nowhere Virginia

I think you should include Bonus Strikes, since most of the time you will be using units that have them.
I've only played a couple games with Creatures, but my current rule of thumb is: don't attack them, just battleback! Or at least if you're going to attack, I'd suggest doing it with a Red unit... (unless you're feebly trying with archers.)
I could totally see a hit point system, though, use a d10 or something.
Using "x2 Lore costs, Lore on a 46", at least it's more likely that you'll get Lore when fighting creatures. (A failed Critical Hit roll nets you a Lore token, which is "half a normal Lore", as well.)


Jérôme NIVART
France Reims Marne

Here is a post i've just created for killing creature in Dow's forum :
http://www.daysofwonder.com/index.php?t=msg&th=9337&rid=2555...
Enjoy !


John Lopez
United States Tucson Arizona
Get off my lawn!
The explanation: Impossible Triangle + TW (my company initials) = my logo.

ironcates wrote: 1/6*1/6*2 dice * x = .50, x=9 ... 9x2 = 18hp
Boy I wish it was that easy to calculate. However, you can't simply multiply across and solve for X.
1/6 * 1/6 = 1/36 (the odds of a single green die repeating the face and getting a critical). That is true. However, two dice are not 1/36 * 2 = 2/36. If that was true, we could say that 36 dice ensure that you will get a hit at 100% odds (36/36).
You can multiply across in the *negative* direction (not getting hits). But even then you must use the number of dice as an exponent, not a multiplier. In that case you take the 1/36 chance, flip it to the chance of failure (35/36) and then raise that to the power of the number of involved dice. For 36 dice, that would be (35/36)^36 = 0.363. This is the chance of failure, not the chance of success, so our chance of success is 63.7% with 36 dice. Not bad, but not 100% either.
There is *no* number of dice that ensure 100% odds of success. 100 dice does give a 94% chance of success.


Steve Cates
United States Visalia California

Thanks John, I was thinking this was the quick and dirty way to get a decent estimate. It seems right to me that 3 attacks with red banner units would take down the beast half the time. I got 2.25 and that seemed right, but let try it differently.
So correct me if I'm wrong for red banner units (4 dice):
[([1(1/6)*(2/6)])^4]^(x) = 0.5 {failure probability} ... x = 3.03 tries ... HP = 4 dice * 3 tries = 12.1 hp
blue banner x = 4.04 tries ... 12.1 hp
green banner, no bonus strike (2/6) becomes (1/6) x = 12.3 tries ... 24.6 hp
You could modify failure probability with 1(0.5) for success probability as a trick I believe but .5 is the same for success or failure.


John Lopez
United States Tucson Arizona
Get off my lawn!
The explanation: Impossible Triangle + TW (my company initials) = my logo.

ironcates wrote: Thanks John, I was thinking this was the quick and dirty way to get a decent estimate. It seems right to me that 3 attacks with red banner units would take down the beast half the time. I got 2.25 and that seemed right, but let try it differently.
So correct me if I'm wrong for red banner units (4 dice):
[([1(1/6)*(2/6)])^4]^(x) = 0.5 {failure probability} ... x = 3.03 tries ... HP = 4 dice * 3 tries = 12.1 hp
blue banner x = 4.04 tries ... 12.1 hp
green banner, no bonus strike (2/6) becomes (1/6) x = 12.3 tries ... 24.6 hp
You could modify failure probability with 1(0.5) for success probability as a trick I believe but .5 is the same for success or failure.
Yes, this seems accurate. I punched
find_root(((1(1/6)*(2/6))^4)^x  1/2, x, 1, 10);
into Maxima and got 3.03, as you found. However, with a red unit normally dealing 1/3 of the dice they roll as hits, these three attempts would cause an "expected" 12 * (1/3) = 4 hits against a normal unit (not mounted).
Blue units will use the same number of dice over time (and the same "expected" hits), but will take an extra turn to do so.
Green banners do worse, taking the 12.3 tries you point out, or 24 dice at 1/6 expected damage. This again is 4 "expected" points of damage.
Since all of these methods generate 4 "expected" points of damage to equal a 50/50 chance of killing a giant, I would probably put a creatures hit points at 8, making them effectively two units. So if Matt Price is still with us, his "gut" instinct was pretty darn good.
Of course, none of this takes into account that the creature will be dealing out damage (and special effects). It would seem that greens might as well pack it in unless you can a darken the skies or a lore card that seriously boosts damage output. On the other hand, the creatures don't put out *that* much damage...


