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This is a known problem (one of my favourite)

Three men, let's call them A, B, and C, engage in a duel.

The rules:

In sequence they can shoot only once, aiming at any target of their choice

C shoots first (privilege given because he is the less skilled), then B, then A, then again C and so on, until only one survives.

The skills (average values):

A is the best, he centers the target 9 times out of 10

B centers the target 8 times out of 10

C centers the target 7 times out of 10.

Which is the best strategy for each one of them?

If they adopt the best strategy, who has the higher chance of being the survivor?

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Duellum/Duo

Private battle between two people.

There isn't a simailar term tor trio - triellum - i think not, more like a melee

However in the perimeters of the prob.

C will win if he drinks from the palace with the challis that has the brew that is true,

or he will have had to aim at B 1st as will A if he survives, so they are in a shoot out and deteriating in condition till C wins cos he has more health and now is aim is best, then Lara Croft comes along and does them all in

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Well any way you look at it, A is absolutely getting hosed in this deal.

To start off with, C should misfire on purpose to let A and B go at it by themselves. If he chose to shoot one of them (and actually hit), then the other man standing has a good chance of hitting C on his shot. There is a good chance that either A or B will die before it's C's turn again. In this case, C will hit the man standing 7 out of 10 times.

B should try to shoot A, and will do so 8 out of 10 times. If he hits, then C will try to shoot him, and will do so 7 out of 10 times.

B misses A 2 out of 10 times. If he does miss then A will hit him 9 out of 10 times.

If A is still alive on his turn, then he should try to shoot B. He will hit 9 out of 10 times, but then C will shoot him 7 out of 10 times. If A misses B, then C will misfire again and B will have another 8 out of 10 chance of hitting him. This will keep going until A or B is dead.

So overall C has the best chance of survival, and A is the least probable to survive.

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Duel etimologically derives from Latin duellum, akin to bellum, meaning war, fight.

I wanted to title truel, ma this word is not in the dictionary.

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Duel etimologically derives from Latin duellum, akin to bellum, meaning war, fight.

I wanted to title truel, ma this word is not in the dictionary.

My first thought was truel also. Word or not, let's use it. ;)

I've seen this problem before, but not solved it.

I'll give it a shot. [npi]

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im going to have to agree with enlightened here. his strategy is the best ive seen.

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This is a known problem (one of my favourite)

Three men, let's call them A, B, and C, engage in a duel.

The rules:

In sequence they can shoot only once, aiming at any target of their choice

C shoots first (privilege given because he is the less skilled), then B, then A, then again C and so on, until only one survives.

The skills (average values):

A is the best, he centers the target 9 times out of 10

B centers the target 8 times out of 10

C centers the target 7 times out of 10.

Which is the best strategy for each one of them?

If they adopt the best strategy, who has the higher chance of being the survivor?

The best strategy is to shoot the opponent with the highest accuracy. (This is the only Nash equilibrium for this version of the game)

from the numbers I ran,

Probability C wins = 146776/455900 (approx 32.2%)

Probability B wins = 15440/23359 (approx 66.1%)

Probability A wins = 1 - P© - P(b) (approx 1.71%)

If people can miss intentionally....then whenever it is someone's turn (assuming infinite ammo) they should miss.

If they hit, then the odds of losing is greater than 50% (only one other person and it is their turn!), and if someone else is the one to hit first then that person will likely lose.

So the duelists will miss forever and everyone will die of old age or hunger or thirst or cancer or boredom or.......something else. :)

This is one Nash equilibrium of the game.

This reminds me of the movie WarGames......"the only way to win is not to play."

If on A's turn he thinks B will shoot him, he should shoot B and if he hits he will then have a 27/97 chance of winning (C has a 70/97 chance)

If B thinks A will shoot, then he should shoot A on his turn. If he hits then he has a 12/47 chance of winning (C has a 35/47 chance)

C never has any incentive to shoot first (unless he believes he will be shot at while A and B are alive....which goes against the numbers).

