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How often on average - shooting arrows into 2 targets

  1. Mar 3, 2014 #1
    how often on average -- shooting arrows into 2 targets


    I was playing a game the other day as an archer. There were two possible setups and ever since i have been trying to decide which one is going to be the best. So here goes.

    option A:
    You fire 3 arrows, they have a chance x to pass through the target and keep on flying towards another target where you have again the same x chance for it to pass through.

    option B:
    You fire 9 arrows, they all hit a target once, dealing half the damage of an arrow in option A.

    at which point for x does it become more beneficial to go for option A?

    Any and all help would be appreciated
  2. jcsd
  3. Mar 3, 2014 #2


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    Gold Member

    Show your attempt.
    What have you done so far?
    -PF rules-
  4. Mar 3, 2014 #3

    Stephen Tashi

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    I assume this means that they "each have a chance x to pass through the target".

    Is there supposed to be a chance than an arrow misses the target completely? You didn't say how the damage done by the arrows in this case is assessed.
  5. Mar 3, 2014 #4
    in reply to Stephen Tashi:
    Yes I ment, that each arrow has its own individual chance to pierce a target.

    For simplicity sake, lets assume they always hit a target.

    The damage for the 9 arrows seems quite obvious to me, that is just 4.5 times the damage of a base arrow
    For option A however there is
    (1 + x + x^2 + x^3 + ... ) * 3 times the base damage of the arrow or [itex]^{\infty}\sum_{n=0}[/itex]3*xn
    Which, im sure, will converge to a certain number, and that number I should probably compare to
    4.5 and then from there extract x to find the minimum chance to pass through for it to be worth it
  6. Mar 3, 2014 #5
    So I'm assuming you're playing a video game and not some kind tabletop RPG or card game?

    Before you are going to be able to apply anything that looks like math to your question, you are going to need to clearly define what "more beneficial" means; best potential dps, best realistic (whatever that might mean) dps, best max damage per shot, etc.

    You also need to take into consideration a few more parameters than you seem to be looking at. For instance, the reduction in damage per shot that you get from B might be meaningless if that is still enough to one-hit everything. Also the percent chance that the arrow passes through something won't likely be the same as the percent chance that a second enemy is hit by the arrow unless you're playing and essentially 1-D game. Do you have finite ammo, or is your supply of arrows unlimited?

    I wouldn't get my hopes up for an easy/straightforward computation if I were you. If computing these things were easy, you wouldn't see the balance issues that most (all?) character-building games have.
  7. Mar 3, 2014 #6
    its an aRPG, so not a tabletop no.
    the 3 and 9 arrows are fired in the same timeframe. So the damage per cast directly translates to similar dps.
    Assume that the reduction is not meaningless and you need at least the full damage equivalent of 1 arrow to kill a monster.
    While it is correct that not every arrow that pierces an enemy will find a next target, it is likely that it will find a second and third target. Supply is unlimited (though option B has a higher mana cost arguably, though this is a rather separate matter).

    What i am searching for, I think, is y <the average amount of monsters hit> if there is an x chance to pierce for each hit.
    That way i can see where the break even point lies for x to hit more than 1.5 monster on average ( and thus surpassing 3 times the hits with half the damage).
  8. Mar 3, 2014 #7

    Stephen Tashi

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    I think it would be 9 times the average damage of one arrow. I don't know what you mean by "damage of a base arrow".

    (1 + x + x^2 + x^3 + ...) sums to 1/( 1-x) if x < 1.

    But if x is the probability of passing through a target, the probability that the arrow hits exactly n targets and fails to pass through the next target is (x^n)(1-x). If B is the mean damage of an arrow per target, the mean damage for one arrow is:

    m = (1)(1-x) B + x (1-x) (2B) + + x^2(1-x)3B + ....
    = (1-x) B ( 1 + 2x + 3x^2 + ....)

    We can look up how to sum an "arthmetico-geometric" series.
  9. Mar 3, 2014 #8
    Since in this case each arrow deals only half the damage. 9 arrows deal the same amount of damage as 4.5 arrows in any scenario where there is no reduced damage per arrow.

    If I read this correctly you are saying that if there is a 1/3 chance to pass through and infinite line of enemies each arrow will hit 1,5 targets on average. Correct ?
    Last edited: Mar 3, 2014
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