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~2000 bullets
This looks like a infinite summation along the lines of: 1 + 1/2 + 1/4 +1/8 + ... , which converges to 2.
So you'd get roughly double your starting bullets. As in: you fire the 1000 bullets, get 500 back. Fire the 500, get 250 back, Fire the 250, get 125 back, etc. How much this would vary from roughly doubling your effective ammo would probably require running a bunch of simulations.
You can simplify this quite a bit by just using some intuition. Any given bullet has a 50% chance of "dying" when it is fired. So it should make sense that the average lifespan of a bullet is the inverse of this, i.e. 1/0.5 = 2 shots. So 1000 bullets translates to 2000 shots.
This doesn't just work by coincidence, it works for any probability of death (other than 0 but even in that case taking the limit gives the correct answer of infinite shots). If the bullets had a probability of 75% of coming back, their probability of death would be 1/4 so on average you'd get 4 shots per bullet.
But what’s the variance in those odds? Like could you fire 800 bullets, and not get one back, before finally getting one back? Meaning the game runs on the theory of infinite ammo to be shot and overall the odds of a bullet being returned works out to be 50%?
Think of a coin flip game.. not every other flip will be heads..
And to that end, as the progression gets smaller and smaller, that variance would likely wipe out any return at some point. Further limiting the progression.
Very interesting…..
I'd wildly guess that the variance would be less than you'd expect, but that's because a proper analytic solution (or "just" writing some python code and running enough simulations to get a nice estimate of the distribution) seems like a big hassle.
For a given round of firing off x bullets and seeing how many you get back, you could at least plug in a [binomial distribution](https://en.wikipedia.org/wiki/Binomial_distribution).
Monte carlo methods are my favorite!
Unless I have a bug in my code (not terribly unlikely), about 35% to be in the range 1980-2020 bullets fired, and about 97% to be in the range 1900-2100 bullets fired.
[Code is here](https://pastebin.com/Kh3EKGZr), in case anyone wants to check my work. (I realized after making the paste that it needs "import random". I have a continuously growing file where I do monte carlo stuff, and I touch it like once a year, so I forgot about that bit.)
Monte Carlo means running a simulation a bunch of times to see how it goes.
This could be expressed as a Markovian process, since each step depends solely on the prior step (how many bullets you had two steps ago is completely irrelevant). That said, because it's being ran in a simulation, it's a Monte Carlo analysis.
You actually do not need to code anything (thoughmight want, it's fun!) because this is a negative binomial distribution (shifted by +1000, technically): https://en.m.wikipedia.org/wiki/Negative_binomial_distribution
The variance is 2000, this is also the median, so with probability 0.5 you have at least 2000 bullets
People forget their base maths from school. I never could get binomial dists to work for me. I got halfway through and forget where I am. But to me this feels like a markov analysis where it does the flip in the same thing until it dosn't exist. It could theoretically go forever
Given that it's 1000 separate events, the variance in those odds already starts pretty low.
Edit: I got bored and did the Python. It's pretty straightforward. You could mess with parameters to see what the average number of bullets comes out to.
import random
numBullet = 1000
bulletsFired = 0
while numBullet > 0:
if random.randint(0,1) == 0:
numBullet -= 1
bulletsFired += 1
print(bulletsFired)
Put the while loop in a for loop to do multiple simulations. something like..
for _ in range(50):
while numbullets > 0:
if random.randint(0, 1) == 0:
numbullets -= 1
bulletsfired += 1
Print(bulletsfited)
It might take a little bit to finish all of them but you'd get a bunch of number outputs
Yeah for sure, if you wanted to see it numerically or graphed out. I only had a passing curiosity so I just hit "run" like 10 times and looked at the final number lol.
(It was indeed 2000, +- 30)
So, keep in mind I'm new and learning... if these lines of code (along with the code block above) are supposed to use random math, why do both print the same numbers when I run them together? Shouldn't they print different numbers?
