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A. B, and C are to fight a three-cornered pistol duel. All know that A's chance of hitting his target is 0.3, C is 0.5, and B never misses. They are to fire at their choice of target in succession in the order A, B, C, cyclically (but a hit man loses further turns and is no longer shot at) until only one man is left unhit. What should A's strategy be?
Can you check my solution of this task (Three-Cornered Duel) in Mathcad Prime 2.0!
See the attach!
Duelist A should shoot into the ground and let B and C duel.
This gives A the first shot at who ever is left which gives A a better chance than .3 to win the duel.
Patrick Maxfield wrote:
Duelist A should shoot into the ground and let B and C duel.
This gives A the first shot at who ever is left which gives A a better chance than .3 to win the duel.
Thanks, Patrik!
It is a well-known answer.
I have asked - is the attached sheet (the Monte Carlo Method) correct.
Let's see: A shoots the ground. (We'll assume he can hit that.)
That leaves B. B will hit something. Most of the time (if I'm B) I'm going to shoot C, because C will hit me five times in 10 and A will only hit me 3 times in 10. B shoots, and C is eliminated.
Now it's A's turn again, and he has a 30% chance of getting B before he gets shot.
Sounds about right!
But: What if B shoots A instead of C?
But: What if B shoots A instead of C?
Indeed. Maybe B is an excellent shot even though he is a complete idiot. Or maybe B is an adrenalin junkie, and likes to live dangerously
Fred Kohlhepp wrote:
Let's see: A shoots the ground. (We'll assume he can hit that.)
Sounds about right!
Thanks, Fred!
Can you please check and edit in the attach names of variables and functions and others form.
Thanks,
Valery
Sorry, Valery, I don't havePrime available right now.
Your "beat most accurate" algorithm does not take in account that two persons could have the same chance of hitting. You MostAccurate-function returns the first of them every time.
For example, if all three are perfect shooters with probability 1, the result should be (0,50%,50%). A decides by chance (50%) whome to kill. Then he will be killed by the other of the two. Your program returns (0;33%;67%) but with your MostAccurate-function I would have expected (0%;0%;100%) as A will shoot at B every time and then C will kill A. So it seems there is something else wrong, too.
Same, if A decides to shoot at the ground with his first shot (tricky?). Then the chances should be (50%; 0; 50%) if B and C follow the "beat most accurate" strategy. Your program again yields (0;33%;67%).
Would be interesting to expand your program to allow for an arbitrary number of participants.
Added: I just see, that in contrary to the description you determine the first shooter by chance.
BTW: You should add a forced abortion of the duel after a decent number of shots - just in case a user choses an accuracy vector of (0;0;0) 😉
winfrod rager wrote:
Your "beat most accurate" algorithm does not take in account that two persons could have the same chance of hitting. You MostAccurate-function returns the first of them every time.
For example, if all three are perfect shooters with probability 1, the result should be (0,50%,50%). A decides by chance (50%) whome to kill. Then he will be killed by the other of the two. Your program returns (0;33%;67%) but with your MostAccurate-function I would have expected (0%;0%;100%) as A will shoot at B every time and then C will kill A. So it seems there is something else wrong, too.
Same, if A decides to shoot at the ground with his first shot (tricky?). Then the chances should be (50%; 0; 50%) if B and C follow the "beat most accurate" strategy. Your program again yields (0;33%;67%).
Thanks, rager!
I know it and will correct the program (new random).
But I do not know two duelist with same level of an accuracy - 0.5 and 0.5 for example. I know with 0.4999999999 and 0.50000000001
What if B is suicidal, and shoots himself?
Richard Jackson wrote:
What if B is suicidal, and shoots himself?
Sorry, it is a Paint Ball Duel.
One duelist:
Richard Jackson wrote:
What if B is suicidal, and shoots himself?
"Do you know, Totski, this is all very like what they say goes on among the Japanese?" said Ptitsin. "The offended party there, they say, marches off to his insulter and says to him, 'You insulted me, so I have come to rip myself open before your eyes;' and with these words he does actually rip his stomach open before his enemy, and considers, doubtless, that he is having all possible and necessary satisfaction and revenge. There are strange characters in the world, sir!"