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Funny Brain Teaser

  1. Aug 20, 2007 #1
    My Friend told me this one and i'm stump.

    Well anyway here it is and have fun!!

    You're a modified human with robotic parts (aka the 'cyborg') in a pistol duel with two other cyborg, making a three-way free-for-all. You have been programmed to fire pistols with an accuracy of 33%. The other two cyborgs shoot with accuracies of 100% and 50%, respectively. The rules of the duel are as follows. Each cyborg gets one shot per round. The shooting order goes from worst shooter to best shooter. Thus, you go first, the 50% cyborg goes second, and the 100% cyborg goes third; repeat until one cyborg remains. If a cyborg is killed, we skip his or her turn, obviously. What would you shoot at in round 1 to maximize your potential chances of survival over time? (Do explain your logic)

    Again i don't have an answer srry :tongue:
  2. jcsd
  3. Aug 20, 2007 #2
    you would shoot at the 100% robot. if you were to shoot at the 50% robot and succeed then you have no chance of survival beyond the 1st round!
  4. Aug 20, 2007 #3


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    Remember that if you can't shoot that well, you might not kill either one of them in the first round, and Mr 100% would use his turn (#3) to shoot at Mr 50% instead of you if he is still alive.
  5. Aug 20, 2007 #4
    he will shoot at either me or the 50% robot with a 1 in 2 probability. at worst, my odds of surviving the first round are 1 in 4. can you find an alternative that is any better? are we assuming that each robot follows its optimal strategy?

    also keep in mind: if the 50% robot is playing out his optimal strategy, then he will also use his turn to shoot at the 100% robot since he is in the same boat that I am. this makes my odds even better if i should miss (at worst, odds of survival 1/2 vs. 1/4).
    Last edited: Aug 20, 2007
  6. Aug 21, 2007 #5
    Isn't this something out of a game theory textbook?
  7. Aug 21, 2007 #6
    The question of "which robot to shoot" is pretty simple. You'll probably miss, which means it won't matter, so the question is, "what happens if I hit?" If you hit the 50% 'bot, that leaves only the 100% bot, who will simply kill you. So you'd be dumb to shoot at the 50% bot. If you hit the 100% bot, however, then you and the 50% bot have a shootout until one of you hits the other, which you at least have a CHANCE of winning.

    Question is, assuming all the cyborgs choose to shoot at a target who gives them the most favorable odds, which one has the best chance of winning the 3-way duel? (A triel?)
    Answer: 50% bot! 100% has a 2/9 chance, 50% has a 5/12 chance, and 33% has a 13/36 chance.
  8. Aug 23, 2007 #7
    I wrote a little program. I might have made an error so if someone disagrees then please say so! :)

    Assuming that the 33% and 50% bots will always try to shoot the 100% bot first, and the 100% bot will shoot the 50% bot first.

    In 10,000,000 fights,
    33% bot wins about 23.11% of the time
    50% bot wins about 36.14% of the time
    100% bot wins about 40.75% of the time

    sound about right? comments?
    Last edited: Aug 23, 2007
  9. Aug 23, 2007 #8
    no the question was right the way it was posed... the answer is that the 33% guy has the best chance of survival if he gives up his first shot (shoot at the ground or something), probability-wise his best shot is that one of the stronger robots takes out another stronger robot, and then he gets the first shot at the bot thats left... this is better for him than hitting the 100% bot, and having to duel the 50% bot 1 on 1 with the 50% bot getting the first shot.

    btw, how in the world do you get that davee?
    Last edited: Aug 23, 2007
  10. Aug 23, 2007 #9
    That's right. If the 33% bot shoots the 100% bot first, his chance of winning is 13/36 (about 36.11%); if he gives up, his chance of winning is 5/12 (about 41.67%).

    Assuming the same, the exact values are:
    33% bot wins 13/36 of the time (~36.11%)
    50% bot wins 5/12 of the time (~41.67%)
    100% bot wins 2/9 of the time (~22.22%)

    But, if the 33% bot gives up his first shoot, then:
    33% bot wins 5/12 of the time (~41.67%)
    50% bot wins 1/4 of the time (~25.00%)
    100% bot wins 1/3 of the time (~33.33%)

  11. Aug 23, 2007 #10
    As explained, the optimal strategy for every player is to shoot at whoever has the highest hit ratio. So, if the 33% bot has a choice, he should shoot the 100% bot. If the 50% bot has a choice, he should shoot the 100% bot. And if the 100% bot has a choice, he should shoot the 50% bot.

