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Math Help - Three Revolvers and One Bullet

  1. #1
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    Three Revolvers and One Bullet

    Hi everyone,

    I'm new here. My background is in both pure and applied mathematics with a strong emphasis on differential equations, algebra, group theory and number theory however I never studied too much probability theory. Me and some friends (mathematicians and engineers) discussed a problem out of a video game we had all played and for the life of us could not come to a satisfying conclusion. Imagine the following problem.

    A man has three revolvers. In case you don't know what I'm on about let me explain. A revolver in the context of this question is a gun which holds six bullets. As one chamber is fired it moves along and there are now 5 more chambers to go. I suppose most of us already knew that but anyway, the important thing is that we've got 6 chambers and that there's an order to them.

    Now, the man with the three revolvers picks one of them randomly and places one bullet into one of its chambers randomly. To ensure there's no bias he then spins the cylinder around before locking it in. It is now unknown exactly which chamber of the cylinder the bullet is in.

    Now, he begins juggling the three guns. Whilst juggling them he randomly selects one from the air, points and pulls the trigger at an unfortunate person. In total, he does this six times.

    What is the probability of the man being shot with the single bullet?

    The problem seemed simple at first but we soon realised it wasn't so... at least we think. Sure there's a 1/3 chance of selecting the loaded gun and the loaded gun has a 1/6 chance of having the bullet in the chamber set to fire... but if it doesn't then the next time we fire it we'll have a 1/5 chance, then a 1/4 and so on. However, that increasing probability is affected by an external random choice.

    Can this problem simply be solved by using expectation? That just doesn't seem very satisfying somehow.

    Me and my friends would love to see any nice neat way to figure this out without resorting to a computer simulation (I don't want to have to remind myself how to write stuff like this in Mathematica).

    Any takers?

    The game was Metal Gear Solid 3: Snake Eater.
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    Quote Originally Posted by RevolverOcelot View Post
    Hi everyone,

    I'm new here. My background is in both pure and applied mathematics with a strong emphasis on differential equations, algebra, group theory and number theory however I never studied too much probability theory. Me and some friends (mathematicians and engineers) discussed a problem out of a video game we had all played and for the life of us could not come to a satisfying conclusion. Imagine the following problem.

    A man has three revolvers. In case you don't know what I'm on about let me explain. A revolver in the context of this question is a gun which holds six bullets. As one chamber is fired it moves along and there are now 5 more chambers to go. I suppose most of us already knew that but anyway, the important thing is that we've got 6 chambers and that there's an order to them.

    Now, the man with the three revolvers picks one of them randomly and places one bullet into one of its chambers randomly. To ensure there's no bias he then spins the cylinder around before locking it in. It is now unknown exactly which chamber of the cylinder the bullet is in.

    Now, he begins juggling the three guns. Whilst juggling them he randomly selects one from the air, points and pulls the trigger at an unfortunate person. In total, he does this six times.

    What is the probability of the man being shot with the single bullet?

    The problem seemed simple at first but we soon realised it wasn't so... at least we think. Sure there's a 1/3 chance of selecting the loaded gun and the loaded gun has a 1/6 chance of having the bullet in the chamber set to fire... but if it doesn't then the next time we fire it we'll have a 1/5 chance, then a 1/4 and so on. However, that increasing probability is affected by an external random choice.

    Can this problem simply be solved by using expectation? That just doesn't seem very satisfying somehow.

    Me and my friends would love to see any nice neat way to figure this out without resorting to a computer simulation (I don't want to have to remind myself how to write stuff like this in Mathematica).

    Any takers?

    The game was Metal Gear Solid 3: Snake Eater.
    Computer simulation is not necessary by any means. If there are too many steps to work out on paper in reasonable time, you could write a computer program that is *not* a simulation to output the answer (for an exact answer, use fractions; if you're comfortable with less precision, you can use floats). By the way in case there's any confusion, simulation means using random number generators, etc.

