It's funny, I've heard this one before so I know what the "thinking outside the box" correct solution is supposed to be, but the math isn't matching up. So far I get the same answer as forkedknight, that Poor (P) should fire at Excellent (E) first, instead of firing at Average (A) or the "thinking outside the box" solution. Let me explain:

**CASE 1: P shoots at E first**

In order for E to win this round, P must miss E on his first shot, and then A must miss E on his first shot. (If A decided to shoot at P instead, E would maximize his chances of survival by eliminating A next, so A shooting at P is a bad idea here.) If this happens, E will eliminate A next, and so he must hope that P misses his subsequent shot, after which E will eliminate P and win the round. The chance of all this happening is (2/3)*(1/3)*(1)*(2/3)*(1) = 4/27.

There are several ways for A to win this round, but they can be simplified into 2 branches:

(1) P shoots and kills E. After this, A can win by shooting and killing P on his first shot, or by missing, hoping that P misses on his second shot, then shooting and killing A, or by missing, hoping, missing, hoping, killing, or by missing, hoping, missing, hoping, missing, hoping, killing, or by etc... The chance of this happening is calculated using the formula for the sum of an infinite geometric series S = a/(1-r) with a = (1/3)*(2/3) and r = (1/3)*(2/3). Therefore, S = (1/3)*(2/3)/(1 - (1/3)*(2/3)) = (2/9)/(7/9) = 2/7.

(2) P shoots and misses E. After this, A can win by shooting and killing E, hoping that P misses on his next shot, then shooting and killing P, or by killing E, hoping, missing, hoping, killing P, or by etc... This is another sum, this time with a = (2/3)*(2/3)*(2/3)*(2/3) and r = (2/3)*(1/3). Therefore, S = (2/3)*(2/3)*(2/3)*(2/3)/(1-(1/3)*(2/3)) = (16/81)/(7/9) = 16/63.

Adding these two branches together, we get P(A) = 2/7 + 16/63 = 34/63.

The probability that P wins the round is just 1 - P(A) - P(E), or 1 - (34/63) - (4/27) = (189 - 102 - 28)/189 = 59/189

Therefore, P(P) = **59/189**, P(A) = **102/189** and P(E) = **28/189**.

**CASE 2: P shoots at A first**

Following an argument similar to the one above, we find that:

P(E) = (1/3)*1 + (2/3)*(1/3)*(1)*(2/3)*1 = 13/27

P(P) = (2/3)*(2/3)*(1/3)/(1-(1/3)*(2/3)) = (4/27)/(2/9) = 4/21

P(A) = 1 - P(E) - P(P) = 1 - 13/27 - 4/21 = (189 - 91 - 36)/189 = 62/189

Converting these results to compare to Case 1, we have P(P) = **36/189**, P(A) = **62/189** and P(E) = **91/189**.

**CASE 3: P misses on purpose!**

This is the "thinking outside the box" solution. The reasoning goes that by deliberately missing on his first shot, P shifts the burden of having to kill E onto A, who is a better shot. If A kills E, then P gets the first shot against lesser opponent A. If A misses E, E will kill A still leaving P with the first shot against the strongest opponent E. However, the math works out as follows (unless I made a mistake somewhere):

P(E) = (1/3)*(1)*(2/3)*(1) = 2/9

P(P) = (2/3)*(1/3)/(1-(1/3)*(2/3)) = (2/9)/(7/9) = 2/7

P(A) = 1 - P(E) - P(P) = 1 - 2/9 - 2/7 = (189 - 42 - 54)/189 = 93/189

Converting these results to compare with Cases 1 and 2, we have P(P) = **54/189**, P(A) = **93/189** and P(E) = **42/189**.

**Results and discussion**

Here are the results for all three cases tabulated for easy reading:

Case 1

P(P) = **59/189**, P(A) = **102/189** and P(E) = **28/189**

Case 2

P(P) = **36/189**, P(A) = **62/189** and P(E) = **91/189**

Case 3

P(P) = **54/189**, P(A) = **93/189** and P(E) = **42/189**

As you can see (unless I made a big mistake in my algebra, which is entirely possible), P maximizes his chance of winning by shooting at E first instead of missing deliberately. I think the explanation is that by having P take a shot at E instead of missing deliberately, P and A have 2 chances instead of 1 to kill the strongest opponent before he goes on a killing spree. This seems to slightly outweigh the benefit of having A do the heavy lifting and having P keep the first shot against a stronger opponent. However, someone PLEASE check the math above to see if I made any errors...I always second-guess myself when trying to disprove a classic solution! ðŸ™‚