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I was torn about whether to post this as a game problem or as a maths problem. My maths is a little shaky, so thought maybe best to post here. The game, for anyone who is familiar with it, is War and Order a mobile RPG / MMMORPG (I was never quite sure on the definitions).

In the game, cavalry troops can learn a skill named "Dodge". When maxed, the skill is:

Cavalry have a 16% chance to dodge enemy attacks.

My question is how that 16% might be applied. Obviously the actual mechanics are under the hood so this is all guesswork.

I dont think a great deal of detail around troop formations here is necessary. But each battle formation can hold up to 10 troop types, and subsequently, 10 "pods" of troops.

A common troop formation is like below. The troops walk towards the enemy and when they get in range they start attacking. Troops attack at the same rate, so its reasonable to assume that in most cases, each troop pod will have an equal number of attacks.

(troops facing down)

  1
  2
  3

Say over 10 seconds, this formation produces 30 attacks, all focused on enemy cavalry with 16% "Dodge". Probability suggests you would expect 4.8 of those attacks to be dodged. Because it's chance based, if you were to rerun the fight a a few times. Maybe sometimes it was 3 attacks dodged, and maybe sometimes it is 7, but overtime it should average to 4.8, or 1.6 per pdo

Now is the time to point out, that while Dodge treats all attacks as equal (or so we assume), I, as the attacker, care more about the attack coming from troop pod 1 (which has higher attack) than troop pod 3 ( which has lower attack but higher health and defense).

Now take this new formation. I have included two new troop pods. Their numbers are small and they exist only to attempt to trick "Dodge".

Comparing to the first scenario, we are now producing 5 attacks at a time, so over that same 10 second time period, 50 attacks are produced.

The basic maths, to me, suggest that now averaged out over time, 8 attacks will be missed. Which comes to the same 1.6 per pod.

(troops facing down)

4  1  5 
   2
   3

Remembering that we care most about the attack from pod 1 not being dodged, because it does the most damage

My theory / question

Is there maths to suggest that formation 2 would perform better than formation 1? Or to state more concretely, in formation 2, given a

At first glance, I would have expected 16% to mean 16% and move on. However:

  • Scenario 1: the attacks we care about make up 33% of the total attacks
  • Scenario 2: the attacks we care about make up 20% of the total attacks

Given that these fights only last a certain amount of time. Because 16% of all attacks are dodged, and in Scenario 2, the attacks we care about (pod 1) make up a smaller percentage of the total attacks is there a greater chance that fewer attacks from pod 1 are dodged?

I've dragged myself into a hole of poisson and binomial distributions, but my maths basics is not strong enough to come up with a working theory on my own.

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  • \$\begingroup\$ What you're describing here is just a variant of the gambler's fallacy. You've tricked yourself into believing independent events are dependent. \$\endgroup\$
    – Cubic
    Commented Jul 7, 2023 at 10:58

2 Answers 2

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No, there is not a greater chance that the attacks from pod 1 will be dodged as each dodge is an independent check (unless there is some feature you haven't described). Pod 1 still gets 10 attacks, and an average 16% of the attacks from each pod will be dodged, completely independently of each other and the other pods. Same for pod 2,3,4, etc.

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No

unless dodge is implemented in a very weird and sophisticated way (see the evasion system in path of exile) AND you know how that implementaion works, there is no way to "game the system".

If it is actually just a random chance applying to each attack individually, then there is literally nothing you can do to "game the system".

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