[Prime]

A, B, and C have a three-way duel.

The rules are as following: A shoots first, then B, then C, and so on. Each man can shoot at whoever he wants on his turn. However, when one of them gets hit, he does not shoot anymore, and no one shoots at him. The duel continues until only one man is left standing. It is well-known fact that A hits his intended target 30% of the time, C – 50%, whereas B never misses.

What is the best strategy for A?

Small hint:

This variation of the duel puzzle actually requires some calculation.

[Guest]

A, B, and C have a three-way duel.

The rules are as following: A shoots first, then B, then C, and so on. Each man can shoot at whoever he wants on his turn. However, when one of them gets hit, he does not shoot anymore, and no one shoots at him. The duel continues until only one man is left standing. It is well-known fact that A hits his intended target 30% of the time, C – 50%, whereas B never misses.

What is the best strategy for A?

Small hint:

This variation of the duel puzzle actually requires some calculation.

If A shoot B and misses does he hit c?

[Guest]

A, B, and C have a three-way duel.

The rules are as following: A shoots first, then B, then C, and so on. Each man can shoot at whoever he wants on his turn. However, when one of them gets hit, he does not shoot anymore, and no one shoots at him. The duel continues until only one man is left standing. It is well-known fact that A hits his intended target 30% of the time, C – 50%, whereas B never misses.

What is the best strategy for A?

Small hint:

This variation of the duel puzzle actually requires some calculation.

... A doesn't want to hit C, cos then B will shoot A. So A has to aim at B...

Lets see...

A shoots at C

- A hits C => B hits A (30%)

- A misses C => B hits C (70%) => A shoots at B

-- A hits B (30%)

-- A misses B => B hits A (70%)

So... if A shoots at C first, chance of surviving is 70% x 30% = 21%

Now... If A shoots at B:

- A misses B then we're in the same situation as before: B hits C, and A has a 30% chance of hitting B with his second shot.

If A hits B, then we have a two way shoot out between C and A... possibly going on forever. I won't do the maths, but there is a chance A will win.

So, if A shoots at B, chance of survival is 21% plus the chance that A will beat C. And my gut instinct was right.

[

• 
  

[Guest]

Calculation? Pish Posh, common sense. A should shoot at B, unless by missing the target, you meant, hit the other person by default.

[Guest]

A should shoot away and not try to hit anything first. Then if B is smart, he will shoot C as C poses the biggest threat to him. Then A should shoot B. This way (assuming B chooses to shoot C) A has a 30% chance of survival. Again, this only works if B and C shoot depending on their best chances for survival and not randomly.

[Guest]

he should aim for b and ask for c to double team with him

[Guest]

A shoots C but misses, his next turn is 50% chance for him to get his target

B Shoots C and Kills C because C has more accuracy than A

A shoots B

[Prime]

If A shoot B and misses does he hit c?

No. Good point, though.

[Prime]

• 
  

... A doesn't want to hit C, cos then B will shoot A. So A has to aim at B...

Lets see...

A shoots at C

- A hits C => B hits A (30%)

- A misses C => B hits C (70%) => A shoots at B

-- A hits B (30%)

-- A misses B => B hits A (70%)

So... if A shoots at C first, chance of surviving is 70% x 30% = 21%

Now... If A shoots at B:

- A misses B then we're in the same situation as before: B hits C, and A has a 30% chance of hitting B with his second shot.

If A hits B, then we have a two way shoot out between C and A... possibly going on forever. I won't do the maths, but there is a chance A will win.

So, if A shoots at B, chance of survival is 21% plus the chance that A will beat C. And my gut instinct was right.

[

Some of the variations are explained well here. However, you have missied the right answer.

[Prime]

A should shoot away and not try to hit anything first. Then if B is smart, he will shoot C as C poses the biggest threat to him. Then A should shoot B. This way (assuming B chooses to shoot C) A has a 30% chance of survival. Again, this only works if B and C shoot depending on their best chances for survival and not randomly.

You have guessed the correct answer, but did not solve the problem.

[Prof. Templeton]

A should shoot at the ground. If A shoots at B and misses, B would shoot at A instead of C, because A would get 2 30% chances and C would get 1 50% chance. So A has a 30% chance of winning, B has a 70% chance and C has no chance.

[Prime]

A should shoot at the ground. If A shoots at B and misses, B would shoot at A instead of C, because A would get 2 30% chances and C would get 1 50% chance. So A has a 30% chance of winning, B has a 70% chance and C has no chance.