Steve Cates
United States Visalia California

Quote: Since all of these methods generate 4 "expected" points of damage to equal a 50/50 chance of killing a giant, I would probably put a creatures hit points at 8, making them effectively two units. So if Matt Price is still with us, his "gut" instinct was pretty darn good.
I think in my last attempt I made one error that the hp is the total dice rolled but instead it's the hits that would be gotten by that many dice against a normal unit. John picked up on that, but why make the creature equal to two units?
Conclusion, make it 4 hit points.


John Lopez
United States Tucson Arizona
Get off my lawn!
The explanation: Impossible Triangle + TW (my company initials) = my logo.

ironcates wrote: Quote: Since all of these methods generate 4 "expected" points of damage to equal a 50/50 chance of killing a giant, I would probably put a creatures hit points at 8, making them effectively two units. So if Matt Price is still with us, his "gut" instinct was pretty darn good. I think in my last attempt I made one error that the hp is the total dice rolled but instead it's the hits that would be gotten by that many dice against a normal unit. John picked up on that, but why make the creature equal to two units? Conclusion, make it 4 hit points.
At this point it is preference, but 4 hit points for a guaranteed kill means that the "average" lifespan of the creature is considerably shorter than "average" lifespan under the traditional rule.
Remember that it has taken the equivalent to 4 hits to reach the *50/50* point. However, if you reach that point in the traditional critical method, it is no more likely to fall over anytime soon than it was previously. The critical hit method gives creates a "long tail" expected lifespan. At the equivalent of 8 hit points the creature would be dead only 75% of the time using traditional rules. Taking this long tail into account, the "average" number of equivalent hit points absorbed by a creature is closer to 8 than 4 (I will have to compute it to be sure what the exact value is...)
http://timewarptechnologies.com/images/DeathCurve.png shows this long tail clearly. You can see where the 50% line drops of the number of attacks by a red unit, and you can see that even with 12 attacks the creature might still be around. (My anti leech code prevents a direct link: cut and past the URL).


Mark McEvoy
Canada Mountain Ontario

In a fantasy game with mythical creatures, I rather *like* the capriciousness of creature battle. I like the theory that a single strike or a single arrow might be all it takes to take down an enormous force. It may be an extremely *unlikely* event, but it is possible. I'm not sure I like an idea that a troop may go into a combat that it is 100% guaranteed impossible to "win". It's fantasy. Fantastic things happen.
Smaug wasn't felled by a thousand papercuts. He was felled by one especially accurate arrow.


John Lopez
United States Tucson Arizona
Get off my lawn!
The explanation: Impossible Triangle + TW (my company initials) = my logo.

Interesting. We reach the 50/50 point at 3 attacks by a red banner unit. Yes, taking the areas under the curve indicates that 4.05 attacks divides the curve equally. (1.74 units under each segment). My intuition failed me there, I thought it would be more than that.
Even more interesting is after whipping up a Monte Carlo simulation of a red banner attacking a creature, I get ~3.5 attacks as the "average" lifespan for the creature. Thinking about this, taking the areas under the curves wasn't such a great idea: it should have been at the integer grain.
So, even taking the "high" result of 4 attacks, that's only another 1.33 additional "expected" hits. So perhaps 5 hits, if you care to rationalize the creature to the same standard as the units in the game?
Quote: Smaug wasn't felled by a thousand papercuts. He was felled by one especially accurate arrow.
Yes, this was just me goofing off. I don't actually plan on varying from the printed rules on this issue. A red unit has a 1 in 5 chance of victory, but it also has a good chance of suffering several rounds of bone crushing. That variability isn't really a problem for me, I was just interested in a rough "equivalence" factor.
As an aside, during 2000 runs of the Monte Carlo sim, 101 of those runs had the creature lasting over 10 rounds against red banners (or 13 "expected" hits), and at the extreme end it lasted 18 rounds (or 24 "expected" hits) four times. So there is a chance that the creature will tank it out with the best and keep on going.


Steve Cates
United States Visalia California

That's cool. I wish I knew how to do those Monte Carlo sims, I'm an engineer and not much of a statistician (sp) I love putting togethter those excel spreadsheets necessary for my work and I even do some on the side for personal finances but I know there's a (whole) nother world out there in probability and stats.
I think 5 is justifiable but remember that a unit of 4 could tank it out as well ... it's of course possible that a unit of four could survive 24 rolls the same as the creature. I guess 5 would be fun just because no other unit has five (for now). (To think I was pushing 9 hp a while back).
I agree on the rules... I probably won't play hit points except maybe when I want to try something new.



