My Wikipedia Problem

Some months back I started getting a lot of traffic from the Wikipedia page on Lanchester's Laws. It seems that someone had noticed my efforts on the subject and linked to me as a reference. It wasn't much really, just a reference in support of a single sentence - Here it is, and [5] links to my first post on Lanchester's Laws:

In modern warfare, to take into account that to some extent both linear and the square apply often an exponent of 1.5 is used.^[4][5][6]

Citation: Lanchester's laws. (2011, December 12). In Wikipedia, The Free Encyclopedia. Retrieved 03:02, January 30, 2012, fromhttp://en.wikipedia.org/w/index.php?title=Lanchester%27s_laws&oldid=465382667

I've put a lot of study into Lanchester's Laws, so I was happy that someone thought I was worth a reference, but I have a problem with that sentence. It's wrong.

To be fair, perhaps I ought to say it is incomplete statement on a complex topic, and the complete explanation would be much, much longer. It is true that an exponent between 1 and 2 is often used to approximate situations where both Linear and Square laws are in effect, but this exponent represents (in a very abstract sort of way) that a portion of each force are subject to the Linear law (exponent of 1), while the remainder is subject to the Square law (exponent of 2). There is no mathematical rule that makes any other values for the exponent correct, it just sort of works to describe how battles actually play out in an average sort of way. The article makes no previous mention of exponents at all, so it's hard to see how anyone could come away with a proper understanding of the statement. To my mind that makes it wrong.

Fixing it though, is another matter. I've thought about fixing it myself, and even contacted a Wiki editor about it, but I haven't had time or energy to take on the task. I still hope to get back to writing regularly again, but I have a stack of other topics to address, and I am not sure I really want to spend my time fixing someone else's problem. 

Tangent: If you are new to the subject of Lanchester's Laws, the Linear law (exponent 1) describes combat attrition in a one-on-one combat setting, such as might occur with archaic weapons or between aircraft in air-to-air dogfights. The Square law (exponent 2) applies when multiple combatants can attack the same target, and vice-versa, such as a naval gunnery battle. Those are ideas though, and in practice there is almost always some complex mixture of these situation.

The Grinder

Assorted bits of internet, selected from a stratified sample, sorted in ascending sequence, collated, assembled in logical order, then just sort of thrown in a pot and given a good hard shake. Completely by accident, I seem to have a fine selection of game design articles for this edition. How did that happen?

--- The Grinder - August 2011 ---

PAXsims - The latest Game Design blog I am following. Some good stuff here I need to check out.








IO9: Robot Art Wallpapers

A Fafnir crashes the party

Wargamer's Notebook: The Moment  -- "I wish this was something I experienced more often. The point during a wargame after several plays or turns when things magically click. The moment when the scaffolding of the the rules and latticework of the bits fall away and the narrative zooms into the foreground and you - as the player - are completely absorbed by the game. Events occur that could not have been imagined, but that are totally plausible. Victory hangs in the balance. And you are not moving counters or playing cards, but making choices that give you hope of victory."

Space Disaster!

original

Genius Sir!

original

More irreverent humor to be found at Perry Bible Fellowship.

Desert Scibes brings us a bit bit of gaming history at Supergalactic Dreadnaught:

The Making of Starfleet Wars

NOTE: Scott R. Spicer (credited as "S.R. Spicer, Lt., TFSF"), along with his father, Ron Spicer (R.E. Spicer, Lt. Commander, TFSF), developed the Starfleet Warsminiatures line for Superior Models. As you probably know if you've read this blog before, the SfW universe contains five starfaring factions, each with a distinctive design style. Curious about the origin of this game and its minis, I emailed Scott Spicer about his work on these spaceship models. He was gracious enough to reply to my inquiry, and his response follows in its entirety ...

Make your own custom dice (with a bit of work).

From SMBC: (If only it were this easy!) 

Gamasutra: Game Design Essentials: 20 Real-World Games

Chess, Go, and Life. Which one is the Best. Game. Ever. A nice bit of gamey mathiness from Patterns in the Void.

Finally, this is just cool.

Taking a RISK - the distribution of armies lost

I came across the following question at boardgames.stackexchange.com:

How can I estimate my chances to win a Risk battle?

Elliot Avedon Virtual Museum of Games, Courtesy Canadian Museum of Civilization.

Now there is already plenty of material on the web about probability in the popular boardgame RISK*, but maybe I can add just a little bit more.

Most people already know that, given the choice, the Attacker should always 3 dice and the Defender should always roll 2, since this always gives the best results (a Nash equilibrium). Since it's always the Attackers choice to roll an attack or not, the relevant question seem to be "How many armies [X] will the Attacker lose if they make [N] attack rolls?"

Elliot Avedon Virtual Museum of Games, Courtesy Canadian Museum of Civilization.

