They're talking about ratios of unit-strength, which correlate directly to odds of success in a battle, not to numbers-of-units or anything more fuzzy like that. Given which, 20:10 and 2:1 have identical odds of success.
Do they, though? It depends on how the combat works.
Suppose, for example, that each side has 5 hit points, and repeatedly you roll a 2:1 die to decide who gets 1 point of damage, until one side reaches 0 hit points. The chance of an "upset", where the weaker side wins, is not 1/3; I compute it to be roughly 14%. If both sides start with 10 hit points, I compute the chance of an upset to be 6.5%. The law of large numbers means that, the more die rolls the combat involves, the less likely an upset is.
Or. Suppose that, at each step, one side has N soldiers and the other has M, and repeatedly a random soldier gets a kill; so that's an N/(M+N) chance that the first side gets a kill, and M/(M+N) that it's the other side. This would make advantages compound within the battle. Then I compute that a 2:1 initial matchup has a 5/6 chance (83%) of victory, and a 10:5 matchup has a 98.8% chance of victory.
(edit) I guess you could say I'm challenging the idea that "unit strength", such that when strength A fights strength B it's decided in one step with probability A/(A+B), makes sense as an abstract concept.
(defmemo meh (a b p)
(if (is b 0)
1
(is a 0)
0
(+ (* p (meh a dec.b p))
(* (- 1 p) (meh dec.a b p)))))
(defmemo nub (a b)
(if (is b 0)
1
(is a 0)
0
(+ (* (/ a (+ a b))
(nub a dec.b))
(* (/ b (+ a b))
(nub dec.a b)))))
kemayo|2 years ago
waterhouse|2 years ago
Suppose, for example, that each side has 5 hit points, and repeatedly you roll a 2:1 die to decide who gets 1 point of damage, until one side reaches 0 hit points. The chance of an "upset", where the weaker side wins, is not 1/3; I compute it to be roughly 14%. If both sides start with 10 hit points, I compute the chance of an upset to be 6.5%. The law of large numbers means that, the more die rolls the combat involves, the less likely an upset is.
Or. Suppose that, at each step, one side has N soldiers and the other has M, and repeatedly a random soldier gets a kill; so that's an N/(M+N) chance that the first side gets a kill, and M/(M+N) that it's the other side. This would make advantages compound within the battle. Then I compute that a 2:1 initial matchup has a 5/6 chance (83%) of victory, and a 10:5 matchup has a 98.8% chance of victory.
(edit) I guess you could say I'm challenging the idea that "unit strength", such that when strength A fights strength B it's decided in one step with probability A/(A+B), makes sense as an abstract concept.