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...