Lord Inquisitor

21-11-2009, 21:50

Yes, it's another Kill Point thread. I hope this one shows something new, so bear with me.

We've had a thread (http://www.warseer.com/forums/showthread.php?t=230121) on whether KP's are a good mission, and whether they are "balanced." Clearly, on average, the smaller army has an advantage. The pro-KP side will contend that this is the case, but it balances out an equal and opposite advantage for large-unit-count-armies in objective missions, they will have more units to take or contest objectives with, and therefore most armies will gravitate to the same number of units, balancing out the mission.

Theory

This is largely a repost from the previous thread, so feel free to skip this section if you've gone through this (or if you hate algebra!)

Assuming you have two players, X and Y. X has x units, Y has y units and we can assume these are not equal and x>y. X scores b KPs and Y scores a KPs. In order for X to win, b>a. But in order to do so, he must kill the fraction b/y of the enemy army. So assuming that ca=b (where c is a constant), in order to draw a KP mission:

ca/y = a/x

=> c = y/x

Conclusions: player X must kill y/x more of the enemy army than Y.

Now if we look at the objective mission (where things get more fuzzy): Player X has x scoring/contesting units and Player Y has y scoring/contesting units (and x>y again). There are z objectives. Player X loses a units and player Y loses b units. Assuming each unit can only capture/contest 1 objective each (big assumption!), that all units can contest all objectives (another big assumption) and, for simplicity's sake, that one player will win if he has at least one less untaken objective than his opponent, then the number of untaken/uncontested objectives U(x) or U(y) are:

U(x)= (x-a)-z (min 0)

U(y)= (y-b)-z (min 0)

U(x)-U(y)>1 for Y to win

So ((x-a)-z)-((y-b)-z)>1

This can only happen if x-a<z and y-b>z

Assuming that both suffer equal proportions of losses, (a/x and b/y), then x-a>y-b therefore given the assumptions above, yes, X will be expected to win but only if x-a>z. Therefore the optimum number of units is the minimum possible to leave z scoring/contesting units surviving the game, less if they can (a) take more than one objective or (b) they can move freely about the board and the enemy cannot.

Conclusion: an army with a larger unit count has an advantage in objective missions but only a) if the enemy has less than z units by the end of the game, (and remember that on average z=3and z is reduced by 1 for each enemy unit that is in range of multiple objectives); and b) if the numerically superior army has free reign to move about the board.

Statistics

So now to test it!

So, the predictions are:

In Kill Point missions, the smaller army is expected to have an advantage.

In Capture and Control missions, z=2, the advantage to the larger army is negligible, so neither is expected to have an advantage.

In Sieze Ground missions, the larger army is expected to have an advantage.

In order to test the predictions, I took a random sample of battle reports, namely the first 10 pages of the Warseer Battle Reports Forum (as they were November 19th). Each battle report was scrutinised and the following data recorded:

Mission Type

Number of Objectives

Number of units in each army (taking combat-squads or combined infantry squads into account)

Outcome of game

Whether the larger or smaller army won.

Battle reports were screened and only battle reports that met the following criteria were included:

A detailed list of both armies' units

Standard 40K missions (no planetstrike/ard boyz/etc)

An unequal number of units between the armies

Clear, online written format (I excluded all video reports, for example)

Obviously this screening was done before looking at the results, and needless to say all eligible reports were included. Edit: Also, to prevent bias, only one report per player was used (as much as I could tell). So the results weren't biased by a particular player submitting multiple reports with the same army (where the player skill is not independent from the army size). If a player submitted multiple reports, the first eligible one was used.

Results are attached as an excel file [edit: text file as I'm not allowed to attach excel files, if anyone wants the original .xlsx file pm me]. To summarise:

Sieze Ground

Total Games = 18

Won by smaller army = 7

Won by larger army = 7

Tie games = 4

Chi-square = 0

P = 1

Capture and Control

Total Games = 24

Won by smaller army = 9

Won by larger army = 7

Tie games = 8

Chi-square = 0.57

0.5 > P > 0.1

Annihilation

Total Games = 38

Won by smaller army = 24

Won by larger army = 11

Tie games = 3

Chi-square = 15.36

P < 0.001

I did a simple Chi-squared goodness-of-fit, using expected values = number of games played / 2 (ignoring draws). I.e. testing to see if the number of wins/losses for the smaller player versus the larger player deviated significantly from the expected 50-50.

Conclusions.

Sieze Ground and Capture & Control missions did not show any significant deviation from random, although C&C did show a slight trend towards the smaller-unit-count player (quite plausible, as small, elite, fearless armies may have an advantage in holding and keeping a small number of objectives). Annihilation missions showed the expected and highly significant advantage to the smaller player, with over double the number of games being won by the smaller side. Two games were noted by the authors as being a VP win for the side that lost by KP.

