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   Author  Topic: three-way pistol duel  (Read 55564 times)
Keith H
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Re: three-way pistol duel  
« Reply #50 on: Feb 8th, 2005, 6:59pm »
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If intentional missing is allowed, A should shoot at nothing to win with probability 5/12. If it is not allowed, A should shoot at C to win with probability 13/36.
 
First off, let us define some things:
 
A, B and C are as before. X, Y, Z, and W are metasyntactic variables,
and can be any of A, B, or C.
 
P_X(YZ) is the total probability of a win for robot X, given that
there are two robots Y and Z remaining, and it is time for robot Y to
shoot. So, obviously,  
 
P_A(BC) = 0 and so forth. Also,  
P_X(CX) = 0 and
P_C(CX) = 1.
 
Let's work these out.
 
For P_A(AB), we add up the probability of winning on each round (when
we shoot) times the probability of making it to that round. So for
round 1, the probability that we get there is 1 and the probability of
winning is 1/3. For round 2, the probability that we get there alive
is 2/3*1/2 (2/3 of the time B gets a shot, and 1/2 of that time it
misses). The probability of winning on round 2 is again 1/3. For round
3, we have to get to round 2 (probability 2/3*1/2) and then get to
round 3 from there (probability 2/3*1/2 again). Again we win there
with probability 1/3.
 
P_A(AB) = 1/3 + (2/3*1/2)*1/3 + ((2/3*1/)2*(2/3*1/2))*1/3...
   = 1/3 + 1/(3**2) + 1/(3**3)....   # by fraction simplification
 = 1/2      # by infinite summation
P_B(AB) = 1/2      # by subtraction
P_A(BA) = 1/2*P_A(AB) = 1/2*1/2 = 1/4     # should be obvious why
P_B(BA) = 3/4
P_A(AC) = 1/3
P_B(BC) = 1/2
P_C(AC) = 2/3
P_C(BC) = 1/2
 
Now, let's introduce another notation: P_X(WYZ) is the probability
that X will win with three robots left, firing in order W -> Y -> Z.
We cannot calculate these directly without knowing what strategy each
robot will take on his turn.
 
One final quantity: P_X(XYZ, W) is the probability that X will win
given that it is X's turn and he chooses to shoot (W can be X, Y, or
Z). I'll ignore the option to consider survival winning, because it is
trivial to realize that if all robots choose this strategy, they all
have probability 1 of survival by never shooting at any robot. Since
they are smart enough to be crafting strategies, they can figure this
out. So, in other words, let's only talk about WINNING, which requires
being the unique survivor.
 
Now let's start with C's choices:
 
P_C(CAB, A) = 1*P_C(BC) = 1/2
P_C(CAB, B) = 1*P_C(AC) = 2/3
P_C(CAB, NONE) = P_C(ABC)
 
For now, let's assume P_C(ABC) is less than 2/3. We will come back to
this later.
 
So, C will choose to shoot B in that case, to maximize his chance to
survive. Note that this means that  
 
P_B(CAB) = 0
P_A(CAB) = P_A(AC) = 1/3
 
Now let's look at B's choices:
 
P_B(BCA, C) = 1/2*P_B(AB) + 1/2*P_B(CAB) = 1/2*1/2 + 1/2*0 = 1/4
P_B(BCA, A) = 1/2*P_B(CB) + 1/2*P_B(CAB) = 0 + 0 = 0
P_B(BCA, NONE) = 1*P_B(CAB) = 0
 
So B will choose to shoot C, to maximize his chances of
survival. Remember, we are still assuming that P_C(ABC) < 2/3.
 
Finally, we look at A's choices:
 
P_A(ABC, B) = 1/3*P_A(CA) + 2/3*P_A(BCA) = 2/3*P_A(BCA)
P_A(ABC, C) = 1/3*P_A(BA) + 2/3*P_A(BCA)  
     = 1/3*1/4 + 2/3*P_A(BCA)
     = 1/12 + 2/3*P_A(BCA)
P_A(ABC, NONE) = 1*P_A(BCA).
 