    That said:

    Subset 1:
    There's a 1/3 chance that the 33% bot and the 50% bot will miss on thir 1st shots (2/3 * 1/2), leaving the 100% bot alive. 100% bot shoots the 50% bot, guaranteeing that 50% bot is dead. So, if both the 33% and the 50% bots miss, 50% bot is GUARANTEED to lose. After this, 33% bot has a 1/3 chance to hit the 100% bot. If he hits, he wins, if he misses, then 100% bot will kill him no matter what. SO:

    The odds of 100% bot winning are 1/3 * 2/3-- IE, the chance that BOTH 33% and 50% miss their first shots, AND 33% misses his second shot. That's the only way that the 100% bot will win (assuming the other bots follow the logic outlined above): Miss, Miss, Hit, Miss, Hit. So, 100% bot has a 2/9 chance of winning.

    For this subset, 33% bot has a 1/9 chance of winning. BUT, he might win in other ways as well, other than "Miss, Miss, Hit, Hit".

    So what happens if the first two shots DON'T both miss? Well, the end result is the same: that 100% bot is dead, and the 33% bot and 50% bot are left to duel until one of them dies.

    The way I figured it out the probability is like this: I took all the remaining possibilities, figured out how probable they were, and added them up. There's better ways of doing this, I'm sure.

    Subset 2: 33% kills the 100% bot on his first shot

    Hit, Hit (50% wins, 1/3*1/2 probability)
    Hit, Miss, Hit (33% wins, 1/3*1/2*1/3 probability)
    Hit, Miss, Miss, Hit (50% wins, 1/3*1/2*2/3*1/2 probability)
    Hit, Miss, Miss, Miss, Hit (33% wins, 1/3*1/2*2/3*1/2*1/3 probability)
    Hit, Miss, Miss, Miss, Miss, Hit (50% wins, 1/3*1/2*2/3*1/2*2/3*1/2 probability)

    Probability that 33% wins and that subset 2 happens: 1/12
    Probability that 50% wins and that subset 2 happens: 1/4

    Subset 3: 33% misses his 1st shot, 50% bot kills 100% bot on his first shot

    Miss, Hit, Hit (33% wins, 2/3*1/2*1/3 probability)
    Miss, Hit, Miss, Hit (50% wins, 2/3*1/2*2/3*1/2 probability)
    Miss, Hit, Miss, Miss, Hit (33% wins, 2/3*1/2*2/3*1/2*1/3 probability)
    Miss, Hit, Miss, Miss, Miss, Hit (50% wins, 2/3*1/2*2/3*1/2*2/3*1/2 probability)

    Probability that 33% wins and that subset 3 happens: 1/6
    Probability that 50% wins and that subset 3 happens: 1/6

    So, overall:
    33% bot: 1/9+1/12+1/6 = 13/36 (~36.111%)
    50% bot: 1/4+1/6 = 5/12 (~41.667%)
    100% bot: 2/9 (~22.222%)

    Of course, if any of the bots has the *option* of intentionally missing (I was assuming that they don't have that option), that changes things around. I'm not sure if it's in the 50% bot or 100% bot's best interests to miss, but it IS in the 33% bot's best interests to miss.

    If the 33% bot intentionally misses on his first shot (and is guaranteed to successfully intentionally miss), then he's the new likely victor. Instead, the odds come out slightly differently:

    33% bot: 5/12 (~41.667%)
    50% bot: 1/4 (25%)
    100% bot: 1/3 (~33.333%)

  12. Aug 23, 2007 #11
    I see, thank you... that was informative =)
  13. Aug 24, 2007 #12
    wow~, you guys rock, so basically the answer to this is that on the first round the 33% bot should intentionally miss so that his survival rate will go up?
  14. Aug 25, 2007 #13
    Yes, thats it!
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