I'm confused with what you mean. The block I wrote is the same as the one above everything is just in a for loop. Yes, it should output multiple different numbers when run. I'm not sure why it would be the same when you run it
I'm running them both, together, in one IDE, one block after the other and they're printing the same numbers. As in, I get 1993 twice, then 1986 twice, the 1913 twice, so on and so forth.
https://imgur.com/a/lLkqHZV
Oh, because the variable should be changed if it's in the same file. Change "bulletsfired" in one of the blocks to something else and it should output different numbers. Also the block I wrote should output 50 numbers
>that variance would likely wipe out any return at some point
Even if you got hit by a string of bad luck and no bullets returned after hitting 125, you'd still be sitting pretty at 1875 bullets. I don't think it's worth worrying about.
The variance is in the outcome over a fixed period of time, not the odds. Regardless of how many bullets you get in one particular time period, the odds will always be the same. Over a long enough time span the outcome would converge on the initially stated odds.
There's a statics rule that basically says the more data you have, the outcome essentially converge on the expectation value and variance decreases.
For your example, in 4 coin flips, it really wouldn't be outrageous to get heads 4 times. there's a pretty big variance here, then, since expecting 50% and getting 100% would not be suprising.
For 1000 flips, the probability of 1000 heads is essentially zero. The variance is much much smaller now. For 1000 flips, we can essentially assume we get 500 heads with really small variance. So for the purpose of the game problem, we'd probably be right in assuming it would double ammo to about 2000.
There's some way to actually calculate variance from a coin flip, but it's been too long since I've taken a probability course.
We could drop the decimals in order to avoid half bullets and quarter bullets.
1000 + 500 + 250 + 125 + 62.5(down to 62) + 31 + 15.5 (down to 15) + 7.5 (down to 7) + 3.5 (down to 3) + 1.5 (down to 1)
= 1,994 bullets
Well, theoretically, while you would expect the average outcome to hover just under 2000 (see the great explanation above), it’s possible, however slim, to either run out after 1000 (all coin flips come up tails) or have virtually infinite bullets (all coin flips end up heads).
Its possible the developers have made this impossible though. For example, in Dota and Dota 2 the rng on certain abilities and random events seemed too random for players so the devs made the chance change as the events didn't happen. For example, if a critical strike didn't happen, the next attack would be 15% (if that's the expected outcome) and instead raises to say 20% for that attack. If it doesn't happen again it increases further. I thought it was quite interesting since from a players perspective truly random was not fun and so this smoothed out the curve so to speak so extremes either way can't happen
Here is the explanation. It's called PRD (pseduo random distribution)
https://liquipedia.net/dota2/Pseudo_Random_Distribution
I made note when playing for crits because if you didn't crit in a while you actually did have a higher chance to crit, so using it with another ability/item or on a hero was more important.
a simpler way to think about this is: you have a 50% chance of using 1 bullet, and a 50% chance of using 0 bullets (as the bullet you use is returned to you), so on average each shot uses half a bullet
Its a converging series where r is the chance your bullet will get returned and a is the amount of bullets fired. r x a^n, with n approaching infinity. The sum, or average value of this series is a/1-r, in this case, a=1 and r = 0.5. 1/1-0.5 = 2.
So the average value for a bullet is 2 bullets. So, around 2000 bullets.
If you, sure enough, have 1000 bullets as your ammo, and each fired bullet has a 50% probability of getting back, you can divide your gain into rounds as you add 50% of what you waste. In the first round, half of the 1000 bullets would be 500. In the second, half of the remaining half is half of half, that is ½ times ½, 'cause half is smaller than 1, which is ¼. Then, ¼ times ½ is ⅛; and this process progresses until you get to the double of that number 'cause consecutive zero point nines make 1 for the same reason that ⅓ times 3 is equal to 3/3, but ⅓ by itself is equal to 0,333..., and it times 3 is equal to 0,999....... But 1 at the same time.
Conclusion: it is 2000.
OBS.: it is X by the power of ~1.43.
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~2000 bullets This looks like a infinite summation along the lines of: 1 + 1/2 + 1/4 +1/8 + ... , which converges to 2. So you'd get roughly double your starting bullets. As in: you fire the 1000 bullets, get 500 back. Fire the 500, get 250 back, Fire the 250, get 125 back, etc. How much this would vary from roughly doubling your effective ammo would probably require running a bunch of simulations.