What exactly is P_A(BCA)?
 
Well, we know B will shoot C:
 
P_A(BCA) = 1/2*P_A(AB) + 1/2*P_A(CAB)  
  = 1/2*1/2 + 1/2*P_A(AC)
  = 1/4 + 1/2*1/3 = 5/12
 
Well, now we can generate
 
P_A(ABC, B)  = 2/3*5/12 = 5/18  
P_A(ABC, C)  = 1/12 + 5/18 = 13/36
P_A(ABC, NONE)  = 5/12
 
That means A should shoot at nothing, if that is allowed. If A shoots
at nothing, then
 
P_C(ABC) < 7/12 (since P_C(*) + P_A(*) <=1), which satisfies our
assumption. So there is no contradiction generated by that assumption.
 
 
What if P_C(ABC) > 2/3? Then C will shoot the ground, and B has to
reexamine his options. Let's say that B decides to miss
intentionally. Then C must miss intentionally (by assumption) and A
is the only one who can shoot. Will she? Well, if she doesn't, she has
 
P_A(ABC) = 0 because nobody else will shoot either. So she must
shoot. Whom will she shoot? She'll shoot at C (since a hit puts her in
a 1-on-1 and her best odds (by far) are with P_A(BA)). As long as she
misses, B and C will pass. So  
 
P_C(ABC) = 0 with this strategy, since C cannot survive. That's a
contradiction, so we must take back our last assumption that B misses
intentionally. Thus, B must choose between shooting A and C. It is
fairly obvious that shooting at C is the best bet.
 
Uh oh. What does A do? If she does nothing, B will just keep shooting
at C until he hits him, making for  
 
P_C(ABC) = 0 again. So A has to shoot somebody. Obviously, shooting at
C is her best bet. But now C cannot possibly win, since he is just
letting A and B fire at him, paralyzed by our assumption that  
 
P_C(ABC) > 2/3. Obviously, that assumption has led to this
contradiction and must be abandoned. C will choose to shoot at B, B
will choose to shoot at C, and A should miss intentionally. A's chance
of survival is P_A(ABC, NONE) = 5/12.
 
If intentional missing is not allowed, then A should shoot at C to win with probability
 
P_A(ABC, C) = 13/36, which is only 1/18 worse.
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Keith H
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Re: three-way pistol duel  
« Reply #51 on: Feb 8th, 2005, 7:10pm »
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Small correction:
 
 
------------------------
 One final quantity: P_X(XYZ, W) is the probability that X will win
given that it is X's turn and he chooses to shoot W (W can be X, Y,  Z, or NONE).  
 
I'll ignore the option to consider survival winning, because it is
trivial to realize that if all robots choose this strategy, they all
have probability 1 of survival by never shooting at any robot. Since
they are smart enough to be crafting strategies, they can figure this
out. So, in other words, let's only talk about WINNING, which requires
being the unique survivor.
 -----------------------------------
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jackdhammer
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Re: three-way pistol duel  
« Reply #52 on: May 11th, 2005, 11:50am »
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I know I'm kind of rehashing this but I liked this riddle.  
 
It appears to me that in order to maximize your chance of survival over time you need to kill robot C.  
 
He is the only one that hits 100% of the time. So I think your first shot should be at him.  
The problem doesn't say that the cyborg wil shoot back at whoever shot at it. So you really have to assume that each cyborg is trying to maximize his chance of survival. After all, why wouldn't they?  
 
So if you shoot at Cyb C if you miss and cyb B miss you will still have one more shot since he would logically shoot at cyborg B first since he has a greater chance of hitting than you do. If you happen to hit cyborg C than you are in a shoot out with cyborg B who only has a 50% chance of hitting you.  
 
So taking your first shot at C really increases your chance of long term survival instead of taking 2 shots you get to take 3 and you may not have to deal with shooter B at all.  
 
