Originally Posted by

**Archie Meade** 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

$\displaystyle \left(\frac{2}{3}(1)\right)\left(\frac{2}{3}(1)\ri ght)\left(\frac{1}{3}\ \frac{1}{6}\right)$ 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...