3 Man Duel (logic problem)

[Lestat]

Someone sent this to me today and I thought it was kinda fun if not a little on the easy side.

*********

You and two other men are having a duel with pistols.



You are the worst shot and hit your target only one time in three, thus you are allowed to shoot first.



Mr. Second Best hit’s his target one time in two and gets to shoot second.



Mr. Best is a crack shot and hits his target every time.



Shooters will shoot in rotation until one man is alive.



You shoot first. What is your strategy and why?

[cambraceres]

duh

[Uethym]

If I shoot at Mr. Best and hit, Mr. Second has a 50% chance of beating me on the next shot.

If I shoot at Mr. Second and hit, I lose immediately.

And if I miss, well, it didn't really matter who I shot at, did it? Either way, there can only be one opponent by the time it gets back to me (if Best doesn't die, he'll hit Second for sure), and there are no more options.

So it seems clear that I go for Mr. Best.

Edit: Unless...(white text)

<font color="white">I intentionally miss. Then either Second kills Best and it's my turn again, or Second misses, Best kills Second, and it's my turn again. Either way, I'm guaranteed the first shot once I'm down to one opponent.
</font>

Clever puzzle, I like that.

[Phil153]

[ QUOTE ]
duh

[/ QUOTE ]
So what's your answer? I agree it's easy but it's not THAT easy.

<font color="white">
Answer: abstain from shooting until one dies.

Shooting the 1/2 is certain death
Shooting the crack shot will give you 9/12 chance of dying if you hit him, 7/12 chance of dying if you miss him.

Aiming at the crack shot gives:
1/3 of the time we hit, which is followed by a 50% and a shoot-off with #2.
2/3 miss = 0 chance of dying in the first round.
In the second round:
- when crack lives (50%), he shoots the 1/2 and we have a 2/3 chance of dying.
- when crack dies (50%), the chance of dying is:
--0.5*2/3 on the first round
--0.5*2/3*2/3*0.5 on the second
--sum(0-&gt;n)[1/3^n] of dying on the nth round = 0.5 at infinity.

</font>

[J. Stew]

funny how people say to intentionally miss. it's like how if you ask a room full of people if you are a better driver than 50% of the drivers out there, more than 50% of the people will raise their hands. go ahead and aim for mr. 2, you're already a [censored] shooter.

[tolbiny]

[ QUOTE ]


You shoot first. What is your strategy and why?

[/ QUOTE ]



<font color="white"> Miss your first shot. </font>

[kidmarine]

RUN! Yer gonna die!

[Borodog]

[ QUOTE ]
RUN! I'm not an idiot.

[/ QUOTE ]

[TimWillTell]

I ran into this problem like 30 years ago, but I remember it a little different.

You would have the choice of being the best shooter, the second best or the worst.
Its not difficult, but when your young the answer can broad your horizon.

[robert jordan]

So whats the diffinitive correct answer?

[bkholdem]

I say shoot at Best. When second shoots he is likely to shoot at Best as well (or i will yell to him why he should if he doesn't) because best is going to kill him next since he is the bigger threat to best. Also, if 2nd best shoots me he is dead when best gets his turn.

[jogger08152]

If you must shoot first, you must shoot at Mr. Best. (It would be best to deliberately flub the first shot, however.)

Outcomes:

You act first and shoot at Mr. 2. You hit, he dies, then Mr 1 shoots and you die.

You act first and shoot at Mr. 1. You succeed, and Mr. 2 gets a shot at you (50/50). In this instance, he has a 50% chance to kill you outright, and if he fails, he will be a 60/40 favorite to kill you thereafter. Thus you will be about an 80/20 dog to survive.

You act first and miss whoever you shot at, and then...

Mr. 2 shoots at Mr. 1 and hits: you have a 33% chance to win immediately, and if you fail, you are a 40/60 dog to win, thus making you a 60/40 overall favorite to survive, or...

Mr. 2 shoots at Mr. 1 and misses: Mr 1 retaliates, killing Mr. 2. You then have a 33% chance to survive.

Best regards,
Jogger

[jogger08152]

In retrospect, I believe I did this math incorrectly. I'll have to think about it some more, but 60/40 in the best case is striking me as incorrect upon further consideration.

