Brain Teaser: A Duel

Water Cooler
#21

LOL

#22

Just googled it. Yeah, like I said, retarded.

#23

Slow Friday, this is as far as I got:

Good job laying it out, but you need to think it through all the way. One change, it says one shot per shooter per round, so a shooter can’t not shoot. The shooter must shoot and miss. If you hit 100 then “30vs50” is the geometric sum of the probability of being shot by “50”, i.e. 0.5/0.67=0.75. Let PD = Probability of Death. So:

33’s Turn

  1. PD(Shoot 100) = 0.33*0.75 + 0.67*(50’s Turn) = 0.25 + 0.67*(50’s Turn)

  2. PD(Shoot 50) = 0.33 + 0.67*(50’s Turn)

  3. PD(Shoot No one) = (50’s Turn)

Therefore you are bound to choose 1 or 3, since two is clearly inferior to both. The question is what is the probability of you dying given that you shoot no one or me (“50’s Turn” is the same as the probability you die if you miss). “100” is certain to shoot because he can only improve his odds of survival, below:

50’s Turn

  1. PD(Shoot 100) = 0.5*(0.33/0.67) + 0.5*(100’s Turn) = 0.25 + 0.5*(100’s Turn)

  2. PD(Shoot 33) = 0.5 + 0.5*(100’s Turn)

  3. Shoot No one = (100’s Turn)

What would “50” think that the value for “100’s Turn” would be (i.e. what is the probability that he dies if he misses)? “50” would think, correctly:

100’s Turn

  1. PD(Shoot 50) = 0.33

  2. PD(Shoot 33) = 0.5

Therefore “100” will clearly choose 2.

So, “50” would assume the probability of death given he misses is 100%.

50’s Turn – Given Info on “100”

  1. PD(Shoot 100) = 0.5*(0.33/0.67) + 0.5*(100’s Turn) = 0.25 + 0.5*(1) = 0.75

  2. PD(Shoot 33) = 0.5 + 0.5*(100’s Turn) = 1

  3. PD(Shoot No one) = (100’s Turn) = 1

“50” would choose 1. Now, back to 33’s turn…all we were missing was the value for “50’s Turn.”

#24

“I wouldn’t shoot anyone the first round”

since you are only 33% accurate, have you considered the possibility that you may accidentally shoot someone even though you weren’t aiming at them…

#25

It’s a trick question, 50 can’t die unless you shoot him more than 9 times.

#26

In most scenarios, max[P(winning)]=max[P(all competitors are shot and I’m alive)]=max[P(not dying)]. But in the degenerate equilibrium scenario where everyone shoots into the ground, P(winning)=0% and P(not dying)=100%. If the competitors’ goal is to maximize the probability of not dying, you can’t improve upon 100% and one can show even more rigorously that the dominating strategy for all is to shoot into the ground.

If the goal is to maximize the probability of winning, then the 100% guy must shoot at the 50% guy if there are two people left in front of him rather than shoot into the ground. Clearly the only way for him to improve his probability of winning by shooting into the ground is if he expects the 50% and 33% guys to shoot at each other. However, their common dominating strategy is (shoot in the ground, shoot the 100% guy) - i.e. they will never shoot at each other as long as the 100% guy is still around.

It seems that the strategy of the 33% guy is to shoot into the ground regardless of whether the goal of the game is to maximize probability of winning, or probability of not dying. However, the other players’ optimal strategy would depend on which goal they are pursuing…

#27

Well if the first shooter doesn’t shoot at all… I’m pretty sure he won’t hit anyone

#28

You kill your own horse, thus making the lion eat your horse as you run away. You don’t have to be the fastest runner, you just have to be faster than the slowest runner

#29

I guess the constraint with killing your own horse is that your kung fu must be strong enough to kill a horse with your bare hands, but not strong enough to kill a lion. If you had say, a gun which you could use to kill the horse, then you should just shoot the lion or the kangaroo.

#30

well, you don’t have to kill the horse. Just attack the balls or something