The reason I thought about this was something a favorite math teacher said when I was young.  
He said that even though there is a 50% chance to flip a coin and get heads, you could potentially flip the coin a hundred times and get tails. The odds don't really play themselves out over a short span.  
I would rather take the shot at C and get that extra chance to kill him and then flip the coin with B and hope I get a lot of Tails  Wink
« Last Edit: May 11th, 2005, 11:51am by jackdhammer » IP Logged
rmsgrey
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Re: three-way pistol duel  
« Reply #53 on: May 12th, 2005, 5:31am »
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on May 11th, 2005, 11:50am, jackdhammer wrote:
I know I'm kind of rehashing this but I liked this riddle.  
 
It appears to me that in order to maximize your chance of survival over time you need to kill robot C.  
 
He is the only one that hits 100% of the time. So I think your first shot should be at him.  
The problem doesn't say that the cyborg wil shoot back at whoever shot at it. So you really have to assume that each cyborg is trying to maximize his chance of survival. After all, why wouldn't they?  
 
So if you shoot at Cyb C if you miss and cyb B miss you will still have one more shot since he would logically shoot at cyborg B first since he has a greater chance of hitting than you do. If you happen to hit cyborg C than you are in a shoot out with cyborg B who only has a 50% chance of hitting you.  
 
So taking your first shot at C really increases your chance of long term survival instead of taking 2 shots you get to take 3 and you may not have to deal with shooter B at all.  
 
The reason I thought about this was something a favorite math teacher said when I was young.  
He said that even though there is a 50% chance to flip a coin and get heads, you could potentially flip the coin a hundred times and get tails. The odds don't really play themselves out over a short span.  
I would rather take the shot at C and get that extra chance to kill him and then flip the coin with B and hope I get a lot of Tails  Wink

 
If you kill C, then you face a duel with B with him taking the first shot - half the time you're dead; the other half, you get into a duel with him where you take the first shot.
 
If you don't kill C, half the time you end up in a duel with B where you take the first shot; the other half you get a shot at C with a 1/3 chance of killing him and a 2/3 chance of dying.
 
If you compare the two situations, in each case, half the time you have first shot in a duel with B, so it's the other half the time that decides which is better for you - if you kill C, then you die the rest of the time, while not killing C means you only die most of the rest of the time...
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Re: three-way pistol duel  
« Reply #54 on: May 12th, 2005, 10:13am »
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on May 12th, 2005, 5:31am, rmsgrey wrote:

means you only die most of the rest of the time...

lol. thats awesome.
 
 
 
But if you don't kill C and B does'n't kill C than you only have 2/3 of a chance to live. Where as if you do kill C you have a 50/50 chance to live. I see that by shooting at neither you would get a "free" round where no one would shoot at you and your chance of survival is 100% for that round, but for long term survival wouldn't you be better off trying to take out the 100% shooter and take your chances with the 50/50 guy?
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Re: three-way pistol duel  
« Reply #55 on: May 12th, 2005, 1:42pm »
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The enemy of my enemy is my friend. In other words, as long as both B and C are alive, they will only shoot at an enemy of yours, so why kill one of your friends? Let them fight it out, then you begin a duel with the survivor where you get first shot.
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Re: three-way pistol duel  
« Reply #56 on: May 12th, 2005, 2:15pm »
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on May 12th, 2005, 1:42pm, asterex wrote:
The enemy of my enemy is my friend. In other words, as long as both B and C are alive, they will only shoot at an enemy of yours, so why kill one of your friends? Let them fight it out, then you begin a duel with the survivor where you get first shot.

 
Yes but the enemy of your enemy in this case is also YOUR enemy. By not shooting at C you are only buying your self one extra round and reducing the chance that he will be dead before he has a chance to kill you. For long term survivabitiy you have to get rid of the 100% shooter as soon as possible.  
Let me elaborate.  
Worse case senario:
If you don't shoot at C and then B misses C you have a 66% chance of guarnteed death. If you miss C on your turn Game over.  
However, if you shoot C or B shoots C theoretically speaking the game could go on for a while. It may take 1million shots before Bs 50% chance of hitting you plays itself out in his favor. He could go on a run of 1000  misses.  
The point is while it isn't likely, you at least have a chance (a 50% chance to be exact  Wink ) to live long enough to shoot him first, where you only have a very limited number of chances to kill C. So why limit that number (and your chances of survival) even more by not shooting at him with your first shot?
« Last Edit: May 12th, 2005, 2:27pm by jackdhammer » IP Logged
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Re: three-way pistol duel  
« Reply #57 on: May 12th, 2005, 6:46pm »
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on May 12th, 2005, 10:13am, jackdhammer wrote:

lol. thats awesome.
 
