Clever calculation, but I think there's an erroneous assumption there:
The chance that he dies before he pulls the trigger is half the probability that he dies at all.
That's not correct. Him pulling the trigger (early or late/not) and him dying are not independent events. According to me, the ratio between the probability of him dying before he pulls the trigger and the probability that he dies at all is slightly greater than 1/2.
Let's approach it a different way, by listing the different possible events, their outcomes, and their probabilities. Since you said that the outcome is independent of the number of players, I'll work out the case for three players.
=============================
General setup:
Suppose N players (p1, p2, ..., pN) standing in a circle; p1 points gun at p2, p2 points gun at p3, etc. and pN points gun at p1.
Define:
- BV is the "bullet vector", e.g. BV = [0 0 1] means the aligned chamber of the guns of p1 and p2 carry no bullet, the aligned chamber of p3's gun carries a bullet.
- SV is the "shoot vector", e.g. SV = [
a b c] means p1 is the
a'th person to shoot, p2 is the
b'th person to shoot, p3 is the
c'th person to shoot. SV is typically a permutation of the numbers 1 to N. (Ideally
a, b, c should be clock times at which each player pulls or would pull the trigger, but because of the following assumption it's only the order that matters, not the timescale.) We are assuming that bullets have zero airtime, i.e. the moment a person pulls the trigger is the moment the person before him dies (provided there was a bullet in the aligned chamber).
- P is the probability of the event in question.
- RV is the "result vector", i.e. the outcome (Live or Die for each player) of the game round. E.g. RV = [L D L] means p1 and p3 Live, and p2 Dies.
For N players, there are 2^N different BVs, and (N!) different SVs. The probability of a BV occurring is
P(BV(i)) = g^B * (1-g)^(N-B)
where g is the load ratio of the gun (number of bullets divided by number of chambers in a gun) and B is the number of players with a bullet in the aligned chamber, for that particular BV(i).
P(SV(i)) = 1/N! for any SV(i), i.e. all SVs have equal odds.
P = P(BV)*P(SV)
Now generate list of events for three players, and one bullet in each gun:
--------------------------------------------------
N = 3 , g = 1/6 :
(1): BV = [000], SV = {[123], [132], [213], [231], [312], [321]} => P = 125/216, RV = [L L L]
(2): BV = [001], SV = {[123], [132], [213], [231], [312], [321]} => P = 25/216, RV = [D L L]
(3): BV = [010], SV = {
[123], [132],
[213], [231],
[312], [321]} => P = 25/216, RV = [L L D]
(4): BV = [100], SV = {[123], [132], [213], [231], [312], [321]} => P = 25/216, RV = [L D L]
(5.1): BV = [011], SV =
[123] => P = 5/1296, RV = [L L D]
(5.2): BV = [011], SV = [132] => P = 5/1296, RV = [D L D]
(5.3): BV = [011], SV =
[213] => P = 5/1296, RV = [L L D]
(5.4): BV = [011], SV = [231] => P = 5/1296, RV = [D L D]
(5.5): BV = [011], SV =
[312] => P = 5/1296, RV = [L L D]
(5.6): BV = [011], SV = [321] => P = 5/1296, RV = [D L D]
(6.1): BV = [101], SV = [123] => P = 5/1296, RV = [D D L]
(6.2): BV = [101], SV = [132] => P = 5/1296, RV = [D D L]
(6.3): BV = [101], SV = [213] => P = 5/1296, RV = [D D L]
(6.4): BV = [101], SV = [231] => P = 5/1296, RV = [D L L]
(6.5): BV = [101], SV = [312] => P = 5/1296, RV = [D L L]
(6.6): BV = [101], SV = [321] => P = 5/1296, RV = [D L L]
(7.1): BV = [110], SV = [123] => P = 5/1296, RV = [L D L]
(7.2): BV = [110], SV = [132] => P = 5/1296, RV = [L D L]
(7.3): BV = [110], SV =
[213] => P = 5/1296, RV = [L D D]
(7.4): BV = [110], SV = [231] => P = 5/1296, RV = [L D L]
(7.5): BV = [110], SV =
[312] => P = 5/1296, RV = [L D D]
(7.6): BV = [110], SV = [321] => P = 5/1296, RV = [L D D]
(8.1): BV = [111], SV = [123] => P = 1/1296, RV = [D D L]
(8.2): BV = [111], SV = [132] => P = 1/1296, RV = [D D L]
(8.3): BV = [111], SV =
[213] => P = 1/1296, RV = [L D D]
(8.4): BV = [111], SV = [231] => P = 1/1296, RV = [D L D]
(8.5): BV = [111], SV =
[312] => P = 1/1296, RV = [L D D]
(8.6): BV = [111], SV = [321] => P = 1/1296, RV = [D L D]
- In the first case (1) everybody lives regardless of shooting order, since there is no bullet in the chambers.
- In the cases (2)(3)(4), the shooting order doesn't matter, since there's only one player with a bullet in the chamber; as soon as he fires, the person standing before him dies while all the others will live.
- In cases (5.x) (where BV = [011]), p2 will always live (because p1 has no bullet) and hence p3 will always die. p1 lives when p2 shoots before p3, and dies when p3 shoots before p2.
- A similar result for BV = [101] (6.x) and BV = [110] (7.x).
- In case (8.x) (where BV = [111]) the person who is the first to shoot will always die, as well as the person he shoots. The person who stands behind the person who shoots first will always survive.
- We can now calculate the probability of p3 dying, by summing the probabilities of events where p3 dies. (Simply look at the events where the third letter in RV is a "D"; highlighted in blue above.)
For p3, the probability of dying is 25/216 + 6*5/1296 + 3*5/1296 + 4*1/1296 = (150+30+15+4)/1296 = 199/1296
Likewise, the probability of living is 125/216 + 2*25/216 + 6*5/1296 + 3*5/1296 + 2*1/1296 = (750+300+30+15+2)/1296 = 1097/1296
Not surprisingly, the same calculation and the same results go for p2 and p1.
So there we have it. The chance of dying = 199/1296 is close to 2/13 (since 2/13 = 200/1300 = 199/1293.5), but not exact the same. The chances at survival are hence even a bit better than you thought.
- We can now also calculate the ratio between the probability that a player dies before he pulls the trigger and the probability that same player dies at all. E.g. for player p3, this is done by summing up the probabilities of the events where SV is in
bold blue (where p2 shoots before p3) and dividing it by the sum of the probabilities of all events in blue (SV in
bold + non-bold blue).
We find the ratio is actually a bit higher than 0.5; in the case of three players and a gun load ratio of 1/6, it's 102/199 (= approx. 0.5126).
If there are any errors in my approach, feel free to reply; I'm open to any corrections.
~
Everyone is unique, except for me ~
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kElevrA7
10 years ago
Nice thread. Such a rare find on imdb these days.
The two of you killed everything I ever loved. **** you both.
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