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Ok, I talked to the engineer. I got it mostly right but everyone shoots to the right as soon as the lights go out. The question is "what's the probability you

Ok, I talked to the engineer. I got it mostly right but everyone shoots to the right as soon as the lights go out. The question is "what's the probability you die?". Edit: You can also challenge people to think about the problem where everyone fires at exactly the same time OR random order since people have different reaction times. Edit 2: "The probability p(n) that a permutation of n things is a derangement -- which tends to 1/e as n->oo -- satisfies the recurrence relation p(n) = [(n-1)p(n-1)+p(n-2)]/n; is it possible that the correct statement of the problem here leads to that same recurrence?" <--- My engineer says that derangements are the correct key to the convergence.

gjm11 7y ago
I think I still don't understand what's happening in this scenario. If everyone shoots simultaneously (so in particular everyone does get the chance to shoot) then I die iff the one person shooting at me hits me. Probability equals probability that a given shot hits (so in this case 1/6). No dependence at all on the number of people. If everyone shoots sequentially, this seems just like what I described above. Probability of death is now p/(1+p) instead of p, at least if you're first to shoot and n is very large. (Unless something's very broken in the heuristic argument I gave. Let's try another. First approximation says a fraction p of people die. But that's not quite right because people who die don't get to shoot, so next approximation says we get p(1-p). Next approximation says we get p(1-p(1-p)). Etc. We can either solve the obvious equation, or else notice that we're getting more and more terms of the binomial expansion of p/(1+p). I don't see anything here that doesn't look, in a crude approximation, like a fraction p of people dying (p, again, is probability that a given shot hits, which in this case is 1/6). I must be misunderstanding something in the problem statement here. Perhaps it would be clearer if I'd seen the movie? Oh, what about this version? You shoot first, things proceed cyclically, and we keep going until just one person is left. What's the chance that it's you? Naively it seems like this should be approximately 1/n no matter what p is; shooting first could confer some advantage but surely it can't be much for large n. So this can't yield anything like 1/e either. Drat.
- sinsterizme 7y ago
  Yeah I don’t understand the question either :( I’m very interested to here the full problem and solution
  - gjm11 7y ago
    Personally, I'd like to see a careful statement of the full problem but not the solution :-).
    - josh2600 7y ago
      From the engineer: "The current statement of the problem is that you have n participants, with 6 slots (1 loaded) in their revolver, each firing to the person on the right."
- boyobo 7y ago
  How about this: People shoot in a random order (now everyone has the same chance of making it to the next round. Is it 1/e?) import numpy as np class Shooter: def __init__(self): self.dead = False self.right = None # The person to the right def simulate(n): # Simulates a round with n shooters # Returns the ratio of survivors shooters = [Shooter() for i in range(n)] for i, shooter in enumerate(shooters): shooter.right = shooters[(i+1) % len(shooters)] np.random.shuffle(shooters) HIT_PROBABILITY = 1/6 survivors = n for shooter in shooters: if not shooter.dead: if np.random.random() < HIT_PROBABILITY: shooter.right.dead = True survivors = survivors - 1 return survivors/n Simulations suggest that the survival probabilities do not converge to 6/7 if you add shuffling. This makes sense, since the survival probability for the random shuffling version must be strictly less than if you are guaranteed to go first. >>>sum([simulate(10000) for j in range(1000)])/1000 0.8463075999999998

gjm11 7y ago

I think I still don't understand what's happening in this scenario. If everyone shoots simultaneously (so in particular everyone does get the chance to shoot) then I die iff the one person shooting at me hits me. Probability equals probability that a given shot hits (so in this case 1/6). No dependence at all on the number of people. If everyone shoots sequentially, this seems just like what I described above. Probability of death is now p/(1+p) instead of p, at least if you're first to shoot and n is very large. (Unless something's very broken in the heuristic argument I gave. Let's try another. First approximation says a fraction p of people die. But that's not quite right because people who die don't get to shoot, so next approximation says we get p(1-p). Next approximation says we get p(1-p(1-p)). Etc. We can either solve the obvious equation, or else notice that we're getting more and more terms of the binomial expansion of p/(1+p). I don't see anything here that doesn't look, in a crude approximation, like a fraction p of people dying (p, again, is probability that a given shot hits, which in this case is 1/6). I must be misunderstanding something in the problem statement here. Perhaps it would be clearer if I'd seen the movie? Oh, what about this version? You shoot first, things proceed cyclically, and we keep going until just one person is left. What's the chance that it's you? Naively it seems like this should be approximately 1/n no matter what p is; shooting first could confer some advantage but surely it can't be much for large n. So this can't yield anything like 1/e either. Drat.

sinsterizme 7y ago
Yeah I don’t understand the question either :( I’m very interested to here the full problem and solution
- gjm11 7y ago
  Personally, I'd like to see a careful statement of the full problem but not the solution :-).
  - josh2600 7y ago
    From the engineer: "The current statement of the problem is that you have n participants, with 6 slots (1 loaded) in their revolver, each firing to the person on the right."
boyobo 7y ago
How about this: People shoot in a random order (now everyone has the same chance of making it to the next round. Is it 1/e?) import numpy as np class Shooter: def __init__(self): self.dead = False self.right = None # The person to the right def simulate(n): # Simulates a round with n shooters # Returns the ratio of survivors shooters = [Shooter() for i in range(n)] for i, shooter in enumerate(shooters): shooter.right = shooters[(i+1) % len(shooters)] np.random.shuffle(shooters) HIT_PROBABILITY = 1/6 survivors = n for shooter in shooters: if not shooter.dead: if np.random.random() < HIT_PROBABILITY: shooter.right.dead = True survivors = survivors - 1 return survivors/n Simulations suggest that the survival probabilities do not converge to 6/7 if you add shuffling. This makes sense, since the survival probability for the random shuffling version must be strictly less than if you are guaranteed to go first. >>>sum([simulate(10000) for j in range(1000)])/1000 0.8463075999999998