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Old 2021-08-02, 11:42   #1
Jul 2015

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Old 2021-08-02, 12:30   #2
Dr Sardonicus
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Feb 2017

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This reminds me of a problem in The 2nd Scientific American Book of Mathematical Puzzles & Diversions, a collection of Martin Garner's "Mathematical Games" columns from Scientific American, of which a PDF may be found here. It is the ninth of the first set of "Nine Problems."

Different versions may be found online, generally without any attribution or reference to earlier versions. Here is Gardner's version:


SMITH, Brown and Jones agree to fight a pistol duel under the following unusual conditions. After drawing lots to determine who fires first, second and third, they take their places at the corners of an equilateral triangle. It is agreed that they will fire single shots in turn and continue in the same cyclic order until two of them are dead. At each turn the man who is firing may aim wherever he pleases. All three duelists know that Smith always hits his target, Brown is 80 per cent accurate and Jones is 50 per cent accurate.

Assuming that all three adopt the best strategy, and that no one is killed by a wild shot not intended for him, who has the best chance to survive? A more difficult question: What are the exact survival probabilities of the three men?
Gardner also traces the problem further back:
The problem, in variant forms, appears in several puzzle books. The earliest reference known to me is Hubert Phillip's Question Time, 1938, Problem 223. A different version of the problem can be found in Clark Kinnaird's Encyclopedia of Puzzles and Pastimes, 1946, but the answer is incorrect. Correct probability figures for Kinnaird's version are given in The American Mathematical Monthly, December 1948, page 640.
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Old 2021-09-06, 07:17   #3
Oct 2017

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The publication of the solution wasn’t helpful for me.

Can anyone explain the strategy: “All players are perfectly rational - they always choose which player to remove in a way that maximizes their chance to win.” ?
I wasn’t able to solve the challenge, because I couldn’t reproduce the values of the second example. The strategy of the first example was: “Always remove the best”. The values of the challenge were the exact values and could be computed with pencil and paper.
Using this strategy for the second example (5 players) yielded approximately:
So the simple strategy (removing the best) is wrong.
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Old 2021-09-06, 11:24   #4
Jan 2017

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Originally Posted by Dieter View Post
So the simple strategy (removing the best) is wrong.
Yes, that is not always the correct strategy for the participants. If you haven't seen the triangular duel puzzle before, see that for analysis ("deliberately miss" may not be a directly valid option here, but can be simulated by adding dummy participants).
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