    EDIT: I originally misread the problem. The below is written thinking that each gun has one bullet in it. Sorry! I'll probably make another post dealing with the actual problem.

    But the first three steps aren't too bad on paper.

    The probability of a bullet on the first shot is clearly 1/6.

    So far we have P(guy gets shot) = 1/6 + something.

    The complement is 5/6, which we'll use in just a bit.

    After this we have 2/3 probability that the probability will be 1/6, and a 1/3 probability that the probability will be 1/5.

    So now we have P(guy gets shot) = (1/6) + (5/6)[ (2/3)(1/6) + (1/3)(1/5) ] + something.

    Let x = (2/3)(1/6) + (1/3)(1/5) for ease.

    It's getting a bit hairy now. Given that no bullet has been fired yet, there is a 2/3 probability that we have 1/6, 1/5, 1/5 as our next probabilities, and a 1/3 probability that we have 1/4, 1/6, 1/6.

    Anyway what I come up with is

    P(guy gets shot) = (1/6) + (5/6)(x) + (5/6)(1-x)[ (2/3) ((1/3)(1/6) + (2/3)(1/5)) + (1/3) ((1/3)(1/4) + (2/3)(1/6)) ] + something


    As you can see it just gets more and more cumbersome, but it's doable.

    Edit 2: While writing one of the below posts, I realized that the methodology for getting the third shot above is flawed, although the underlying reasoning is sound. So I grayed it out.
    Last edited by undefined; May 27th 2010 at 10:21 AM.
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    Okay, so now I'll pretend that I have the ability to read.

    Well my above approach would still work, but I'm beginning to think this is a lot simpler than all of that.

    Juggling the guns means that essentially we are just pulling bullets randomly out of a bin, without replacement. Since we only shoot six times, it's impossible to start re-using empty barrels, which works out very well for us. So I think the probability the guy gets shot is simply

    (1/18) + (17/18)(1/17) + (17/18)(16/17)(1/16) + ... (six terms total)

    = 6/18 = 1/3.

    I'm still thinking it over, but I think this makes perfect sense.

    Imagine if we didn't have the 6-barrel aspect and we were just drawing balls out of a bin, without replacement. There are 17 blue balls and 1 red ball. So imagine taking the balls all out, and placing them in a line from left to right. The red ball is equally likely to be in any spot as any other. So the probability of drawing it first is 1/18, of drawing it first or second is 2/18, etc.
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    Yes, this is pretty much where I got to. There's a problem though. Assuming that it's like pulling one red ball out of a bin of 18 where only one is red is not an accurate analogy in my opinion.

    It's more like having three bins each with 6 balls and in one of them there is one red ball. So initially the very first pick we make (first shot fired) is a 1/18 chance of getting the red ball (single bullet) but then the second choice gets tricky.

    We're left with a situation of if...

    (a) The gun with the bullet was not chosen then we still have a 1/18 chance.

    OR

    (b) The gun with the bullet was chosen but the bullet was not fired in which case the next time that gun is selected at random the chance of the bullet firing will now rise to 1/5.

    To go back to the ball analogy, I don't think it's as simple as placing them all in one long line of 18 balls since if they act like the guns then when we choose a non-red ball from the red-ball-carrying bin then we would not replace it but if we choose a non-red ball from either of the other two bins we would replace it.

    That's what's getting me. The fact that we replace some choices when they come up but not all of them.
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    Quote Originally Posted by RevolverOcelot View Post
    Yes, this is pretty much where I got to. There's a problem though. Assuming that it's like pulling one red ball out of a bin of 18 where only one is red is not an accurate analogy in my opinion.

    It's more like having three bins each with 6 balls and in one of them there is one red ball. So initially the very first pick we make (first shot fired) is a 1/18 chance of getting the red ball (single bullet) but then the second choice gets tricky.

    We're left with a situation of if...

    (a) The gun with the bullet was not chosen then we still have a 1/18 chance.

    OR

    (b) The gun with the bullet was chosen but the bullet was not fired in which case the next time that gun is selected at random the chance of the bullet firing will now rise to 1/5.