So, A missing on purpose and letting C and B kill each other is another Nash equilibrium.

Those may be the only 2 equilibria....

"Best strategy" = best response = dependent upon the actions the others will take.

What do you guys think?

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I think the old rule was that C intentionaly missed then went on to win, C is likely to miss due to nerves beeing the poorest shot.

I suggest dueling banjos, thus find DELIVERANCE

Der der dah dun dun DANG!

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If people can miss intentionally....then whenever it is someone's turn (assuming infinite ammo) they should miss.

If they hit, then the odds of losing is greater than 50% (only one other person and it is their turn!), and if someone else is the one to hit first then that person will likely lose.

So the duelists will miss forever and everyone will die of old age or hunger or thirst or cancer or boredom or.......something else. :)

This is one Nash equilibrium of the game.

This reminds me of the movie WarGames......"the only way to win is not to play."

If on A's turn he thinks B will shoot him, he should shoot B and if he hits he will then have a 27/97 chance of winning (C has a 70/97 chance)

If B thinks A will shoot, then he should shoot A on his turn. If he hits then he has a 12/47 chance of winning (C has a 35/47 chance)

C never has any incentive to shoot first (unless he believes he will be shot at while A and B are alive....which goes against the numbers).

So, A missing on purpose and letting C and B kill each other is another Nash equilibrium.

Those may be the only 2 equilibria....

"Best strategy" = best response = dependent upon the actions the others will take.

What do you guys think?

I totally agree with the reasonning here and I'm very curious to see how you got your numbers...

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This puzzle results are very interesting and explains a lot of what's happening in the world today!!

Can you guys see where I'm going with this?

Edited by roolstar

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The OP states that survival is the goal. In that case ...

Not being the first to kill someone is the best strategy.

If everyone adopts it, there is a stalemate.

If killing your opponents had been the goal, then ...

Shooting the better of your two opponents is best.

First to shoot likely will not win [1 kill but dies],

second to shoot is the likely winner [1 kill but survives] and

third to shoot [he probably won't get a chance] is the likely loser [0 kills and dies].

Can you guys see where I'm going with this?

I can only guess ... anti-war argument?

Otherwise known as the "let's you and him fight" strategy.

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This puzzle results are very interesting and explains a lot of what's happening in the world today!!

Can you guys see where I'm going with this?

WMD.. Saddam H was gonna put a shot across our bows then we all blow ourselves up.

Ossama BL should have missed!

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Being the biggest power on earth suddently seems like a big disadvantage...

But if the aim is to survive (bonanova), the best strategy would then be for ALL of them to miss ALL the time!!!! This will give them 100% chances of survival!! And we go back to the prisonners dilemna implications! And A should not return fire even if shot at...

So if we suppose that missing intentionally is an option, they need to keep this a COLD WAR!!!!!!!

And if they cannot miss intentionally, well, always shoot the strongest! Too bad!

Edited by roolstar

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I agree, but there should be a delimination of chances,

some people will say that someone could miss the first time only to increase their chances the second time Same kind of thinking in the post about twins and the odds of having a B/G

round 1

C shoots in the air, knowing he will miss. 7/10

B Shoots at A, knowing A will shoot at him Misses at 8/10

A shoots at B since its only fair to return fire. misses at 9/10

round 2

C still not savy enough to hit anyone shoots in the air again, 1 down at 7/ of the next nine

B Shoots at A hoping to hit but misses at 8/9

A Shoots at B again while cursing, and hits him since he is down to 9/9

Round 3

C has to shoot A for atleast chance, sitting at 7/8 hes got a good chance

B dead

A shoots at C and scores big.

Round 3

A goes to the bar and has a drink.

granted, some others woudl say that each time they shoot they have the same amount of chance.

but this would still bring A to win every time. He still has the best odds.

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Well any way you look at it, A is absolutely getting hosed in this deal.