You must have misread the statement of the problem.

[Guest]

A should shoot himself, making sure he hits a non-lethal area.

Edited August 11, 2008 by speropotus

[Prof. Templeton]

You must have misread the statement of the problem.

How so? You already said this is the correct solution, so why?

[Prime]

How so? You already said this is the correct solution, so why?

I said "correct answer" -- not "correct solution". You seem to imply that a man may get two shots in one on one duel against B.

[Prime]

A should shoot himself, making sure he hits a non-lethal area.

I would have to accept your solution as the only correct one, if I were any less specific in my statement of the problem. (And Taliesin helped me to make it even more specific with his question.) I clearly stated that A has only 30% chance of hitting his target. So what happens if he aimes at non-lethal area of his body and misses?

Edited August 11, 2008 by Prime

[Guest]

A=>B 

-hit> 30%

   C=>A

   -hit> 50% DEAD

   -miss> 50%

      A=>C

      -hit> 30% ALIVE ---> 4.5% total (30% * 50% * 30%)

      -miss>

         Vanishingly small chance of survival

-miss> 70%

   B=>C

   -hit-> 100%

      A=>B

      -hit> 30% ALIVE ---> 21% total (70% * 30%)

      -miss> 70%

         B=>A

         -hit> 100% DEAD


Chance of survival: 21% + 4.5% + small fraction ~= 26%


A=>C

-hit> 30%

   B=>A

   -hit> 100% DEAD

-miss> 70%

--See table above.


Chance of survival: 21%


A=>Foot

-miss> 100% ???

   B=>C

   -hit> 100%

      A=>B

      -hit> 30% ALIVE ---> 30% total (100% * 100% * 30%)

      -miss>

         B=>A

         -hit> 100% DEAD


Chance of survival: 30%[/codebox]

If you can't guarantee a miss on demand, I agree with the first explanation and claim it is around 26%, otherwise it is 30%.

Edited for formatting.

Edited August 12, 2008 by itsclueless

[Prime]

A=>B 

-hit> 30%

   C=>A

   -hit> 50% DEAD

   -miss> 50%

      A=>C

      -hit> 30% ALIVE ---> 4.5% total (30% * 50% * 30%)

      -miss>

         Vanishingly small chance of survival

.......

[/codebox]

That's the first solid attempt at estimating probabilities here. Although 26% is not precise enough.

Here is the way to calculate probabilities:

[spoiler=Geometric series.]Here is how to calculate probability for A’s victory in a one on one duel between A and C, where C shoots first.

The chance for C to miss is 0.5, the chance for A to hit is 0.3, and the chance for A to miss is 0.7.

In order for A to win on his first shot C must miss and A must hit. The probability of that is 0.5*0.3.

For A to win on his second turn C must miss, A – miss, C – miss, A – hit. The probability: 0.5*0.3*0.5*0.7.

For A to hit on the third shot probability is (0.5*0.3)(0.5*0.7)^2.

Thus overall probability for A to prevail over C is:

0.5*0.7 + (0.5*0.7)*(0.5*0.3) + (0.5*0.7)*(0.5*0.3)^2 + . . . an so on to infinity.

This is geometric series with common ratio (0.5*0.7) = 0.35 and scale factor (0.5*0.3) = 0.15. Since common ratio is less than 1, the sum of the infinite series equals:

0.15/(1-0.35) = 0.230769.

Now if A shoots at B and hits (the probability of that is 30%, or 0.3) then the duel between C and A takes place and the overall probability for A’s victory in that scenario is 0.3*0.230769=0.069231 (just under 7%).

If A misses B on his first shot (70%), then B kills C and A gets another chance to hit B (30%). Thus A’s probability to win in that scenario is 0.7*0.3 = 0.21.

Overall probability for A’s triumph if he aims his first shot at B is 0.21 + 0.069 = 0.279, or just under 28%.

Whereas, if A shoots in the air, then B kills C and A has his 30% chance to hit B.

So it is slightly better for A to shot in the air than at B on his first turn (30% vs. 28%).

The strategy for A to aim at C on his first shot is obviously wrong – then his chance is just 21%.

[/spoiler]

[Guest]

The chance for C to miss is 0.5, the chance for A to hit is 0.3, and the chance for A to miss is 0.7.