    To go back to the ball analogy, I don't think it's as simple as placing them all in one long line of 18 balls since if they act like the guns then when we choose a non-red ball from the red-ball-carrying bin then we would not replace it but if we choose a non-red ball from either of the other two bins we would replace it.

    That's what's getting me. The fact that we replace some choices when they come up but not all of them.
    All right, so I'm claiming that pulling balls out of a single bin is an equivalent problem, and you're claiming it's not. (Or maybe you're claiming that it may or may not be an equivalent problem, and you require a proof.)

    Besides what I wrote in the previous post, I'm not sure there's much I can do to convince you other than proving the probability comes out the same in each case.

    As you noticed, things get complicated just for the first two shots.

    You'll agree that the probability that the guy gets shot on the first shot fired is 1/18, right? This is equal to (2/3)(0) + (1/3)(1/6).

    So that I don't have to keep writing "guy gets shot," let's define X_n as the event that the guy gets shot on the nth shot fired, and let Y_n be the event that the guy gets shot when n shots are fired. So the original problem is to find P(Y_6).

    I'll be using this:

    P(Y_6) = P(X_1) + (P(\overline{X_1}))(P(X_2|\overline{X_1})) + (P(\overline{X_1}))(P(\overline{X_2}))(P(X_3|\over  line{X_1} \cap \overline{X_2})) + \cdots (6 terms total).

    So P(X_1) = \frac{1}{18}.

    Now see the tree I attached. The first level of branching is which revolver is chosen; the second level is whether or not the guy gets shot. Given that \overline{X_1}, there is a \frac{\frac{5}{18}}{\frac{5}{18}+\frac{2}{3}}=\fra  c{5}{17} probability of having 1/5, 0, 0; and there is a \frac{\frac{2}{3}}{\frac{5}{18}+\frac{2}{3}}=\frac  {12}{17} probability of having 1/6, 0, 0.

    So P(X_2|\overline{X_1}) equals (5/17)(1/3)(1/5) + (12/17)(1/3)(1/6) = 1/17.

    So P(Y_2) = \frac{1}{18} + \frac{17}{18}\cdot\frac{1}{17} = \frac{2}{18}.

    I could continue... but it gets really cumbersome. Maybe there's an easier way to write it all out. If we wanted to be really clear, we could define Z_n as the event that the gun with the bullet gets chosen on the nth shot, but hopefully you were able to read what I wrote above without this.

    I could also write a simulation, although I'm not sure you'd be convinced by that either.

    Edits:

    Actually perhaps the easiest way to show that the numbers come out the same is to take a huuuge piece of paper and draw a tree that is 12 levels deep. Whenever a probability is 0 or the guy gets shot, we can terminate our branching, so it will be a rooted binary tree that is full but not perfect (terms from graph theory, for anyone who's not familiar, see here). Or, if we wanted to simply not draw the branches with 0 probability, it will no longer be full either... but this might make it harder to read, I guess it's up to preference.

    The tree would get very large quickly, and at each depth n=2k, there would be 2^k nodes that will have branches. So drawing this tree could take quite a while.

    Anyway, if you want a computer simulation, this one would be super easy, just ask and I'll write you one.

    I could also write a program to calculate the probability using the tree. This approach would have the advantage of being able to know P(Y_7) for which the single bin problem is (according to my claim) no longer equivalent.
    Attached Thumbnails Attached Thumbnails Three Revolvers and One Bullet-revolvertree1.png  
    Last edited by undefined; May 27th 2010 at 11:03 AM.
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    Quote Originally Posted by RevolverOcelot View Post
    Hi everyone,

    I'm new here. My background is in both pure and applied mathematics with a strong emphasis on differential equations, algebra, group theory and number theory however I never studied too much probability theory. Me and some friends (mathematicians and engineers) discussed a problem out of a video game we had all played and for the life of us could not come to a satisfying conclusion. Imagine the following problem.

