Mixing
Expected value of Russian roulette at a firing range played on end with finite total shots
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Rather than shooting yourself in the head, you are playing Russian roulette and shooting at a target. You put a bullet in the chamber and spin it and pull the trigger until the bullet is fired.
Suppose you pull the trigger a total of N times, reloading and spinning the chamber each time the bullet is shot. What is the expected number of bullets that end up being shot from the gun?
Slightly reworded: in a round of 100 total trigger pulls with a chamber size of 10, the odds of shooting 100 bullets (landing on the bullet every single time) is orders of magnitude less likely than shooting 10 bullets (the bullet being at the "end" of the chamber every single time), because the likelihood in a vacuum of the bullet landing anywhere in the chamber is equal, but stacking 10 events on end is a lot easier than stacking 100. How can we calculate a probability distribution of shooting n total bullets in a total of N trigger pulls?
My intuition tells me that the most likely result is shooting n=N/(c/2) where c is the size of the chamber, but I'm unsure how to get there or whether that's even correct.
probability probability-distributions
asked Jan 25 at 18:48
Gerrit PostemaGerrit Postema
1
$endgroup$
add a comment |
$begingroup$
Rather than shooting yourself in the head, you are playing Russian roulette and shooting at a target. You put a bullet in the chamber and spin it and pull the trigger until the bullet is fired.
Suppose you pull the trigger a total of N times, reloading and spinning the chamber each time the bullet is shot. What is the expected number of bullets that end up being shot from the gun?
Slightly reworded: in a round of 100 total trigger pulls with a chamber size of 10, the odds of shooting 100 bullets (landing on the bullet every single time) is orders of magnitude less likely than shooting 10 bullets (the bullet being at the "end" of the chamber every single time), because the likelihood in a vacuum of the bullet landing anywhere in the chamber is equal, but stacking 10 events on end is a lot easier than stacking 100. How can we calculate a probability distribution of shooting n total bullets in a total of N trigger pulls?
My intuition tells me that the most likely result is shooting n=N/(c/2) where c is the size of the chamber, but I'm unsure how to get there or whether that's even correct.
probability probability-distributions
asked Jan 25 at 18:48
Gerrit PostemaGerrit Postema
1
$endgroup$
$begingroup$
It depends on how sophisticated you want the model to be. The most basic one is the Binomial distribution with parameter $N$ (total trigger pulls) and $p = 1/n$ where $n$ is the chamber size.
$endgroup$
– Lee David Chung Lin
Jan 25 at 23:10
add a comment |
$begingroup$
Rather than shooting yourself in the head, you are playing Russian roulette and shooting at a target. You put a bullet in the chamber and spin it and pull the trigger until the bullet is fired.
Suppose you pull the trigger a total of N times, reloading and spinning the chamber each time the bullet is shot. What is the expected number of bullets that end up being shot from the gun?
Slightly reworded: in a round of 100 total trigger pulls with a chamber size of 10, the odds of shooting 100 bullets (landing on the bullet every single time) is orders of magnitude less likely than shooting 10 bullets (the bullet being at the "end" of the chamber every single time), because the likelihood in a vacuum of the bullet landing anywhere in the chamber is equal, but stacking 10 events on end is a lot easier than stacking 100. How can we calculate a probability distribution of shooting n total bullets in a total of N trigger pulls?
My intuition tells me that the most likely result is shooting n=N/(c/2) where c is the size of the chamber, but I'm unsure how to get there or whether that's even correct.
probability probability-distributions
asked Jan 25 at 18:48
Gerrit PostemaGerrit Postema
1
$endgroup$
Rather than shooting yourself in the head, you are playing Russian roulette and shooting at a target. You put a bullet in the chamber and spin it and pull the trigger until the bullet is fired.
Suppose you pull the trigger a total of N times, reloading and spinning the chamber each time the bullet is shot. What is the expected number of bullets that end up being shot from the gun?
Slightly reworded: in a round of 100 total trigger pulls with a chamber size of 10, the odds of shooting 100 bullets (landing on the bullet every single time) is orders of magnitude less likely than shooting 10 bullets (the bullet being at the "end" of the chamber every single time), because the likelihood in a vacuum of the bullet landing anywhere in the chamber is equal, but stacking 10 events on end is a lot easier than stacking 100. How can we calculate a probability distribution of shooting n total bullets in a total of N trigger pulls?
My intuition tells me that the most likely result is shooting n=N/(c/2) where c is the size of the chamber, but I'm unsure how to get there or whether that's even correct.
probability probability-distributions
probability probability-distributions
asked Jan 25 at 18:48
Gerrit PostemaGerrit Postema
1
asked Jan 25 at 18:48
Gerrit PostemaGerrit Postema
1
asked Jan 25 at 18:48
Gerrit PostemaGerrit Postema
1
asked Jan 25 at 18:48
Gerrit PostemaGerrit Postema
1
asked Jan 25 at 18:48
Gerrit PostemaGerrit Postema
1
1
$begingroup$
It depends on how sophisticated you want the model to be. The most basic one is the Binomial distribution with parameter $N$ (total trigger pulls) and $p = 1/n$ where $n$ is the chamber size.
$endgroup$
– Lee David Chung Lin
Jan 25 at 23:10
add a comment |
$begingroup$
It depends on how sophisticated you want the model to be. The most basic one is the Binomial distribution with parameter $N$ (total trigger pulls) and $p = 1/n$ where $n$ is the chamber size.
$endgroup$
– Lee David Chung Lin
Jan 25 at 23:10
$begingroup$
It depends on how sophisticated you want the model to be. The most basic one is the Binomial distribution with parameter $N$ (total trigger pulls) and $p = 1/n$ where $n$ is the chamber size.
$endgroup$
– Lee David Chung Lin
Jan 25 at 23:10
$begingroup$
It depends on how sophisticated you want the model to be. The most basic one is the Binomial distribution with parameter $N$ (total trigger pulls) and $p = 1/n$ where $n$ is the chamber size.
$endgroup$
– Lee David Chung Lin
Jan 25 at 23:10
add a comment |
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It depends on how sophisticated you want the model to be. The most basic one is the Binomial distribution with parameter $N$ (total trigger pulls) and $p = 1/n$ where $n$ is the chamber size.
$endgroup$
– Lee David Chung Lin
Jan 25 at 23:10