 
 
But if you don't kill C and B does'n't kill C than you only have 2/3 of a chance to live. Where as if you do kill C you have a 50/50 chance to live. I see that by shooting at neither you would get a "free" round where no one would shoot at you and your chance of survival is 100% for that round, but for long term survival wouldn't you be better off trying to take out the 100% shooter and take your chances with the 50/50 guy?

Scenario A - Do not kill C immediately
1/2 - B kills C. Scenario C starts.
1/6 - C kills B and you kill C.
1/3 - C kills B and C kills you.
 
Scenario B - Do kill C immediately
1/2 - B kills you immediately.
1/2 - B misses you initially. Scenario C starts.
 
Scenario C - C is gone and it's your shot
1/3 - you kill B immediately.
1/3 - you miss B and B kills you.
1/3 - both miss. Scenario C restarts.
 
 
In scenario B, half the time you die before you can find yourself in scenario C.
 
In scenario A, 1/3 of the time you die and 1/6 of the time you win before you can find yourself in scenario C.
 
In both scenarios A and B, you find yourself in scenario C half the time.
 
 
Essentially, you'd rather have B use his first shot trying to kill C than spend it trying to kill you - if B fails to kill C (worst case), you still have a chance to kill C yourself, while if B fails to miss you (worst case) you have no chance to win. Since the worst case outcomes of B's first shot are equally likely between the two scenarios, and the best case outcome is the same situation and equally likely between the two scenarios, you prefer the scenario where you can survive the worst case...
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Re: three-way pistol duel  
« Reply #58 on: May 13th, 2005, 9:17am »
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on May 12th, 2005, 6:46pm, rmsgrey wrote:

Scenario A - Do not kill C immediately
1/2 - B kills C. Scenario C starts.
1/6 - C kills B and you kill C.
1/3 - C kills B and C kills you.
 
Scenario B - Do kill C immediately
1/2 - B kills you immediately.
1/2 - B misses you initially. Scenario C starts.
 
Scenario C - C is gone and it's your shot
1/3 - you kill B immediately.
1/3 - you miss B and B kills you.
1/3 - both miss. Scenario C restarts.
 
 
In scenario B, half the time you die before you can find yourself in scenario C.
 
In scenario A, 1/3 of the time you die and 1/6 of the time you win before you can find yourself in scenario C.
 
In both scenarios A and B, you find yourself in scenario C half the time.
 
 
Essentially, you'd rather have B use his first shot trying to kill C than spend it trying to kill you - if B fails to kill C (worst case), you still have a chance to kill C yourself, while if B fails to miss you (worst case) you have no chance to win. Since the worst case outcomes of B's first shot are equally likely between the two scenarios, and the best case outcome is the same situation and equally likely between the two scenarios, you prefer the scenario where you can survive the worst case...

 
Really good answer (imo) but this is starting to feel like a high road low road type of deal.
 
I feel that the only way to have any long term survival is to kill C. So in my opinion I would rather have as many shots as I can at him since neither I nor the other cyborg have 100% accuracy. I know those guys said that mathmatically speaking you should shoot at neither but when it comes to odds numbers don't always play themselves out like you think they should. Anyone who has been to Vegas and played black jack knows that. Hell, you can flip a coin a bunch of times to see how 50/50 doesn't guarantee even results. However the more times you flip, the better your chances of getting what you need.  
 
