Three Revolvers and One Bullet

May 2010
4
1
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.
 

undefined

MHF Hall of Honor
Mar 2010
2,340
821
Chicago
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. (Itwasntme)

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.
 
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MHF Hall of Honor
Mar 2010
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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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May 2010
4
1
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.
 

undefined

MHF Hall of Honor
Mar 2010
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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.
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 \(\displaystyle X_n\) as the event that the guy gets shot on the nth shot fired, and let \(\displaystyle Y_n\) be the event that the guy gets shot when n shots are fired. So the original problem is to find \(\displaystyle P(Y_6)\).

I'll be using this:

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

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

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 \(\displaystyle \overline{X_1}\), there is a \(\displaystyle \frac{\frac{5}{18}}{\frac{5}{18}+\frac{2}{3}}=\frac{5}{17}\) probability of having 1/5, 0, 0; and there is a \(\displaystyle \frac{\frac{2}{3}}{\frac{5}{18}+\frac{2}{3}}=\frac{12}{17}\) probability of having 1/6, 0, 0.

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

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

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 \(\displaystyle Z_n\) 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 \(\displaystyle n=2k\), there would be \(\displaystyle 2^k\) 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 \(\displaystyle P(Y_7)\) for which the single bin problem is (according to my claim) no longer equivalent.
 

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Dec 2009
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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

\(\displaystyle 2^0,\ 2^1,\ 2^2,\ 2^3,\ 2^4\ or\ 2^5\)

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

To calculate the probabilities

1 shot

\(\displaystyle \frac{1}{3}\ \frac{1}{6}\) right gun, 1 in 6

2 shots

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

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

3 shots

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

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

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

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

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

Then there's the fast way...
 

undefined

MHF Hall of Honor
Mar 2010
2,340
821
Chicago
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

\(\displaystyle 2^0,\ 2^1,\ 2^2,\ 2^3,\ 2^4\ or\ 2^5\)

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

To calculate the probabilities

1 shot

\(\displaystyle \frac{1}{3}\ \frac{1}{6}\) right gun, 1 in 6

2 shots

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

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

3 shots

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

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

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

\(\displaystyle \left(\frac{1}{3}\ \frac{5}{6}\right)\left( \frac{1}{3}\ \frac{4}{5}\right)\left( \frac{1}{3}\ \frac{1}{4}\right)\) 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.
 
May 2010
4
1
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.
 
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Oct 2009
340
140
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:

\(\displaystyle K \ge 1, B = 1\)
\(\displaystyle K \ge 2, B = 2\)
.
.
.
\(\displaystyle K \ge 6, B = 6\)

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.\(\displaystyle B = 2\) requires \(\displaystyle K \ge 2\). Clearly, \(\displaystyle P(B = b) = 1/6\), \(\displaystyle K \sim Bin(6, 1/3)\). Calculating the probability from here is trivial: letting \(\displaystyle F\) be the cdf of the binomial, this gives

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

so I suspect this is wrong if you guys went to all this trouble. For what it's worth, the probability works out to \(\displaystyle 1/3\) 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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Oct 2009
303
33
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 \(\displaystyle \frac{1}{3}\) chance that the person will be shot..