If I get the first shot heads up, it would be:
33% - contest over and I win, or
67% - contest continues, and then, if it does,
50% it ends and I lose, 16.7% it continues and I win (this is the 60/40 I mentioned above: (33.3% + 16.7% vs. 33%)) without having to brave another shot, 33% it continues with me a 3-to-2 dog. Ugh.

[robert jordan]

[ QUOTE ]
If you must shoot first, you must shoot at Mr. Best. (It would be best to deliberately flub the first shot, however.)

Outcomes:

You act first and shoot at Mr. 2. You hit, he dies, then Mr 1 shoots and you die.

You act first and shoot at Mr. 1. You succeed, and Mr. 2 gets a shot at you (50/50). In this instance, he has a 50% chance to kill you outright, and if he fails, he will be a 60/40 favorite to kill you thereafter. Thus you will be about an 80/20 dog to survive.

You act first and miss whoever you shot at, and then...

Mr. 2 shoots at Mr. 1 and hits: you have a 33% chance to win immediately, and if you fail, you are a 40/60 dog to win, thus making you a 60/40 overall favorite to survive, or...

Mr. 2 shoots at Mr. 1 and misses: Mr 1 retaliates, killing Mr. 2. You then have a 33% chance to survive.

Best regards,
Jogger

[/ QUOTE ]



I like.

[Silent A]

You act first and hit best shooter. 2nd best aims at you. If he hits you die (50%) if he misses you shoot. Your chances of winning are now:
1/3 + 1/9 + 1/27 ... = 1/2. Total chance of survival for you = 1/4. Chance of survival for him = 3/4.

You act first and hit 2nd shooter. 1st kills you. Best always wins.

You act first and miss. 2nd aims as 1st and kills 50%. You then aim at him and win 1/3 + 1/9 + ... = 1/2 of the time. In this scenario you win 50% and 2nd wins 50%.

You act first and miss, and 2nd misses 1st. 1st kills 2nd. You have 1 shot to kill 1st and win 1/3. Best wins 2/3.

If you aim at #1, you win = (1/3)(1/4) + (2/3)(1/2)(1/2) + (2/3)(1/2)(1/3) = 13/36
#1 wins (2/3)*(1/2)*(2/3) = 8/36
#2 wins (1/3)(3/4) + (2/3)(1/2)(1/2) = 15/36

If you aim at #2, you win = (2/3)(1/2)(1/2) + (2/3)(1/2)(1/3) = 10/36
#1 wins = 1/3 + (2/3)(1/2)(2/3) = 20/36
#2 wins = (2/3)(1/2)(1/2) = 6/36

If you deliberately miss, you win = (1/2)(1/2) + (1/20)(1/3) = 15/36
#1 wins = (1/2)(2/3) = 12/36
#2 wins = (1/2)(1/2) = 9/36

Therefore, you should deliberately miss and you have the best chance to win when you follow the correct strategy.

[J. Stew]

If you can deliberately miss your first shot, why can't others do the same, and keep missing until all bullets run out and everybody stays alive? Or you could just not shoot and run away giggling like a sissy.

You must aim naturally . . .

[vhawk01]

[ QUOTE ]
If you can deliberately miss your first shot, why can't others do the same, and keep missing until all bullets run out and everybody stays alive? Or you could just not shoot and run away giggling like a sissy.

You must aim naturally . . .

[/ QUOTE ]


Because that would be stupid. Try doing the optimal strategy calculation for Guy #2 or Guy #3. If Guy#2 intentionally misses, he is dead, because there is no incentive whatsoever (and a strong disincentive) for #3 to ever try to miss. #2 must shoot at #3 in order to have any chance of survival.

[Lestat]

You guys are good. Shoot your first shot into the air is the correct answer. Just like in poker, being second best sucks.

[rebuyboy]

[ QUOTE ]
If you can deliberately miss your first shot, why can't others do the same, and keep missing until all bullets run out and everybody stays alive? Or you could just not shoot and run away giggling like a sissy.

You must aim naturally . . .

[/ QUOTE ]

if you must aim naturally, i think it is well established that the answer is shoot #3.