    A man has three revolvers. In case you don't know what I'm on about let me explain. A revolver in the context of this question is a gun which holds six bullets. As one chamber is fired it moves along and there are now 5 more chambers to go. I suppose most of us already knew that but anyway, the important thing is that we've got 6 chambers and that there's an order to them.

    Now, the man with the three revolvers picks one of them randomly and places one bullet into one of its chambers randomly. To ensure there's no bias he then spins the cylinder around before locking it in. It is now unknown exactly which chamber of the cylinder the bullet is in.

    Now, he begins juggling the three guns. Whilst juggling them he randomly selects one from the air, points and pulls the trigger at an unfortunate person. In total, he does this six times.

    What is the probability of the man being shot with the single bullet?

    The problem seemed simple at first but we soon realised it wasn't so... at least we think. Sure there's a 1/3 chance of selecting the loaded gun and the loaded gun has a 1/6 chance of having the bullet in the chamber set to fire... but if it doesn't then the next time we fire it we'll have a 1/5 chance, then a 1/4 and so on. However, that increasing probability is affected by an external random choice.

    Can this problem simply be solved by using expectation? That just doesn't seem very satisfying somehow.

    Me and my friends would love to see any nice neat way to figure this out without resorting to a computer simulation (I don't want to have to remind myself how to write stuff like this in Mathematica).

    Any takers?

    The game was Metal Gear Solid 3: Snake Eater.
    Here's a way to look at this....

    The target could be hit on the 1st, 2nd, 3rd, 4th, 5th or 6th shot
    or all 6 shots could miss.

    To calculate the probability of a hit, we can sum the probabilities of
    1. a hit on the 1st shot
    2. a hit on the 2nd shot predeced by an empty chamber on the 1st
    3. a hit on the 3rd shot preceded by 2 empty chambers
    4. a hit on the 4th shot preceded by 3 empty chambers
    5. a hit on the 5th shot preceded by 4 empty chambers
    6. a hit on the 6th shot preceded by 5 empty chambers.

    Assuming the gunman is an expert marksman, his probability of being off-target is negligible.

    Hence, if we denote a gun without a bullet as W (for wrong gun)
    and the gun with the bullet as R (for right gun),
    then we can write the sequences using the fact that to hit the target, the gunman must have the right gun on the final shot.

    Hence the options are...

    R................... one shot probability

    WR
    RR..................two shots probability

    WWR
    WRR
    RWR
    RRR................three shots probability

    WWWR
    WWRR
    WRWR
    WRRR
    RWWR
    RWRR
    RRWR
    RRRR..............four shots probability

    continue for five and six shots..

    Since the gunman must have the right gun to fire the last shot successfully,
    then there are

    2^0,\ 2^1,\ 2^2,\ 2^3,\ 2^4\ or\ 2^5

    possible sequences for 1, 2, 3, 4, 5, 6 shots in that order.

    To calculate the probabilities

    1 shot

    \frac{1}{3}\ \frac{1}{6} right gun, 1 in 6

    2 shots

    \left(\frac{2}{3}(1)\right)\left(\frac{1}{3}\ \frac{1}{6}\right) WR

    \left(\frac{1}{3}\ \frac{5}{6}\right)\left(\frac{1}{3}\ \frac{1}{5}\right) RR

    3 shots

    \left(\frac{2}{3}(1)\right)\left(\frac{2}{3}(1)\ri  ght)\left(\frac{1}{3}\ \frac{1}{6}\right) WWR

    \left(\frac{2}{3}(1)\right)\left( \frac{1}{3}\ \frac{5}{6}\right)\left(\frac{1}{3}\ \frac{1}{5}\right) WRR

    \left(\frac{1}{3}\ \frac{5}{6}\right)\left( \frac{2}{3}(1)\right)\left(\frac{1}{3}\ \frac{1}{5}\right) RWR

    \left(\frac{1}{3}\ \frac{5}{6}\right)\left( \frac{1}{3}\ \frac{4}{5}\right)\left( \frac{1}{3}\ \frac{1}{4}\right) RRR

    continue to sum the probabilities for 4, 5 and 6 shots.

