The Triangular Duel

**Soroban** · May 28th 2009, 11:01 AM

Three men $\{A,B,C\}$ 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.

$A$ always hits his target.
$B$ hits his target with probability 4/5.
$C$ 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.

**aidan** · May 28th 2009, 11:23 AM

Originally Posted by Soroban

Three men $\{A,B,C\}$ 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.

$A$ always hits his target.
$B$ hits his target with probability 4/5.
$C$ 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.

Without working out the probabilities:
A would tend to shoot at the better of the other 2 or at B.
B would tend to shot at A, because he's the best.
C would tend to shot at A, same reason.
Looks like C might survive.
Now to the math...

**chisigma** · May 28th 2009, 11:07 PM

Sorry... I don't understand a little detail: who does shoot first?... and who does shoot second [if he survives...]?...

Kind regards

$\chi$ $\sigma$

**Soroban** · May 29th 2009, 07:11 AM

Hello, chisigma!

The shooting order is set up by a random drawing.

So we must consider all $3! = 6$ shooting orders.
. . (Well, sort of . . .)

**Deadstar** · May 29th 2009, 01:26 PM

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...

**chisigma** · May 29th 2009, 10:26 PM

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 $P_{a}$ , $P_{b}$ and $P_{c}$ the probabilities that a, B and C survive. Since at the end only one survives is...

$P_{a} + P_{b} + P_{c} = 1$ (1)

Now we examine the three cases...

a) A shoots first. In this case A kills B at first, so that C has $\frac{3}{5}$ of probability to kill A and survive. In this case is...

$P_{a}= \frac{2}{5}$ , $P_{b} = 0$ , $P_{c}= \frac{3}{5}$

b) B shoots first. In this case B shoots to A with probability $\frac{4}{5}$ 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...

$P_{a}=0$ , $P_{c}= \frac{3}{5} (1+\frac{2}{25} + (\frac{2}{25})^{2} + ...)= \frac{3}{5}\cdot \frac{1}{1- \frac {2}{25}} = \frac{15}{23}$ , $P_{b}= 1-\frac{15}{23}= \frac{8}{23}$

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 $\frac{1}{3}$ we arrive to...

$P_{a}= \frac{1}{3}\cdot {2}{5} + \frac{2}{3}\cdot \frac{1}{5}\cdot \frac{2}{5} = \frac{14}{75} \approx .1866666...$

$P_{c}= 2\cdot (\frac{1}{3}\cdot {3}{5} + \frac{1}{3}\cdot \frac{4}{5}\cdot \frac{15}{23}) = \frac{54}{115} \approx .46956521739...$

$P_{b} = 1-P_{a} - P_{c} \approx .3437681159...$

The conclusion is a little suprising: C has the best chanche of survive and A the last!... so is life

!...

Kind regards

$\chi$ $\sigma$

**CaptainBlack** · May 29th 2009, 11:29 PM

Originally Posted by Soroban

Three men $\{A,B,C\}$ 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.

$A$ always hits his target.
$B$ hits his target with probability 4/5.
$C$ 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.

If this seems a bit too contrived consider the voting in the last vote-off in the TV game Weakest-Link.

CB

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