Question of probability

>>7696773
.000793

>>7696784
No working out and no contact info Im dumb i need the working out.

i simulated the situation n times for k=1 000 000 and k=100 000 000, my laptop almost melted while doing so

k=1 000 000
0, 0, 0, 0, 3, 436, 73988, 925573

k=100 000 000
0, 0, 0, 1, 278, 42438, 7403501, 92553782

nth number corresponds to the number of simulations where (n-1) patrons stayed on the field
using interpolation it's visible that the probability of no patrons staying on the field is approximately of the order of 10^(-15) or lower

This is a hell of a question tb.h.
For non-HSfags who could help, OP basically wants to know the chance that your first 16 damage (although there is also 1 damage spare) does 2 on each dwarf, then the final 7 shots pick them off.

>>7697340
It would be the first 14 damage. The total number of ways you could kill all the dwarves would be the number of ways you could do 2 damage to each dwarf first ((14*13/2)*(12*11/2)*...*(2*1/2)) times the number of ways you could finish the 7 of them off (7!) times the number of ways you could mix in the 3 damage to the hero (24*23*22/3!) since it only takes 21 to kill the dwarves and if you deal more than that to them you wouldn't have enough damage left to finish off any new ones that get made. From there though you have to divide by the total number of ways damage could be dealt, which is fucking ridiculously complicated.

This would be a fun Project Euler type problem. Write an algorithm to calculate the probability (not estimate, calculate). Can't brute force something that has about 8^24 possibilities (obviously less technically, but around that maginitude), but can't realistically calculate by hand.

Unfortunately I don't have the time and I'm out of experience.

File: patron.jpg (17KB, 360x527px) Image search: [Google]

17KB, 360x527px

>>7697486
I can simplify it a bit. Start at the 7 at the top and move downward until you get to the 1 on the bottom. Take the product of the 14 numbers you passed through. Do this for every possible path, sum the results, and divide by 8^15 and that's the probability of wiping the Patrons.

>>7697882
So you could write this up as a Markov chain with a large, sparse, and triangular (acyclic) transition matrix.

>>7697464
Consider the following three cases:

Case I: Your first 14 shots hit each of the 7 dwarves twice
Case II: Your first 15 shots hit all hit each of the 7 dwarves twice and the hero once, AND the last shot is not at the hero (else we'd be in case I)
Case III: Your first 16 shots hit each opponent 8 times, AND the last shot is not at the hero (else we'd be in case II)

If you reach attain any of these 3 'victory' conditions, you know you've won regardless of how your remaining shots turn out.
On the other hand, if you finish 16 shots without attaining any of these goals, then there will be at least one empty space (by pigeonhole principle, 16 shots for 8 opponents and none of the three cases hold) AND there will be at least one dwarf with 2 health (discarding the opponent killed, there are 13 shots for 7 opponents, and that opponent can't be the hero else you would have reached one of the cases), so you know you've lost.

Each of the three cases is a simply probability exercise: since no one dies the denominator is always a power of 8, and the numerator is the number of ways to target.

The final probability should be the sum of these 3 mutually exclusive and exhaustive cases, unless I've missed something?

>>7698539
Op here, the problem with this is you need to also sum up the probability of each way the first 14 shots COULD hit each of the 7 dwarves twice.

Here's an example for Case I only --- Each indices represents how many times a bolt hit a minion and each power step down represents a case where it has hit one twice and must move on to the next one, the chances start at 7/8 for case one due to there being 8 targets and for case one to be true it must hit only the trolls.

I first picked randomly:

(7/8)^3
(6/8)^4
(5/8)^3
(4/8)^1
(3/8)^1
(2/8)^1
(1/8)^1

For 0.00030322189.

Then pick in order to get them all down by 1 then the remaining 1:

(7/8)^8
(6/8)^1
(5/8)^1
(4/8)^1
(3/8)^1
(2/8)^1
(1/8)^1

For 0.00094375007.

Then randomly again:

(7/8)^4
(6/8)^2
(5/8)^2
(4/8)^1
(3/8)^3
(2/8)^2
(1/8)^1

For 0.00002653191.

All possible outcomes for you to wipe the board clean for case one ALONE, good lucky defining all the possibility manually this way, while correct Im looking for a formula and the exact percentage

>>7698539
>>7699373
Sorry i skimmed this you said "exhaustive cases"

Never the less now all we need are three of these >>7697882
as you said we now only need to work out the first 3 outcomes involving the first possible 14 15 and 16 shots.

>>7698539
Oh fuck you have tho because the last 10 shots have a chance to all hit the hero but thats easy right? aint it just ((8/8^3)*7/8*6/7*5/6*4/5*3/4*2/3*1/2)

>>7699430
i gota think and post in one go but to correct myself again its the sum of

((8/8^3)*7/8*6/7*5/6*4/5*3/4*2/3*1/2)
((8/8^2)*7/8*6/7*5/6*4/5*3/4*2/3*1/2)
(8/8*7/8*6/7*5/6*4/5*3/4*2/3*1/2)
(7/8*6/7*5/6*4/5*3/4*2/3*1/2)