To start off with, C should misfire on purpose to let A and B go at it by themselves. If he chose to shoot one of them (and actually hit), then the other man standing has a good chance of hitting C on his shot. There is a good chance that either A or B will die before it's C's turn again. In this case, C will hit the man standing 7 out of 10 times.

B should try to shoot A, and will do so 8 out of 10 times. If he hits, then C will try to shoot him, and will do so 7 out of 10 times.

B misses A 2 out of 10 times. If he does miss then A will hit him 9 out of 10 times.

If A is still alive on his turn, then he should try to shoot B. He will hit 9 out of 10 times, but then C will shoot him 7 out of 10 times. If A misses B, then C will misfire again and B will have another 8 out of 10 chance of hitting him. This will keep going until A or B is dead.

So overall C has the best chance of survival, and A is the least probable to survive.

The solution given by Enlightened is the one I found when I first came across this puzzle, which I think was devised to show that although in a one-to-one competition usually the best wins, when the competitors are more than two, paradoxically the less skilled might prevail. In a sense it's a metaphor of life in a high competitive society, where not always the best wins (politicians?).

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was devised to show that although in a one-to-one competition usually the best wins...

In this case if it were a one-to-one competition the less skilled player would win most of the time (that is if his chance of hitting is greater than 1/2). It does, however, show that the person given the chance to go first usually wins. So in other words, "The first shall be first, and the last shall, well, die."

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I will assume that they all miss their first shots and get better as time goes on.

So, C shoots at A (the biggest threat). B shoots at A (the biggest threat). A shoots at B (the next biggest threat).

Then C again shoots at A. B again shoots at A. A kills B.

So, C shoots at A again. And A shoots at C, but misses because he changed targets.

C kills A.

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C must keep shooting in the air. This way A and B will keep shooting at each other and then C will get first shot at the winner, and, therefore, better than 70% chance of survival.

On the other hand, if he shoots at A, for example, then if he misses -- it's the same as above. But if he hits him -- then B gets the first shot at C and C has got less than 20% chance of surviving that duel.

I'll post another version of this puzzle, which actually requires some calculations.

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The meaning of life, and all that is wonderful with the world:

As previously mentioned, of course C shoots into the air. What changes is what comes next.

B realizes what A already knows: A is totally screwed. Even if he survives until the next round, he still will have someone shooting at him with at least 7/10 chance of killing him. B knows that the most important thing for A is surviving (the problem asks who has the best chance of doing so) and so he shoots into the air as well.

This makes sense, because B knows that A won't try to shoot at anyone for fear of getting shot at right back, and will instead shoot into the air, knowing full well that C sure as heck isn't aiming anywhere lethal.

So then A shoots into the air as expected, since he doesn't want to die.

After going around again, they realize this isn't going anywhere, so they lay down their arms and together begin a glorious commune based on peace, harmony, and the mutual satisfaction of not dying. All is well with the world, and war has come to an end.

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The meaning of life, and all that is wonderful with the world:

As previously mentioned, of course C shoots into the air. What changes is what comes next.

B realizes what A already knows: A is totally screwed. Even if he survives until the next round, he still will have someone shooting at him with at least 7/10 chance of killing him. B knows that the most important thing for A is surviving (the problem asks who has the best chance of doing so) and so he shoots into the air as well.

This makes sense, because B knows that A won't try to shoot at anyone for fear of getting shot at right back, and will instead shoot into the air, knowing full well that C sure as heck isn't aiming anywhere lethal.

So then A shoots into the air as expected, since he doesn't want to die.

After going around again, they realize this isn't going anywhere, so they lay down their arms and together begin a glorious commune based on peace, harmony, and the mutual satisfaction of not dying. All is well with the world, and war has come to an end.

The meaning of life 2: The firm hand of democracy.click here and read.

Disparaging ethnic and political content removed by bonanova August 12 2008 2:35 am EDT

Clueless - before you post again,

- bn

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