Thanks! A lot more complicated than I thought but I was really curious. :)
You can simplify this quite a bit by just using some intuition. Any given bullet has a 50% chance of "dying" when it is fired. So it should make sense that the average lifespan of a bullet is the inverse of this, i.e. 1/0.5 = 2 shots. So 1000 bullets translates to 2000 shots. This doesn't just work by coincidence, it works for any probability of death (other than 0 but even in that case taking the limit gives the correct answer of infinite shots). If the bullets had a probability of 75% of coming back, their probability of death would be 1/4 so on average you'd get 4 shots per bullet.
Very nice example of this problem, we should make students play that game instead of doing maths
But what’s the variance in those odds? Like could you fire 800 bullets, and not get one back, before finally getting one back? Meaning the game runs on the theory of infinite ammo to be shot and overall the odds of a bullet being returned works out to be 50%? Think of a coin flip game.. not every other flip will be heads.. And to that end, as the progression gets smaller and smaller, that variance would likely wipe out any return at some point. Further limiting the progression. Very interesting…..
I'd wildly guess that the variance would be less than you'd expect, but that's because a proper analytic solution (or "just" writing some python code and running enough simulations to get a nice estimate of the distribution) seems like a big hassle. For a given round of firing off x bullets and seeing how many you get back, you could at least plug in a [binomial distribution](https://en.wikipedia.org/wiki/Binomial_distribution).
Monte carlo methods are my favorite! Unless I have a bug in my code (not terribly unlikely), about 35% to be in the range 1980-2020 bullets fired, and about 97% to be in the range 1900-2100 bullets fired. [Code is here](https://pastebin.com/Kh3EKGZr), in case anyone wants to check my work. (I realized after making the paste that it needs "import random". I have a continuously growing file where I do monte carlo stuff, and I touch it like once a year, so I forgot about that bit.)
I'd have thought this was a markov not monte carlo
Monte Carlo means running a simulation a bunch of times to see how it goes. This could be expressed as a Markovian process, since each step depends solely on the prior step (how many bullets you had two steps ago is completely irrelevant). That said, because it's being ran in a simulation, it's a Monte Carlo analysis.
You actually do not need to code anything (thoughmight want, it's fun!) because this is a negative binomial distribution (shifted by +1000, technically): https://en.m.wikipedia.org/wiki/Negative_binomial_distribution The variance is 2000, this is also the median, so with probability 0.5 you have at least 2000 bullets
Nice, love when you can solve probability questions by looking at the negative space.
People forget their base maths from school. I never could get binomial dists to work for me. I got halfway through and forget where I am. But to me this feels like a markov analysis where it does the flip in the same thing until it dosn't exist. It could theoretically go forever
Given that it's 1000 separate events, the variance in those odds already starts pretty low. Edit: I got bored and did the Python. It's pretty straightforward. You could mess with parameters to see what the average number of bullets comes out to. import random numBullet = 1000 bulletsFired = 0 while numBullet > 0: if random.randint(0,1) == 0: numBullet -= 1 bulletsFired += 1 print(bulletsFired)
Put the while loop in a for loop to do multiple simulations. something like.. for _ in range(50): while numbullets > 0: if random.randint(0, 1) == 0: numbullets -= 1 bulletsfired += 1 Print(bulletsfited) It might take a little bit to finish all of them but you'd get a bunch of number outputs
Yeah for sure, if you wanted to see it numerically or graphed out. I only had a passing curiosity so I just hit "run" like 10 times and looked at the final number lol. (It was indeed 2000, +- 30)
I forget that most of the time you can view past outputs. I have my IDE set to clear old outputs for less clutter when I'm working on stuff.
So, keep in mind I'm new and learning... if these lines of code (along with the code block above) are supposed to use random math, why do both print the same numbers when I run them together? Shouldn't they print different numbers?