Thanks for humoring me on this one guys/gals. I haven't had any intelligent conversation with anyone in sometime so this is quite refreshing.
« Last Edit: May 13th, 2005, 9:24am by jackdhammer » IP Logged
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Re: three-way pistol duel  
« Reply #59 on: May 13th, 2005, 9:42am »
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Hmm, the way I interpret it it's not about "playing the odds" Vegas style so much as about getting the first shot when it's down to two.
 
You're the least-accurate dueler, so as long as all three remain the other two will duke it out and leave you alone--I think everybody agrees on that much at least.
 
Now if you shoot at C and hit him, it's down to just you and B, and B has the first shot, giving him a three-to-one advantage over you.
 
If, however, you intentionally miss and leave B and C to duke it out first, you have the first shot against whoever survives.  That gives you even-odds against B and C is only two-to-one over you--either way better than your chances with B (or, worse, C) firing first against you.
 
The advantage of having the first shot in the two-way duel outweighs the advantage of firing at C as soon as possible.
 
--SMQ
« Last Edit: May 13th, 2005, 9:45am by SMQ » IP Logged

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Re: three-way pistol duel  
« Reply #60 on: May 16th, 2005, 4:55am »
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In the possible duel scenarios:
 
Taking first shot against C, you die on his first shot 2/3 times, and win the rest.
 
Taking second shot against B, you die on his first shot only half the time, but after he's had two shots at you, your chances are down to 1/3 of even being alive - you only have a 1/6 chance of having won before B gets a second shot, and only a 1/12 chance of winning at some point after B's second shot (or a 1/4 chance of winning overall)
 
Taking first shot against B, you win on the first shot as often as against C (1/3) and have a good chance of winning later (1/6 - 1/2 overall)
 
 
Even if each of you only has 2 bullets, your chances of survival are no better if you kill C than if C kills B - and if you don't kill C, you also have the chance of getting the best-case result of B killing C.
 
If you have more than 2 bullets each, your chances of survival with C dead go down, while your chances of survival with C killing B remain the same - either C gets his second shot and wins, or you win with your second shot - either way, having spare bulets doesn't change anything, so shooting C looks worse the more bullets you have.
 
 
 
The trouble with having a long shoot-out with B is that, while it gives you more chance to get lucky and hit, it also gives him more chances to get lucky and hit, and he's more likely to get lucky than you are - the only way you can negate that advantage is by getting more shots off than him - since you can have at most one extra shot (by shooting first), the longer the duel, the less that advantage counts for - your 3 shots to his 2 early on is a lot more of an advantage than 1001 to his 1000 if it drags on a bit. If you and B were to replace your lethal weapons with paintball guns and fire an agreed number of rounds at each other, the winner being the one who scores the most hits, then your best chance would be to make the number of rounds as small as possible - the more rounds you play, the closer the end results become to the "average" ones of him hitting three times for every twice you manage...
 
Another possible variant to give you a feel for how things work out: you and B duel with paintball guns again, but one of you gets an extra shot every time you hit - whoever would have had first shot in the live-fire duel. It should be fairly obvious that if B gets the extra shots, he absolutely slaughters you, while, if you get the extras, things are a lot closer. This effectively breaks up the paintballing into repeated "first to hit" duels with the same person starting each one - effectively playing out the same duel many times - and the end result is the same as the result of trying the normal duel that many times.
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Re: three-way pistol duel  
« Reply #61 on: May 19th, 2005, 10:42am »
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But thats my point exactly, you said "a long shoot out with B" I understand it gives him more chances to hit you. But at least you have a chance and a "long shoot out with B" is possible. There is no chance for a long shoot out with C. From the way the riddle is worded you feel like you really are SOL but you are looking for the shot that will give you the best chance of survivng longer. If that is the case then getting the extra shot on C is your best bet. If B misses you have a 1/3 chance of living because on the next turn you die. If you happen to kill C then you have a 50/50 chance of living and then a 1/3 chance of winning.
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Re: three-way pistol duel  
« Reply #62 on: May 20th, 2005, 6:32am »
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on May 19th, 2005, 10:42am, jackdhammer wrote:
But thats my point exactly, you said "a long shoot out with B" I understand it gives him more chances to hit you. But at least you have a chance and a "long shoot out with B" is possible. There is no chance for a long shoot out with C. From the way the riddle is worded you feel like you really are SOL but you are looking for the shot that will give you the best chance of survivng longer. If that is the case then getting the extra shot on C is your best bet. If B misses you have a 1/3 chance of living because on the next turn you die. If you happen to kill C then you have a 50/50 chance of living and then a 1/3 chance of winning.