In order for A to win on his first shot C must miss and A must hit. The probability of that is 0.5*0.3.

For A to win on his second turn C must miss, A – miss, C – miss, A – hit. The probability: 0.5*0.3*0.5*0.7.

For A to hit on the third shot probability is (0.5*0.3)(0.5*0.7)^2.

Thus overall probability for A to prevail over C is:

0.5*0.7 + (0.5*0.7)*(0.5*0.3) + (0.5*0.7)*(0.5*0.3)^2 + . . . an so on to infinity.

This is geometric series with common ratio (0.5*0.7) = 0.35 and scale factor (0.5*0.3) = 0.15. Since common ratio is less than 1, the sum of the infinite series equals:

0.15/(1-0.35) = 0.230769.

By my calculations, upon reaching a shootout with C and having a shot at him (15% chance of happening), the equation looks like below.

.3 + .3(.7*.5) + .3(.7*.5)^2 + .3(.7*.5)^3 ... = .3/(1 - .7*.5) = 6/13

So, when A shoots B, he has a 21% chance of winning before a shootout and a (15% * 6/13) chance if he gets into one. Add them together and I get a 27.923% chance of survival. This is in agreement with your numbers.

[Prof. Templeton]

I said "correct answer" -- not "correct solution". You seem to imply that a man may get two shots in one on one duel against B.

Maybe I'm missing something.

If A shoots at B and misses and B shoots at C, than A shoots again. If he hits 30% of the time wouldn't his chance of hitting B after 2 shots be .3+.3 - (.3*.3) or 51%. Greater then C's one 50% chance( if nobody takes out B he only gets one chance because he goes last). But if A intentionally shoots away, C's 50% would be greater than his remaining 30% chance(and he only gets one more chance if nobody takes out B).

A and C would only get into a shootout if A were to shoot at B and score a hit on his first shot. Otherwise B will take out one of them.

[Guest]

Maybe I'm missing something.

If A shoots at B and misses and B shoots at C, than A shoots again. If he hits 30% of the time wouldn't his chance of hitting B after 2 shots be .3+.3 - (.3*.3) or 51%. Greater then C's one 50% chance( if nobody takes out B he only gets one chance because he goes last). But if A intentionally shoots away, C's 50% would be greater than his remaining 30% chance(and he only gets one more chance if nobody takes out B).

A and C would only get into a shootout if A were to shoot at B and score a hit on his first shot. Otherwise B will take out one of them.

Where is the .3 + .3 thing coming in? You already stated that he missed his first shot, so the probability of hitting on that first shot is nto a factor in his chances of killing on the second shot.

[Prof. Templeton]

Where is the .3 + .3 thing coming in? You already stated that he missed his first shot, so the probability of hitting on that first shot is nto a factor in his chances of killing on the second shot.

Probability of hitting with either of his 30% chances. Isn't the probability that either occurs P(A1 ∪ A2) = P(A1) + P(A2) - P(A1) x P(A2) Where A1 is the first shot and A2 is the second? It's been a long time.

[Guest]

Probability of hitting with either of his 30% chances. Isn't the probability that either occurs P(A1 ∪ A2) = P(A1) + P(A2) - P(A1) x P(A2) Where A1 is the first shot and A2 is the second? It's been a long time.

Correct, but if he hits with his first shot, C isn't dead, and A isn't home free yet. You have another of those geometric sequences.

[Prof. Templeton]

Correct, but if he hits with his first shot, C isn't dead, and A isn't home free yet. You have another of those geometric sequences.

I'm only referring to B's viewpoint. 2 shots from A are worse than 1 shot from C. If B shoots C first A gets 2 shots, but if B shoots A first C get 1 shot. Provided B is still around.

I see the error in my first post so all this is moot. Although the "answer" was correct if B shot A first he would still have weathered 1 shot from A and 1 shot from C before he could win, which is worse than 2 shots from A. So my logic was incorrect.

[Guest]

A shoots at B (30% chance of a hit). If A shoots at and hits C, then A guarantees that he dies when B shoots.

B will shoot at C because C has a greater chance of hitting B than A does.

Then A has another shot at B (giving him 60% chance overall -- 2 tries with 30% chance) before B kills A.

[Guest]

Why is everyone forgetting that C has to shoot?

I think that (Forgot the name sorry) the person who said about them not shooting randomly, was correct but i can't seem to find the right answer so good luck to you all.