No one has it so far, though one person was very close.
Here are some more hints.
Hint 3: The probability of the defender hitting is worked out as (1000x2000)/(200+2000) (rounded down) out of 1000. The attacker gets the rest of the 1000. (The die has 1000 sides you see.)
So the attacker has exactly 9.1% (or 91/1000) chance each round of making a hit. But if you assumed 10% you would still be pretty close.
Hint 4: The damage dealt per hit by the warrior is 8HP, meaning it will take 13 hits to kill the infantry. Since we want the warrior to not take a single hit, the damage dealt by the infantry per hit is irrelevant.
Hmm, from this we can start working. Let's try with 10%. So, we want a 50% change of not taking any damage. Say we have x first strikes (can you tell I was a math major ?
) First, I'll play with the cases where we kill the infantry within those first strikes (ignoring a case like if we had 40 first strikes, we lose all 40 of those battles, then win the next 13).
So, that means that we want the lowest x such that P(13 hits in x chances) > .5. Uggh, I think this way means we have to do some calculating with a binomial series or something. So, try something like 130. The odds of coming out unscathed will be:
(130 choose 13)*0.1^13*0.9^117 + (130 choose 14)*0.1^14*0.9^116 + ... lots more terms. Using a trusty online binomial calculator, the odds of this work out to be 0.547. If I plug in 0.091 instead of 0.1 it shifts to 0.4036.
Using 0.091 now, let's try values for x:
x = 200: odds unscathed ~92.5%
x = 150: odds unscathed ~61.5%
x = 140: odds unscathed ~51.2%
x = 139: odds unscathed ~50.3%
x = 138: odds unscathed ~49.1%
x = 137: odds unscathed ~48.0%
x = 136: odds unscathed ~46.9%
Now, I did ignore cases before where we didn't kill the infantry within the first strikes. If we actually wanted to calculate that, we'd take the same procedure as above, but add on the odds of getting 12 hits within the first strikes * the chance that we get the next hit, plus the odds of getting 11 hits within the first strikes * the chance of getting the next 2 hits + ... again, lots more terms. Now, since that's the tail end, it diminishes rapidly. So, how much will that extra term account for. Let's work out the next 2 terms for x=137, and then I can estimate after:
P(12 hits in 137 chances)*P(hit on 138th combat) = 0.1186 * 0.091 = 0.0107926
P(11 hits in 137 chances)*P(hits on 138 and 139th combat) = 0.1128 * 0.091^2 = 0.0009340968 which is tiny.
But our 12 hits within the first strikes followed by a hit on the next one actually is about 1% chance of happening, which bumps out threshold to 138 first strikes, instead of the 139 needed without counting that last case. Now, the rounding errors above I doubt will contribute anything more, so my guess would be 138 first strikes, which would be up to what, drill 40?
