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I choose to think of this a duel (or “truel”) with paintball guns, so that once a participant is hit they are eliminated non-fatally.

We then consider the probabilities of winning the truel (i.e. to be the only participant that hasn’t received a “hit”), rather than the probability of “surviving”. If each participant refused to shoot (or deliberately missed), then all three participants would “survive” at the point when they give up (or run out of ammunition), but no-one would win.

The approach I used is similar to the one I used in

Enigma 287.To work out the best strategy, we determine the function

target(X, [Y, Z])where:Using: A = George (60%), B = Fred (75%), C = Ernest (100%).

If any participant is faced with a single opponent, (so a normal duel), then his best strategy is to target that opponent. In the case of C this gives him a 100% chance of winning.

So, we need to determine the best strategies when faced with two opponents.

If C is faced with two opponents (A and B) he can eliminate one of them. If the remaining opponent misses then C subsequently eliminates him. So, faced with two opponents C should eliminate the opponent with the greatest chance of hitting him. i.e. C should target B.

If B is faced with two opponents (A and C). If he targets A and hits, then A is eliminated, and C goes on to eliminate B and win. If he targets A and misses, then C has a choice of two opponents and targets B, eliminating him. So it ends badly for B if he chooses to target A instead of C. If he targets C, then there is a 75% chance he will eliminate C, and a chance that A will miss, giving him at least a chance to eliminate A and survive:

Now we can use the targeting function to work out the following probabilities of winning, where:

solving this last equation gives:

so:

Finally if A is faced with two opponents (B and C), we have the following possibilities:

If A targets B:

If A targets C:

So, if A has to target one of his opponents then his best chance of winning is 36% by targeting C.

However, if A can deliberately miss both B and C, we get:

So A can increase his chance of winning to 65% by deliberately missing (or forfeiting) his turn.

B doesn’t have that option, if he forfeits his turn, then he will be targeted by C and eliminated.

Solution:George is the most likely to win. He has a 65% chance of winning.Fred, the middling shot, has a 25% chance of winning.

Ernest, the best shot, only has a 10% chance of winning.

So, this “truel” is an example of “survival of the weakest”, as the weakest shot has the best chance of winning and the best shot has the lowest chance of winning.

Here is a simulation of the above strategies in Python. It runs 1,000,000 trials in 1.4s and produces results that agree with the above analysis.

Here’s the output of a typical run:

You can play around with alternate targeting strategies by changing the

target()function.The number of trials to run (default: 1,000,000) can be specified as a command-line argument.

This is very reminiscent of one put forward many years ago by Martin Gardner in his column in

Scientific American: see alsoMore Mathematical Puzzles and Diversions(Pelican, 1966) for an analysis. The difference there was that Smith is 100% accurate, Brown 80%, and Jones only 50%; they draw lots to determine the cyclic order of shooting (until one is hors de combat). The problem apparently dates from 1938 or even earlier.Gardner concluded that under those conditions Jones’ best strategy is to shoot into the air until one of his opponents is dead (for they are likely to see each other as much the greater risk and leave him alone). Then he gets first shot at the survivor.

Incidentally, according to the dictionary the Latin word duellum was a poetic variant of bellum and had nothing to do with duo.