If you work it through, after everyone has had up to two shots:
 
Killing C with your first shot:
 
B has won 2/3 of the time
You have won 1/6 of the time
The duel continues 1/6 of the time
 
Missing C with your first shot:
 
C has won 1/3 of the time
B has won 1/6 of the time
You have won 1/3 of the time
The duel continues 1/6 of the time
 
 
If you compare the possible durations:
 
Killing C with your first shot:
 
dead on the 2nd shot 1/2
dead on the 4th shot 1/6
dead on the 6th or later shot 1/12
survive indefinitely 1/4
 
Missing C with your first shot:
 
dead on the 4th shot 1/6
dead on the 5th shot 1/3
dead on the 6th or later shot 1/12
survive indefinitely 5/12
 
Killing C first shot not only means that you're less likely to actually win the duel, but also that you're likely to die sooner than if C lives - half the time you don't even get a second shot.
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Re: three-way pistol duel  
« Reply #63 on: May 21st, 2005, 2:25am »
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Jackdhammer, it seems that you accept the mathematics as correct, but are unwilling to accept it based on a logical argument.  Look at it this way:
 
Let's say instead that C has a 100% chance of killing someone when he shoots, and B has a 99% chance.  Does your argument still hold in this case?  But we can see that, if we shoot at C and kill him, we obviously have a less than 1% chance of surviving the duel (it's actually a little above 0.3 %).  Whereas, if we wait until someone dies, we have at least a 1/3 chance of killing the remaining person and winning the duel.  So there's no question on what you should do here.
 
Now, you might say that this situation is different, since B has a better chance of killing you.  This is true;  but where does that 50% versus 99% chance appear in your argument?  If you insist that your argument applies to 50% and not to 99%, what is the reason for this?  At what percentage does the argument switch from applying to not applying?
 
It should be clear that, since your logical argument is not about the numbers, it has no way of determining exactly the percentage at which C's strategy should change.  So really, you can't determine the correct strategy from an argument like that;  at best you can have a reasonable heuristic.
 
I ran into a similar situation when some people told me the optimal strategy for blackjack couldn't possibly be right;  they tried to convince me with an argument that didn't take into account all the mathematical details, just some round numbers.  But that doesn't work;  the reason it is better to hit on a 12 when dealer shows 2 is that, when you crunch the numbers on the millions of possible outcomes, you get that hitting returns 37% on the average, and standing returns 35%.  No logical argument can deduce that.
 
Quote:
Really good answer (imo) but this is starting to feel like a high road low road type of deal.  
 
I feel that the only way to have any long term survival is to kill C. So in my opinion I would rather have as many shots as I can at him since neither I nor the other cyborg have 100% accuracy. I know those guys said that mathmatically speaking you should shoot at neither but when it comes to odds numbers don't always play themselves out like you think they should. Anyone who has been to Vegas and played black jack knows that. Hell, you can flip a coin a bunch of times to see how 50/50 doesn't guarantee even results. However the more times you flip, the better your chances of getting what you need.  

 
I'm afraid I don't see the point you are trying to make.  It's true that you aren't usually guaranteed to get the expected value;  that's why we're talking about probability in the first place!  The best we could have hoped to do is maximize C's chances of survival, not guarantee his survival.  So how can this be an objection to the solution?  And what do you mean by "high road low road type of deal"?
« Last Edit: May 21st, 2005, 4:16am by Deedlit » IP Logged
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Re: three-way pistol duel  
« Reply #64 on: Apr 25th, 2006, 8:17pm »
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I'm new here, I just stumbled across this thread, I'm sure most of the principles have long gone but I wanted to post an idea:
 
The previous calculations were correct, but it doesn't seem that everyone feels that good about the answer.  If multiplying the probabilities doesn't make intuitive sense to you, try this:
 
Start at the end, and work backwards.  No real math involved.
If I type AB, that means A shoots at B first, BA means it's B's turn to shoot at A.
 
