"Help me understand the math" Topic

Here's the overview: The system uses opposed die rolls from a d4 through d12. Better troops roll larger die types as a reflection of quality. For sake of example, let's say average troops roll d6 for attacking and defending.
Let's say I've determined that fighting in a village should result in the equivalent of a 3:1 advantage for the defender. If both attacker and defender are rolling d6s, is there a way to translate a 3:1 advantage for the defender? I would prefer to move "up" a die type rather than rolling more dice, if that's possible.

One way to get 3:1 is to have attackers rolling a D4 and Defenders a D6. Highest score wins, Ties favour the defender.

rerolls (?)

works with some systems, not with others..

Die type changes give you odd and uneven shifts in odds. For example, opposed D6 will give a win for side A41% of the time, a win for side B 41% of the time and a tie otherwise.

Note: If the defender wins ties then he will win a straight D6 roll off 72% of the time….

But a D8 versus a D6 gives a win for the D8 56% of the time. A D10 versus a D6 gives the D10 a win 65% of the time. A D12 wins 71% of the time versus a D6. If ties go to the defender those odds drop.

You might consider bonus dice. Being in a village gives you 1 or 2 bonus dice of your troop type (D6 in this case) and you pick the best result.

I've just thought of a much simpler way to get your 3:1 – just roll 3D6 against 1D6. Choose the highest of the 3D6 and use that score against the single D6.

Table below give all possible % for opposed die rolls.

Probabilites for win / draw / lose (attacker in column on left, cross-reference with defender in row)

D4 D6 D8 D10 D12
D4 0.38 / 0.25 / 0.38 0.25 / 0.17 / 0.58 0.19 / 0.17 / 0.69 0.15 / 0.10 / 0.75 0.13 / 0.08 / 0.79
D6 0.58 / 0.17 / 0.25 0.42 / 0.17 / 0.42 0.31 / 0.17 / 0.56 0.25 / 0.10 / 0.65 0.21 / 0.08 / 0.71
D8 0.69 / 0.13 / 0.19 0.56 / 0.13 / 0.31 0.44 / 0.13 / 0.44 0.35 / 0.10 / 0.55 0.30 / 0.08 / 0.63
D10 0.75 / 0.10 / 0.15 0.65 / 0.10 / 0.25 0.55 / 0.10 / 0.35 0.45 / 0.10 / 0.45 0.38 / 0.08 / 0.54
D12 0.79 / 0.08 / 0.13 0.71 / 0.08 / 0.21 0.63 / 0.08 / 0.30 0.54 / 0.08 / 0.38 0.46 / 0.08 / 0.46

As you can see, a D6 vs D6 roll means an equal win 42% for either side. If you increase the defender to a D12 (rolling D6 vs D12), that means 21% win for the attacker, 71% for the defender, roughly 3:1.

Now that sounds like Piquet. What opposed die roll does, increasing the die size of the side with greatest chance, is still give a winning chance to the underdog.

E.g. The higher rated side is rolling D12, the underdog D4. The D12 can still roll 1, and the D4 can roll 4. In Piquet, if that was a melee result, the D12 unit would be unralliably routed.

So much for pitting Knights against peasants. Perhaps the knights charged across rabbit warrens, perhaps the peasants were being VERY well led. Who knows, but like this happens in war.

Now that sounds like Piquet. What opposed die roll does, increasing the die size of the side with greatest chance, is still give a winning chance to the underdog.

Stargrunt II also uses opposed die rolls. It's not a new idea.

If you increase the defender to a D12 (rolling D6 vs D12), that means 21% win for the attacker, 71% for the defender, roughly 3:1.

However, my multiple dice solution gives you EXACTLY 3:1.

Olicana -- Piquet and FoB are my favorite rulesets and often my inspiration for my own rules.

Phil -- thanks for the table. Very helpful for a non math person like myself.

For the full math:

Suppose Dy attacks Dx with y >=x: (if y < x we have the symmetric case)

Total number of possible outcomes: x*y

Probability Dy has the highest score: [x*y – x – x*(x-1)/2]/xy = 1 – (x+1)/2*y

Probability both dice roll the same number: x/x*y = 1/y

Probability Dx has the highest score:: [x*(x-1)/2]/xy = (x-1)/2*y

For the case where we roll 3D6 against 1D6, I calculate:
1D6 wins 225/1296 or about 17.4%
Tie 216/1296 or about 16.7%
3D6 wins 855/1296 or about 66%

That is 3.8:1 in favor of the 3D6.

anydice.com
will allow you to calculate any odds like this.

Roll a D6 and add +1 for each positive column shift on the odds ratio.

So accepting opposed die rolls start off at 1:1, going to 3:1 is two column shifts i.e. you go from 1:1 to 2:1 and then to 3:1, so having shifted two columns the player with a 3:1 advantage just adds +2 to their die roll.

This of course could be complete rubbish, as I am rubbish at maths, but like mechanisms.

Roll a D6 and add +1 for each positive column shift on the odds ratio.

Um, no. Adding +1 does not double the odds, and +2 doesn't triple them:

With a straight 1D6 vs 1D6:
Win 41.6%
Tie 16,6%
Lose 41.6%

With a +1:
Win 58.3%
Tie 13.8%
Lose 27.7%

With a +2:
Win 72.2%
Tie 11.1%
Lose 16.6%