Using some probabilities from a RISK FAQ for the probabilities of losses, I calculated the average attacker losses (about 0.921 per roll) and standard deviation (~0.81). It's not possible to lose 0.9 armies in RISK! as the attacker losses vary between 0, 1, and 2 per roll. However, as the results of many attack rolls are added up, the losses will begin to resemble the normal distribution. We can plug the average and standard deviation into a normal approximation formula, and get back a probability for losses in a fairly simple calculation. For a given number of attack rolls "A" and number of attacking armies lost "K",

Z = (K - 0.921*A) / (0.811*sqrt(K))

where Z is a standard normal variable (mean 0, standard deviation 1), and the probability of K-or-fewer losses in A attack rolls can be evaluated with the standard normal probability CDF function, otherwise known as the NORMSDIST(Z) function in Excel. That's pretty much it, except maybe for some graphs to show off the results.

Cumulative probability of K attacking armies lost in 25 attack rolls

The blue line shows the cumulative probability of K losses in A attack rolls (vertical red line). The yellow triangles show an approximate 50% confidence interval, meaning that your actual losses should be within this range 50% of the time. The red diamonds show a 90% interval for the same. These intervals are actually slightly wider than the stated 50%/90%; because I rounded outwards to the nearest whole number of armies, and there is no other good way to do it.

Cumulative probability of K attacking armies lost in 35 attack rolls

Recall this is an approximation, and it depends on there being lots of independent random events (dice rolls!) for the approximation to work well. It should start to work very well somewhere between 20-30 attack rolls, and depending on how fussy you are may give usefully accurate results for as few as 10-15 rolls. For smaller battles, or deciding whether or not you should attack just one more time, you might consider more accurate methods (look here, for starters).

Cumulative probability of K attacking armies lost in 70 attack rolls

You can think of this in terms of Defender losses too. Two armies are lost with every attack (between the attacker and defender), so if there are A attack rolls and K attacking armies are lost, that means the defender will be losing 2*A - K armies. For instance, in 25 attack rolls a total of 50 armies are lost; the probability of the attacker losing 20 armies the same as the probability of the defender losing 30.

Cumulative probability of K attacking armies lost in 50 attack rolls

Nostalgic Tangent: Calculating the probabilities of losses for the attacker and defender was one of my first mathematical efforts to figure out a game. I didn't know how to do the calculation, but I wrote a program on my Apple II+ to roll lots of electronic dice for me, and calculated the probabilities that way. Later I learned that this technique is called Monte Carlo simulation, and statisticians do this regularly to examine the properties of new statistical methods.

*** There WILL BE a link to the spreadsheet for these calculations, but it's getting late, so I'll have to add that in tomorrow. ***


Having written this, I realized that I haven't quite answered the question. I've given the probability for a given number of attacks/losses, but the question is "How many armies will it cost me to win?"

There is another approximation to answer that, but it's much less well known. I guess I'll have to write a part 2. Stay tuned!

---------------------------------------------------------------------------------

* RISK is a registered trademark of HASBRO, Inc., of course.

Footnote: The RISK! game images used in this post are from Elliott Avedon's Virtual Game Museum, and are used with permission of the Canadian Museum of Civilization. The Virtual Game Museum has much more interesting game related information, and I may be posting about it again.

The Grinder - July 4th Edition

A collection of red-glaring rockets and bombs bursting in air, without the rockets and bombs.

The Grinder for 7/4/2011:

 Paint-It-Pink : Ashley has some Battletech math going on --> How I Learned to Stop Worrying and Love the Medium Laser. A good discussion of how to evaluate the relative strength of weapons in Battletech, or any game.

"Can I haz tactix?" 

Found on Operation Odyssey Dawn: If the animation doesn't work, go see it here.

Linkback! --> 程阳：Probability versus Odds


MathOverflow: Which popular games are the most mathematical?

Proof: The 120 cell is a 4 dimensional figure that can be considered the 4 dimensional analog of the dodecahedron. It has 720 five sided faces, 1200 edges, and 600 vertices. This animation shows 3 dimensional cross sections of the 120 cell in a way that is similar to taking 2 dimensional cross sections of a 3 dimensional figure. Translation --> Very Cool animation!

The Number Warior: Q*Bert Teaches the Binomial Theorem (an award winner too). Sort of a long (2-part) video.

Proof-of-False: Do games offer a solution for US Tax Reform?

From doctormatt: A Collection of Dice Problems with solutions and useful appendices

The Math Less Traveled: The Wonderful World of Mike Reilly.

Mike Reilly is a Toy/Puzzle designer and screenwriter, see what he has done at Reilly4Puzzles.

Reinwood's CBT Workbench gives us an AAR for Fourth Succession War: Skondia The Final Battle Kublacon. AND it's got no math in it. Honest!

A small update to my Graph Paper Race post (added link to a relevant article). This continues to be one of my more popular posts.


I didn't plan this, but somehow this has ended up being the most math-heavy edition of The Grinder to date. Oh well, it's all sausage now.

Dice Distributions Revisited

Recent thoughts about calculating the distribution of the maximum sum of several dice (ex: roll 3, sum the highest 2) made me realize I needed a better tool for calculating the distribution of sums of dice in the first place. I first wrote about this some time ago in Dice Distributions, so I knew how to do it better, I just hadn't gotten around to doing it.

Tangent: While researching this I can across a great set of mathematical Dice Problems from Doctormatt (Web Page, Blog).