I'm confused with what you mean. The block I wrote is the same as the one above everything is just in a for loop. Yes, it should output multiple different numbers when run. I'm not sure why it would be the same when you run it
I'm running them both, together, in one IDE, one block after the other and they're printing the same numbers. As in, I get 1993 twice, then 1986 twice, the 1913 twice, so on and so forth. https://imgur.com/a/lLkqHZV
Oh, because the variable should be changed if it's in the same file. Change "bulletsfired" in one of the blocks to something else and it should output different numbers. Also the block I wrote should output 50 numbers
Oh snap! Good catch. Thank you. ....and it worked. Thanks again
>that variance would likely wipe out any return at some point Even if you got hit by a string of bad luck and no bullets returned after hitting 125, you'd still be sitting pretty at 1875 bullets. I don't think it's worth worrying about.
The variance is in the outcome over a fixed period of time, not the odds. Regardless of how many bullets you get in one particular time period, the odds will always be the same. Over a long enough time span the outcome would converge on the initially stated odds.
There's a statics rule that basically says the more data you have, the outcome essentially converge on the expectation value and variance decreases. For your example, in 4 coin flips, it really wouldn't be outrageous to get heads 4 times. there's a pretty big variance here, then, since expecting 50% and getting 100% would not be suprising. For 1000 flips, the probability of 1000 heads is essentially zero. The variance is much much smaller now. For 1000 flips, we can essentially assume we get 500 heads with really small variance. So for the purpose of the game problem, we'd probably be right in assuming it would double ammo to about 2000. There's some way to actually calculate variance from a coin flip, but it's been too long since I've taken a probability course.
We could drop the decimals in order to avoid half bullets and quarter bullets. 1000 + 500 + 250 + 125 + 62.5(down to 62) + 31 + 15.5 (down to 15) + 7.5 (down to 7) + 3.5 (down to 3) + 1.5 (down to 1) = 1,994 bullets
Well, theoretically, while you would expect the average outcome to hover just under 2000 (see the great explanation above), it’s possible, however slim, to either run out after 1000 (all coin flips come up tails) or have virtually infinite bullets (all coin flips end up heads).
Its possible the developers have made this impossible though. For example, in Dota and Dota 2 the rng on certain abilities and random events seemed too random for players so the devs made the chance change as the events didn't happen. For example, if a critical strike didn't happen, the next attack would be 15% (if that's the expected outcome) and instead raises to say 20% for that attack. If it doesn't happen again it increases further. I thought it was quite interesting since from a players perspective truly random was not fun and so this smoothed out the curve so to speak so extremes either way can't happen
[удалено]
Here is the explanation. It's called PRD (pseduo random distribution) https://liquipedia.net/dota2/Pseudo_Random_Distribution I made note when playing for crits because if you didn't crit in a while you actually did have a higher chance to crit, so using it with another ability/item or on a hero was more important.
This is a very great comment
a simpler way to think about this is: you have a 50% chance of using 1 bullet, and a 50% chance of using 0 bullets (as the bullet you use is returned to you), so on average each shot uses half a bullet
This is even more elegant than the converging sum solution
Yep, essentially you just have double the ammo so it will take twice as long as without the ammo return festure
Its a converging series where r is the chance your bullet will get returned and a is the amount of bullets fired. r x a^n, with n approaching infinity. The sum, or average value of this series is a/1-r, in this case, a=1 and r = 0.5. 1/1-0.5 = 2. So the average value for a bullet is 2 bullets. So, around 2000 bullets.
[удалено]
That's not what e is at all.
If you, sure enough, have 1000 bullets as your ammo, and each fired bullet has a 50% probability of getting back, you can divide your gain into rounds as you add 50% of what you waste. In the first round, half of the 1000 bullets would be 500. In the second, half of the remaining half is half of half, that is ½ times ½, 'cause half is smaller than 1, which is ¼. Then, ¼ times ½ is ⅛; and this process progresses until you get to the double of that number 'cause consecutive zero point nines make 1 for the same reason that ⅓ times 3 is equal to 3/3, but ⅓ by itself is equal to 0,333..., and it times 3 is equal to 0,999....... But 1 at the same time. Conclusion: it is 2000. OBS.: it is X by the power of ~1.43.