    Then there's the fast way...
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    Quote Originally Posted by Archie Meade View Post
    Here's a way to look at this....

    The target could be hit on the 1st, 2nd, 3rd, 4th, 5th or 6th shot
    or all 6 shots could miss.

    To calculate the probability of a hit, we can sum the probabilities of
    1. a hit on the 1st shot
    2. a hit on the 2nd shot predeced by an empty chamber on the 1st
    3. a hit on the 3rd shot preceded by 2 empty chambers
    4. a hit on the 4th shot preceded by 3 empty chambers
    5. a hit on the 5th shot preceded by 4 empty chambers
    6. a hit on the 6th shot preceded by 5 empty chambers.

    Assuming the gunman is an expert marksman, his probability of being off-target is negligible.

    Hence, if we denote a gun without a bullet as W (for wrong gun)
    and the gun with the bullet as R (for right gun),
    then we can write the sequences using the fact that to hit the target, the gunman must have the right gun on the final shot.

    Hence the options are...

    R................... one shot probability

    WR
    RR..................two shots probability

    WWR
    WRR
    RWR
    RRR................three shots probability

    WWWR
    WWRR
    WRWR
    WRRR
    RWWR
    RWRR
    RRWR
    RRRR..............four shots probability

    continue for five and six shots..

    Since the gunman must have the right gun to fire the last shot successfully,
    then there are

    2^0,\ 2^1,\ 2^2,\ 2^3,\ 2^4\ or\ 2^5

    possible sequences for 1, 2, 3, 4, 5, 6 shots in that order.

    To calculate the probabilities

    1 shot

    \frac{1}{3}\ \frac{1}{6} right gun, 1 in 6

    2 shots

    \left(\frac{2}{3}(1)\right)\left(\frac{1}{3}\ \frac{1}{6}\right) WR

    \left(\frac{1}{3}\ \frac{5}{6}\right)\left(\frac{1}{3}\ \frac{1}{5}\right) RR

    3 shots

    \left(\frac{2}{3}(1)\right)\left(\frac{2}{3}(1)\ri  ght)\left(\frac{1}{3}\ \frac{1}{6}\right) WWR

    \left(\frac{2}{3}(1)\right)\left( \frac{1}{3}\ \frac{5}{6}\right)\left(\frac{1}{3}\ \frac{1}{5}\right) WRR

    \left(\frac{1}{3}\ \frac{5}{6}\right)\left( \frac{2}{3}(1)\right)\left(\frac{1}{3}\ \frac{1}{5}\right) RWR

    \left(\frac{1}{3}\ \frac{5}{6}\right)\left( \frac{1}{3}\ \frac{4}{5}\right)\left( \frac{1}{3}\ \frac{1}{4}\right) RRR

    continue to sum the probabilities for 4, 5 and 6 shots.

    Then there's the fast way...
    This is the same as my tree method, but not requiring as large a piece of paper! Thanks.
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    Well I'll be darned, it ended up being fairly simple after all. I guess me and my friends just confused ourselves more than we needed to. I'm certain I had tried to do the question the same way or at least a similar way and then some how convinced myself the method would not work! Now I can finally lay this to rest.

    Thanks guys.
    Last edited by mr fantastic; May 28th 2010 at 04:30 PM. Reason: m --> r
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    Maybe this is wrong, but can't you take this approach?