And now back to our story -

I first set up a spreadsheet to give me Pascal's Triangle, which looks like this:

This gets used in lookup functions to calculate (in a second worksheet) what I'm calling the "Dice Triangle", The number of dice [N] rolled is indicated in the column headers, and the sum of N D-sided dice [X] rolled in indicated in the first column. The number of ways to roll that sum is indicated in the table. The number of sides on the die [D] can be changed by entering a different value into the green shaded cell.

The first D rows of the table come directly from Pascal's Triangle. Subsequent rows are calculated from previous rows of this table. A few more details of how this is done in my earlier post (Dice Distributions), otherwise you can ask me, or dig into the spreadsheet for yourself (sorry, in a hurry today).

Here is the spreadsheet: Dice Distribution Calculator
I have not made this public, so you cannot change it directly online. You can download a copy for yourself (under the File dropdown) and play with it to your hearts content.

Maximums and Minimums of Dice Rolls

A week month a while back I received questions from Christian and a read post from Saxywolf on essentially the same question: What is the probability of rolling a given value on an D-sided die, if you roll N dice and take the highest (or the lowest).

Here's the trick:
The probability of rolling a 1 on 1 d6 is 1/6. (regular 6-sided dice)

For 2d6 the probability of rolling a 1 as the maximum is 1/6 times 1/6, or 1/(6*6) = 1/36.

For Nd6 the probability of rolling a 1 as the maximum is 1/6 times itself N times, or (1/6)^N.

For D-sided dice just substitute D for 6 above, so that the probability of rolling a 1 as the maximum of N D-sided dice is 1/D times itself N times, or (1/D)^N.

Now consider the problem of rolling 2-or less as the maximum. The probability of rolling 2-or-less is 2/D, and the probability of rolling a  2-or-less as the maximum is 2/D times itself N times, or (2/D)^N.
That's 2-or-less, but we really just want the probability of rolling 2, not 1 or 2.  BUT we already know the probability of rolling a 1 as the maximum on the same dice, so we can subtract that to get what we want:

The probability of rolling a 2 as the maximum is 2/D times itself N times, minus the probability of rolling 1 as the maximum, or (2/D)^N - (1/D)^N.

And that's it. Using the same math you can work the complete distribution of the maximum for any number of dice with any number of (equally likely) faces. Just start at 1 and work up. For minimums, just turn the problem around and find the probability of X-or-less.

Still too much math? Fear not for there is a spreadsheet to do the calculations for you:
 Link to Google Docs Spreadsheet

See the blue numbers in the green-shaded cells? Change those to the number of side on your dice and the number you want to roll, and it will calculate the distribution for you. It might even work inside the blog?Nope, but it was worth a try. The spreadsheet is now public and can be edited at the link above (also downloaded). Changes made there WILL show up here when the page is reloaded, which means you are looking at whatever was most recently entered.



And a chart to display the results:

This is a bit of an experiment, both linking to a shared spreadsheet, and adding the HTML code to it directly inside my blog post. One upshot of this is that when one person changes the spreadsheet, it will change it for everyone. Play nice! Let me know if it works too. [Fixed!]

The Grinder

[A dazzling display of delightful de ... um ... I need a D-word ... deviations ... detritus ... de-links?

The Grinder -  6/4/2011 edition

And speaking of dazzle, could Dazzle-camouflage make a comeback? This recent research supports says Dazzle Camouflage Affects Speed Perception. [Hat-Tip IO9]

Image: WWIaviation.blogspot.com

This just in! Check out some dazzling WWI aviation paint schemes.

Terrain table pictures ... Shiny!








This could be interesting ...

Invasion3042 is a massive multiplayer online game that is based off the game Battletech. It is a free game and is not for profit.

Has anybody tried it?

Discoblog brings us Tiny Toss-able Robots.

World Peace Games, with teacher John Hunter. Video from TED. It's a bit slow to get started, but gets interesting about 8 minutes in. Never cross a 9-year-old girl with tanks!

[Hat-Tip Greg Laden]


[The Endeavour] There are exactly five platonic solids*, and you can prove it!
* Perhaps more familiar to my readers as dice - the d4, d6, d8, d12, and d20.

Another video, this one with singing and dancing! Roll A D6.

[Hat-Tip: Operation Odyssey Dawn]


[IO9] MTG as an RPG?
The was an MTG computer game, long ago, with very nearly this premise. With a bit of creativity it could be good for multiplayer too. Before that was a great game called Master of Magic. Also, there could be a new Star-Trek animated series!

Play JAM! Can you beat the computer? Can you beat it every time?? Can you figure out the secret??? (without peaking!) Here is a hint - You have almost certainly played this game before, and many times. [Hat-Tip Terrace Tao]

Non-Transitive Dice

Starcraft Humor from Abstruse Goose. I didn't get it until I saw the caption.

Green Cube: The Physics Boardgame

Enough bedazzlement for two sittings, but that's happens when I don;t post for a whole month. Writers-block sucks. Want to preview the next Grinder, before I post it? Check out my Google Reader Shared Links page for GRB.