Three men decide to settle their dispute with a duel.
They stand at the vertices of a large equilateral triangle
and will take turns shooting at each other.
They draw lots to determine who gets to shoot first, second and third.
At his turn, each man gets to shoot at one other man.
always hits his target.
hits his target with probability 4/5.
hits his target with probability 3/5.
(a) Find the probability of each man's survival.
(b) Who has the best chance for survival?
Note: These men are highly intelligent.
. . . . .They will select their targets with great care.
I'm interested in seeing how this is worked out so ill put down my rough idea of whats happening but I'm not sure how to work out the final probabilities... I run into trouble whenever A gets shot...
6 options of order...
ABC (A goes 1st, B 2nd, C 3rd)
A will shoot B since he's the bigger threat.
Then C wins with a prob of 0.6.
A wins with a prob of 0.4 (prob that C misses)
ACB
A shoots B
C wins with prob 0.6
A wins -> 0.4
BAC
B shoots A
--- Scenario 1 ---
B hits A
C wins with prob > 0.6?
B????
--- Scenario 2 ---
B misses A
A shoots B
C wins with prob 0.6
A wins with prob 0.4
BCA
B shoots A
--- 1 ---
B hits A
C wins with prob > 0.6?
--- 2 ---
B misses A
C hits A
B wins with prob > 0.8?
--- 3 ---
B misses A
C misses A
A shoots B
C wins with prob 0.6
A wins with prob 0.4
CAB
C shoots A
--- 1 ---
C hits A
B wins with prob > 0.8?
--- 2 ---
C misses A
A shoots B
C wins with prob 0.6
A wins with prob 0.4
CBA
C shoots A
--- 1 ---
C hits A
B wins with prob > 0.8?
--- 2 ---
C misses A
B hits A
C wins with prob > 0.6?
--- 3 ---
C misses A
B misses A
A shoots B
C wins with prob 0.6
A wins with prob 0.4.
Just how to put this mess all together...
All right!...
... first we have to extablish the best strategy for A, B and C, that means the strategy that maximises their chance of survive. As it will be justified at the end the 'best strategies' are...
... for A : he shoots to B, unless only C is surviving
... for B : he shoots to A, unless only C is surviving
... for C : he shoots in the air, unless only A or only B are surviving
Now we indicate with , and the probabilities that a, B and C survive. Since at the end only one survives is...
(1)
Now we examine the three cases...
a) A shoots first. In this case A kills B at first, so that C has of probability to kill A and survive. In this case is...
, ,
b) B shoots first. In this case B shoots to A with probability of success. There are now two possibilities...
b1) A is kapput so that starts the competition beween B and C with C first shooting. The probabilities in this case are...
, ,
b2) A survives, so that we arrrive in situation A first shooting, allready examined, or C first shooting, that we will to examine now...
c) C shoots first. In this case C shoots in air and we have again the case a) or b).
Combining the situation a) b) and c), each of which has probability we arrive to...
The conclusion is a little suprising: C has the best chanche of survive and A the last!... so is life !...
Kind regards