    The bullet is placed in chamber B of gun 1 (assign the number B after spinning the revolver). Define K = the number of times gun 1 is chosen in the 6 juggling acts. Then someone will be shot under the following circumstances:

    K \ge 1, B = 1
    K \ge 2, B = 2
    .
    .
    .
    K \ge 6, B = 6

    This makes sense, because if the gun is (say) in the second chamber, you must select it twice to ensure the person is shot i.e.  B = 2 requires K \ge 2. Clearly, P(B = b) = 1/6, K \sim Bin(6, 1/3). Calculating the probability from here is trivial: letting F be the cdf of the binomial, this gives

    1 - \frac{F(0) + F(1) + \cdots + F(5)}{6}

    so I suspect this is wrong if you guys went to all this trouble. For what it's worth, the probability works out to 1/3 on the nose. If there are no errors, it appears that most of the statement of the problem is smoke and mirrors, when all you really had to do was say that there are 18 chambers total, and 6 shots are going off...
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    I'm a bit confused here:

    Now, he begins juggling the three guns. Whilst juggling them he randomly selects one from the air, points and pulls the trigger at an unfortunate person. In total, he does this six times.
    Does what six times? Pulls the trigger of the same, randomly selected, gun or goes through the whole routine of juggling the guns, grabbing one firing once and repeating six times?

    Because if he fires the same gun six times, obviously there's a \frac{1}{3} chance that the person will be shot..
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    What I mean is he's juggling three 6-Shooters in the air. One of them has a single bullet loaded and the other two have no bullets. As he juggles he randomly selects a gun, points and fires but then continues juggling all three guns. So whilst juggling he randomly selects and fires 6 times in that fashion.

    @ theodds

    Yes, I suppose much of the question does appear to be smoke and mirrors. It does seem to simplify quite nicely which is what I was hoping for. However, I don't think it's quite the statement which makes it confusing. I think what makes the problem confusing is that it doesn't seem obvious to be able to simplify the problem.

    If you still don't understand then please watch the scene here. Start watching at about 2:00:

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    Quote Originally Posted by RevolverOcelot View Post
    If you still don't understand then please watch the scene here. Start watching at about 2:00:

    Thanks for the illustration, I haven't kept up with gaming and only played the really old Metal Gear.. but from internet searches I see they carried on the tradition of being disguised as a cardboard box. But I guess the truck no longer have started to move.

    By the way, I didn't mean to be off topic, just expressing thanks. On topic, I think it's cool there are many different approaches to the same problem.
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    Quote Originally Posted by RevolverOcelot View Post
    Hi everyone,

    I'm new here. My background is in both pure and applied mathematics with a strong emphasis on differential equations, algebra, group theory and number theory however I never studied too much probability theory. Me and some friends (mathematicians and engineers) discussed a problem out of a video game we had all played and for the life of us could not come to a satisfying conclusion. Imagine the following problem.

    A man has three revolvers. In case you don't know what I'm on about let me explain. A revolver in the context of this question is a gun which holds six bullets. As one chamber is fired it moves along and there are now 5 more chambers to go. I suppose most of us already knew that but anyway, the important thing is that we've got 6 chambers and that there's an order to them.

    Now, the man with the three revolvers picks one of them randomly and places one bullet into one of its chambers randomly. To ensure there's no bias he then spins the cylinder around before locking it in. It is now unknown exactly which chamber of the cylinder the bullet is in.

    Now, he begins juggling the three guns. Whilst juggling them he randomly selects one from the air, points and pulls the trigger at an unfortunate person. In total, he does this six times.

    What is the probability of the man being shot with the single bullet?

    The problem seemed simple at first but we soon realised it wasn't so... at least we think. Sure there's a 1/3 chance of selecting the loaded gun and the loaded gun has a 1/6 chance of having the bullet in the chamber set to fire... but if it doesn't then the next time we fire it we'll have a 1/5 chance, then a 1/4 and so on. However, that increasing probability is affected by an external random choice.

    Can this problem simply be solved by using expectation? That just doesn't seem very satisfying somehow.

    Me and my friends would love to see any nice neat way to figure this out without resorting to a computer simulation (I don't want to have to remind myself how to write stuff like this in Mathematica).

    Any takers?

    The game was Metal Gear Solid 3: Snake Eater.
    The probability that the shot is fired is 1/3.

    To see this, let's say the number of times gun i is chosen is X_i. Then
    P(X_1 = i, X_2 = j, X_3=k) = \binom{6}{i \; j \; k} (1/3)^6
    where i, j, k are non-negative integers with i+j+k=6.