If A is dead:
1. BC = 50% win for B
2. CB = 100% win for C
 
If B is dead:
3. AC 33% win for you
4. CA 100% win for C
 
If C is dead:
5. BA = you win 1/4 of the time
6. AB = you win 1/2 of the time
but even though I calced the #s on this one, you actually don't have to worry about the exact probability!  the only factor is that you can see BA is not as good as AB.  This should be clear because if you go first you win 1/3 plus a chance if B misses.  If B goes first, he wins 1/2 the time, plus half of the times you miss.
 
Ok, back a step.
 
A + B still alive, who does C shoot?  C's decision is easy because he has 100% accuracy, so he just looks at cases 1 and 3 above.  You win 33% and B wins 50%, so C will shoot B for a greater chance.
 
A + C still alive, who does B shoot?  B knows that C goes next, and will 100% shoot him, based on C's decision above.  So B MUST shoot at C.  If B misses, C will shoot B, and you go to case 3.  If B succeeds, you go to case 6.
 
So who do you shoot?  You can see that if you miss, it just devolves into the above case with everyone still alive.  So the only case you have to work out is if you're successful.  If you get B, you're 100% dead.  If you don't, you have a chance.  If you HAVE to shoot someone, it should be C.
 
But if you don't have to shoot someone?  The trick here is that if one of those guys shoots the other, it's YOUR TURN TO SHOOT, with only one other guy.  If you succeed in shooting one of them, it's THE OTHER GUY'S TURN TO SHOOT.
 
So you can see without even looking at the cases above, that you win more often when you shoot first.  It's best not to shoot at all.
 
-t.
 
 
 
 
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Re: three-way pistol duel  
« Reply #65 on: Oct 27th, 2006, 10:00am »
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What is with these 3 pages of long explanations?
Most likely whoever made the joke meant that you have to shoot. The answer is as simple as this:
 
You have only 33% chance of killing either cyborg you shoot, so the most probable outcome will be both cyborgs being alive after your shot regardless of which cyborg you pick to shoot first. The probabilities of your survival from the 50% cyborg's shot will be the same if both cyborgs are alive, regardless of which you shoot first (If you exclude revenge factors).
 
So you only have to consider the chance of dying if you happen to kill whichever cyborg you shoot at.  Simple. If you kill the 50% cyborg, you have a 100% chance of dying. If you kill the 100% cyborg, you have a 50% chance of dying.
 
So shoot the 100% cyborg!
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Re: three-way pistol duel  
« Reply #66 on: Oct 27th, 2006, 10:39am »
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on Oct 27th, 2006, 10:00am, flamingdragon wrote:
Most likely whoever made the joke meant that you have to shoot.

I respectfully disagree.  The riddle (as linked to from the first post of this thread) ends "...what should you shoot at in round 1...".  Not "which other cyborg should you shoot at", but "what should you shoot at" (emphasis added).  I think this phrasing clearly establishes the validity -- if not the correctness -- of "the ground" as an answer.
 
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Re: three-way pistol duel  
« Reply #67 on: Oct 27th, 2006, 3:58pm »
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Once you realize the "shoot the ground" answer, it becomes obvious that this is the intended solution. This is the nature of a good puzzle: the answer is counter-intuitive, but clearly correct. Because the solution runs against your intuition, it both makes the riddle harder, and challenges you to revise your false conceptions that led to the false intuition in the first place.
 
That shooting the ground (or otherwise passing up your chance to shoot someone else) should actually improve your chances of survival is definitely counter-intuitive, but the mathematics to see that it does is fairly straight-forward, so there is no doubt that it is correct.
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Re: three-way pistol duel  
« Reply #68 on: Oct 28th, 2006, 2:59am »
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on Oct 27th, 2006, 10:00am, flamingdragon wrote:
What is with these 3 pages of long explanations?

