RLVG's weird math problem

[RLVG]

There's 1250 birds. You have a gun and get one shot of each bird per visit. Each shot has a chance of killing by 15%.
If you kill 100 birds first visit, there's 1150 for the next visit, kill another 150 and the number drops to 1000 for next visit and so on.

How many visits would it take to kill every single bird in average?
What's the chance of it? How likely would it be at 20 visits VS 200?

In addition, the same thing but with 500 birds instead.



Why am I asking this? Because this genius at work thinks he can do speed math and I thought of something randomly complicated.

[Ash]

1 visit is all that's needed if you're not a fuckin noob

[Firebringer]

There's 1250 birds. You have a gun and get one shot of each bird per visit. Each shot has a chance of killing by 15%.
If you kill 100 birds first visit, there's 1150 for the next visit, kill another 150 and the number drops to 1000 for next visit and so on.

How many visits would it take to kill every single bird in average?
What's the chance of it? How likely would it be at 20 visits VS 200?

In addition, the same thing but with 500 birds instead.



Why am I asking this? Because this genius at work thinks he can do speed math and I thought of something randomly complicated.

So the way I approached this problem was using the 15% as a crutch for figuring it out. The way I did it though assumes that you always are killing 15% (which isn't how it would actually work given that isn't how statistics work since each attempt is a stastical individual event, but w/e)

Answer: 40-43 Times

Specifics: I treated this as I would treat a bond amortization problem. To figure out what a bounds interest are you have to times the carrying value by a percentage, percentage being this 15%. Carrying being the 1250 birds, each time however this decreases amount of carrying value and the interest goes down. Essentially less birds to kill so less chances to kill. You put in a excel document takes about two minutes to do the table you get somewhere between 32-35 times before it goes down to around 5 birds, then it starts going into weird fractions of numbers. But around 40 there are no more birds.

[RLVG]

So the way I approached this problem was using the 15% as a crutch for figuring it out. The way I did it though assumes that you always are killing 15% (which isn't how it would actually work given that isn't how statistics work since each attempt is a stastical individual event, but w/e)

This makes the answer highly inaccurate.

15% in fraction of 100 is in essence 15/100 or 3/20.
With 1250 birds, it means 1250 "coin flips" with 3/20 chance of landing the correct side per bird and a 17/20 chance of failing to shoot a bird.

I have no idea how to actually calculate individual events as a statistic and perfect the answer in total.

In the absolutely worst case, it's possible to not shoot anyone and in the absolutely best case it's possible to shoot all of them in a single go.



Making it 15% constantly on the total number as opposed to each individual one is kind of biased.

[Cryptonic]

Could be infinite

[Orpz]

From what I understand, you shoot each bird and hit with 0.15 probability and thus miss with 0.85.

The problem posed can be phrased in terms of the expected value of a negative binomial random variable: "How many tries should it take until 1250 successes are reached?" The formula for that is r/p, where r is the number of desired successes and p is the probability.

r/p = 1250/0.15 = 8333.33

Also, to find out the average amount of trips, we assume that we cull 15% of the population each visit and don't shoot at dead birds. Thus, the number of birds remaining after each visit are (1250, 0.85 * 1250, 0.85 * 0.85 * 1250, ... ).

So, after n visits, we have 0.85^(n) * 1250 birds remaining. This is an exponential decay function with asymptote 0, so the best way would just be to take the Floor of this function (which should be fine since you cant have fractions of a bird anyway).

So we find out when f(x) = 1, and then take the next integer greater than that (because it's a decreasing function so f(x + ) is strictly less than f(x), and we will get our desired value of 0.
So on average it should take 44 trips.

requesting Mattzed check my work, gonna be late to class I stay longer

[MattZed]

Orpz, we're taking the max of 1250 geometric random variables, not their sum, so this won't follow negative binomial. Just gotta use the CDF and do this one the hard way. Quick check on Stack Exchange says there's no closed form for this, and computing the expected value will require summing an infinite series.

The probability that any given bird is alive after n visits is .85^n. The probability that a given bird is dead at time n is thus 1-.85^n, and then the probability that they're all dead at time n is (1-.85^n)^1250. Thus, the probability that your nth visit is the one you killed the last one is the probability they're all dead after n visits minus the probability they're all dead at n-1 visits. (meaning there was at least one still to kill after your last visit) That is, (1-.85^n)^1250 - (1-.85^(n-1))^1250. The expected value of this random variable is then the sum from 1 to infinity of n*((1-.85^n)^1250 - (1-.85^(n-1))^1250))

(which is approximately 47.9) For 500 birds, we replace 1250 with 500 and get 42.3 expected tries.

The probability you would do it in either exactly 20 or 200 tries is so small as to be basically 0.

[Cryptonic]

But every time you miss, that stupid dog laughs at you. Then you get angrier and angrier and try to shoot the dog when he laughs, only to find that you can't shoot when that scene is happening. The anger then gets in the way of the calmness required to shoot properly, lowering your accuracy. No one took this into consideration.

[Brendan]

Fking nerds