    Each gun has the bullet with probability 1/3. So

    P(shot) = (1/3) P(shot | gun 1 has bullet) + (1/3) P(shot | gun 2 has bullet) + (1/3) P(shot | gun 3 has bullet)

    but all three of the conditional probabilities are the same by symmetry, so

    P(shot) = 3 (1/3) P(shot | gun 1 has bullet) = P(shot | gun 1 has bullet)

    I.e., we may as well assume gun 1 has the bullet. If so, and gun 1 is
    chosen i times, then the probability that the bullet is shot is i/6. So

    [1] P(\text{shot}) = \sum_{i+j+k=6} \binom{6}{i \; j \; k} (1/3)^6 (i/6).

    To simplify this sum, observe that
    (x+y+z)^6 = \sum_{i+j+k=6} \binom{6}{i \; j \; k} x^i y^j z^k
    Differentiating with respect to x,
    6 (x+y+z)^5 = \sum_{i+j+k=6} \binom{6}{i \; j \; k} i x^{i-1} y^j  z^k
    Multiply by x:
    6x (x+y+z)^5 = \sum_{i+j+k=6} \binom{6}{i \; j \; k} i x^i y^j z^k
    Now let x=y=z=1/3:
    2 = \sum_{i+j+k=6} \binom{6}{i \; j \; k} i (1/3)^6
    Divide by 6:
    1/3 = \sum_{i+j+k=6} \binom{6}{i \; j \; k} (i/6) (1/3)^6

    Comparing with [1], we see
    P(\text{shot}) = 1/3.

    The simplicity of this result suggests that there is a simpler method of deriving it.
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  14. #14
    MHF Contributor
    Joined
    Dec 2009
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    We could also sum the probabilities for the bullet being in any of the 6 chambers of a particular gun and hitting the target.

    We can see sequentially how it works out shot by shot.
    That allows you to answer numerous questions regarding the scene.

    The fast way in this case, of course is to imagine this showdown being enacted day by day.

    At most, 6 chambers will be fired.

    We will not always have 6 shots.

    If the bullet is not released, 6 shots will have been taken.
    There are 18 chambers, the bullet could be in any of them.

    Sometimes the bullet is released, on average 1 in every 3 showdowns, since the gunman has at least 6 shots from 18 available chambers.

    The gunman does not shoot a third of the 18 chambers on average,
    if the scene was being re-enacted day by day,
    since he stops if he hits the target on any of his first 5 available.

    When he does shoot 6 times, he has a 1 in 13 chance of firing the bullet on average on his 6th shot.
    He has a 1 in 14 chance of firing the bullet on his 5th shot,
    a 1 in 15 chance on his 4th,
    a 1 in 16 on his 3rd,
    a 1 in 17 on his 2nd,
    a 1 in 18 on his first

    all on average, over a span of time.

    Hence, the probability is....starting with 6 shots

    \frac{17}{18}\frac{16}{17}\frac{15}{16}\frac{14}{1  5}\frac{13}{14}\frac{1}{13}+ \frac{17}{18}\frac{16}{17}\frac{15}{16}\frac{14}{1  5}\frac{1}{14}+ \frac{17}{18}\frac{16}{17}\frac{15}{16}\frac{1}{15  }+ \frac{17}{18}\frac{16}{17}\frac{1}{16}+ \frac{17}{18}\frac{1}{17}+ \frac{1}{18}

    =\frac{1}{18}+\frac{1}{18}+\frac{1}{18}+\frac{1}{1  8}+\frac{1}{18}+\frac{1}{18}

    Each time however, we may consider that he does have 6 chambers from the 18 available to try, giving a 1 in 3 chance of releasing the bullet,
    though the "choice" is random.
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  15. #15
    Newbie
    Joined
    May 2010
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    Is this question not much simpler working out the probability of him not being shot, i.e (17/18)(16/17)(15/16)(14/15)(13/14)(12/13) = 2/3 and hence has 1/3 chance of being shot.
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