 
Quite often, the explanation is there because people have asked questions about the riddle, or the answer was not believed to be immediately obvious to all who saw it. Also, from what I remember of the thread (I didn't bother re-reading it for this post) it is not all explanation, but involves some side-tracking based on semantics, creating different versions of the problem to be discussed and solved.
 
In general, long discussions on this board are based around exploring alternatives, rather than just immediately jumping to an answer an concluding that it is correct. If everyone had accepted the first answer which gave a solution in the "100 Prisoners and a lightbulb" thread, then it would be a whole lot shorter and much less interesting or innovative. Yes, a lot of the time the discussions lead nowhere, but that's part of being creative.
 
Sorry for the rant, just felt this was something I wanted to explain.
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Re: three-way pistol duel  
« Reply #69 on: Oct 28th, 2006, 8:12am »
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Actually, the three pages of explanation are here to disprove processes like the one dragon gave, as these were soon shown to be the wrong ones.
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Re: three-way pistol duel  
« Reply #70 on: Nov 16th, 2006, 8:14pm »
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The ground would be the best answer, but it really all depends on wether the riddler meant for u to be able to not shoot a cyborg. The only way to know would be to ask him.
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Re: three-way pistol duel  
« Reply #71 on: Nov 17th, 2006, 2:24pm »
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on Nov 16th, 2006, 8:14pm, flamingdragon wrote:
The ground would be the best answer, but it really all depends on wether the riddler meant for u to be able to not shoot a cyborg. The only way to know would be to ask him.

I'm pretty sure W.Wu intended for the deliberate miss to be an option - otherwise the puzzle isn't that interesting.
 
Incidentally, my personal favourite framing of the riddle is for 3 perfectly accurate gunmen to have different ammo loadouts - one has 100% live ammo; one has 50% live ammo, 50% blanks/tracers; and the last has a third live ammo, two-thirds assorted duds. The duds and the live ammo are thoroughly mixed and indistinguishable except by firing them, and the duds are harmless. Starting with the 1/3 live guy, followed by the 1/2, they take turns in firing one round of ammo from their supply in any direction with the aim of being the sole survivor.
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Re: three-way pistol duel  
« Reply #72 on: Aug 15th, 2007, 3:06pm »
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Chronos nailed it when he said "don't shoot at all". (Ever see the movie "The Duelists"?)
 
But if you do shoot, aim for the 100%er. This is NOT a mathmatic problem, it's a LOGIC puzzle! Cyborg C presents the greatest danger and must be taken out first. Sure, you only hit the target once in three times, but all you need is that ONE CHANCE! If you miss, the 50%er tries his hand. If he misses, the 100%er kills him and you get another shot. If the 50%er kills Cyborg C, you face a less dangerous opponant.
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Re: three-way pistol duel  
« Reply #73 on: Aug 15th, 2007, 3:20pm »
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on Aug 15th, 2007, 3:06pm, Zatanna wrote:
But if you do shoot, aim for the 100%er. This is NOT a mathmatic problem, it's a LOGIC puzzle!
It's both, I'd say.
Why wouldn't it be a mathematics problem? If you calculate your probability of survival given the options, that should tell you which choice is the logical one to make.
 
Quote:
Cyborg C presents the greatest danger and must be taken out first.
That being said, it is a matter of time before someone constructs a similar puzzle where taking out the cyborg which poses the greatest danger is the wrong choice.
 
Let's say we also have a cyborg D which misses everything he shoots. Voila, C still poses the greatest threat, but shooting D is the better option.
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Re: three-way pistol duel  
« Reply #74 on: Aug 15th, 2007, 3:59pm »
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Quote:
It's both, I'd say.  
Why wouldn't it be a mathematics problem? If you calculate your probability of survival given the options, that should tell you which choice is the logical one to make.

OK, you got me there. Roll Eyes
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