So: is there an advantage to high-unit-count armies for objective missions that counterbalances the clear advantage for small armies in annihilation missions? Answer: no.

We've had a thread (http://www.warseer.com/forums/showthread.php?t=230121) on whether KP's are a good mission, and whether they are "balanced." Clearly, on average, the smaller army has an advantage. The pro-KP side will contend that this is the case, but it balances out an equal and opposite advantage for large-unit-count-armies in objective missions, they will have more units to take or contest objectives with, and therefore most armies will gravitate to the same number of units, balancing out the mission.

Theory

This is largely a repost from the previous thread, so feel free to skip this section if you've gone through this (or if you hate algebra!)

Assuming you have two players, X and Y. X has x units, Y has y units and we can assume these are not equal and x>y. X scores b KPs and Y scores a KPs. In order for X to win, b>a. But in order to do so, he must kill the fraction b/y of the enemy army. So assuming that ca=b (where c is a constant), in order to draw a KP mission:

ca/y = a/x

=> c = y/x

Conclusions: player X must kill y/x more of the enemy army than Y.

Now if we look at the objective mission (where things get more fuzzy): Player X has x scoring/contesting units and Player Y has y scoring/contesting units (and x>y again). There are z objectives. Player X loses a units and player Y loses b units. Assuming each unit can only capture/contest 1 objective each (big assumption!), that all units can contest all objectives (another big assumption) and, for simplicity's sake, that one player will win if he has at least one less untaken objective than his opponent, then the number of untaken/uncontested objectives U(x) or U(y) are:

U(x)= (x-a)-z (min 0)

U(y)= (y-b)-z (min 0)

U(x)-U(y)>1 for Y to win

So ((x-a)-z)-((y-b)-z)>1

This can only happen if x-a<z and y-b>z

Assuming that both suffer equal proportions of losses, (a/x and b/y), then x-a>y-b therefore given the assumptions above, yes, X will be expected to win but only if x-a>z. Therefore the optimum number of units is the minimum possible to leave z scoring/contesting units surviving the game, less if they can (a) take more than one objective or (b) they can move freely about the board and the enemy cannot.

Conclusion: an army with a larger unit count has an advantage in objective missions but only a) if the enemy has less than z units by the end of the game, (and remember that on average z=3and z is reduced by 1 for each enemy unit that is in range of multiple objectives); and b) if the numerically superior army has free reign to move about the board.

Statistics

So now to test it!

So, the predictions are:

In Kill Point missions, the smaller army is expected to have an advantage.

In Capture and Control missions, z=2, the advantage to the larger army is negligible, so neither is expected to have an advantage.

In Sieze Ground missions, the larger army is expected to have an advantage.

In order to test the predictions, I took a random sample of battle reports, namely the first 10 pages of the Warseer Battle Reports Forum (as they were November 19th). Each battle report was scrutinised and the following data recorded:

Mission Type

Number of Objectives

Number of units in each army (taking combat-squads or combined infantry squads into account)

Outcome of game

Whether the larger or smaller army won.

Battle reports were screened and only battle reports that met the following criteria were included:

A detailed list of both armies' units

Standard 40K missions (no planetstrike/ard boyz/etc)

An unequal number of units between the armies

Clear, online written format (I excluded all video reports, for example)

Obviously this screening was done before looking at the results, and needless to say all eligible reports were included. Edit: Also, to prevent bias, only one report per player was used (as much as I could tell). So the results weren't biased by a particular player submitting multiple reports with the same army (where the player skill is not independent from the army size). If a player submitted multiple reports, the first eligible one was used.

Results are attached as an excel file [edit: text file as I'm not allowed to attach excel files, if anyone wants the original .xlsx file pm me]. To summarise:

Sieze Ground

Total Games = 18

Won by smaller army = 7

Won by larger army = 7

Tie games = 4

Chi-square = 0

P = 1

Capture and Control

Total Games = 24

Won by smaller army = 9

Won by larger army = 7

Tie games = 8

Chi-square = 0.57

0.5 > P > 0.1

Annihilation

Total Games = 38

Won by smaller army = 24

Won by larger army = 11

Tie games = 3

Chi-square = 15.36

P < 0.001

I did a simple Chi-squared goodness-of-fit, using expected values = number of games played / 2 (ignoring draws). I.e. testing to see if the number of wins/losses for the smaller player versus the larger player deviated significantly from the expected 50-50.

Conclusions.

Sieze Ground and Capture & Control missions did not show any significant deviation from random, although C&C did show a slight trend towards the smaller-unit-count player (quite plausible, as small, elite, fearless armies may have an advantage in holding and keeping a small number of objectives). Annihilation missions showed the expected and highly significant advantage to the smaller player, with over double the number of games being won by the smaller side. Two games were noted by the authors as being a VP win for the side that lost by KP.

So: is there an advantage to high-unit-count armies for objective missions that counterbalances the clear advantage for small armies in annihilation missions? Answer: no.