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Topic: three-way pistol duel (Read 63080 times) |
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Eric Yeh
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Re: three-way pistol duel
« Reply #25 on: Jul 24th, 2003, 3:30pm » |
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ye, but as long as theres any marginal utility to actually winning, this is no longer a nash equilibrium.
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Peter Seebach
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Re: three-way pistol duel
« Reply #26 on: Mar 27th, 2004, 11:37pm » |
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Other options:
1. Fire at "the set of all things which are not the 100% accurate robot". You are 33% likely to hit, and 67% likely to miss, killing the accurate robot.
2. Fire at "nothing". 
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rmsgrey
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Re: three-way pistol duel
« Reply #27 on: Mar 28th, 2004, 5:55am » |
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on Mar 27th, 2004, 11:37pm, Peter Seebach wrote:Other options:
1. Fire at "the set of all things which are not the 100% accurate robot". You are 33% likely to hit, and 67% likely to miss, killing the accurate robot. |
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That still leaves you in a duel with the 50% robot, with him having the first shot - if you have to be in a duel with either of the other robots, your chances are better if you shoot first rather than second. Once you get into a duel, firing at "the set of all things which aren't your opponent" then becomes a good idea, though it does rest on an unrealistic model of shot accuracy...
Quote:2. Fire at "nothing".
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This solution (or at least "deliberately miss") was suggested in the very first post in the thread, and is generally accepted as the intended solution to the puzzle.
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Leon
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Re: three-way pistol duel
« Reply #28 on: Apr 16th, 2004, 2:41pm » |
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Even if A has 50% accuracy rate, A should still miss on purpose.
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If A hits C (50%), B will fire at A (50%) = 25% of death, plus
If A misses C (50%), C may fire at and kill A (50% since both A and B are equal threats) = 25% death = 50% total death rate
If A fires at nothing, B fires at C. If he hits C (50%), A gets first crack at B. If B misses, then it all starts over again. 0% chance of death (same idea as original). If B misses C, C may fire at and kill A (50% since both A and B are equal threats) = 25% death = 25% total death rate.
25% < 50%, A should miss.
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If A is at 50% accuracy, what do you think B should do?
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Leon
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Re: three-way pistol duel
« Reply #29 on: Apr 26th, 2004, 10:42am » |
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Am I wrong in thinking that in a 3 way situation, regardless of anyone else's accuracy (even if B and C are 100%), unless A has a higher chance of getting shot at that B, A should always miss on purpose?
i.e. A and B have 99% accuracy and C has 100% - A miss on purpose
A = 99%, B and C = 100%, A miss on purpose
Am I missing soemthing?
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King_T
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Re: three-way pistol duel
« Reply #30 on: May 4th, 2004, 1:32pm » |
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A = you, 33% shooter
B = 50% shooter
C = 100% shooter
At first I did all the fractions, percentages, etc but I'm misremembering something - so I attacked this problem mechanically and followed each scenario to your death or completion of the first round, whichever came first:
Shoot at C and there are 8 outcomes:
1 - hit C, B hits you
2 - hit C, B misses you
3 - miss C, B hits you
4 - miss C, B misses you, C hits you
5 - miss C, B misses you, C hits B
6, 7, 8 - (same as 4, 5, 6)
6,7,8 are repeated to reflect your shooting percentage of 1 in 3. So there are 3 happy endings and 5 unhappy endings for you. If you shoot at C, you will survive the first round (3/8) 37.5% of the time.
Shoot at B and there are 13 outcomes:
1 - hit B, C hits you
2 - miss B, B hits you
3 - miss B, B misses you, C hits you
4 - miss B, B misses you, C hits B
5 - miss B, B hits C
6 - miss B, B misses C, C hits you
7 - miss B, B misses C, C hits B
8 to 13 - (same as 2 to 7)
You will survive the first round (6/13) 46% of the time.
If you miss on purpose there are 6 outcomes:
1 - B hits you
2 - B misses you, C hits you
3 - B misses you, C hits B
4 - B hits C
5 - B misses C, C hits you
6 - B misses C, C hits B
You will survive the first round (3/6) 50% of the time.
Interestingly, if you shoot at yourself with your same 33% accurracy (if that's possible), then the following 13 outcomes are possible:
1 - you shoot yourself
2 to 6 (above, like missing on purpose)
7 to 13 (same as 2 to 6)
You will survive (6/13) 46% of the time.
Much as the NRA would like to disagree - your best chance is to miss on purpose (or don't shoot), followed by shooting at B or yourself, followed distantly by shooting at C, the most accurate shooter.
In all cases, after the first round - blaze away at the remaining robot.
K
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ThudnBlunder
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Re: three-way pistol duel
« Reply #31 on: May 5th, 2004, 8:48am » |
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Assume the classical interpretation of the problem.
Let probability that A hits his target = a
Let probability that B hits his target = b
Let probability that C hits his target = 1
Let 0 < a < b < 1
Assume that there is only A and B left, and that it is A to shoot.
Let probability that A survives in this case = aB
Then aB = a / [1 - (1 - a)(1 - b)]..................[1]
A Shoots First at B:
Let probability of A surviving in this case = AxB
Then AxB = (1 - a)[a(1 - b) + baB]...............[2]
A Shoots First at C:
Let probability that A survives in this case = AxC
Missing C first go leaves exactly the same probability of survival as missing B first go.
This probability was previously found to be AxB.
Hitting C first go leaves A and B, with B to shoot.
Therefore chance of survival = a(1 - b)aB
Hence AxC = AxB + a(1 - b)aB (thus AxC is always greater than AxB)
= a(1 - a)(1 - b) + [a(1 - b) + b(1 - a)]aB...............[3]
A Shoots First at Nobody:
Let probability that A survives in this case = AxN
Shooting at nobody leaves B to shoot at C.
This is the same case as comes about when A shoots first at B and misses.
That is, AxB = (1 - a)AxN
Hence AxN = a(1 - b) + baB.........................[4]
A Shoots First at Himself:
Let probability that A survives in this case = AxA
I found it very amusing to read in the previous post that A has a better chance of surviving by trying to blow his own brains out than by having a pop at C, the sure shot!
In fact, AxA = AxB ...................................[5]
because if you kill yourself you are dead and if you kill B you are equally dead.
--------------------------------------------
For example,
let
a = 1/3
b = 1/2
Then
aB = a / [1 - (1 - a)(1 - b)] = (1/3) / [1 - (2/3)(1/2)] = 1/2
AxB = (1 - a)[a(1 - b) + baB] = (2/3)[(1/3)(1/2) + (1/2)(1/2)] = 10/36
AxC = AxB + a(1 - b)aB = 10/36 + (1/3)(1/2)(1/2) = 13/36
AxN = a(1 - b) + baB = (1/3)(1/2) + (1/2)(1/2) = 15/36
If A shoots at nobody,
B's chances of survival = b(1 - aB) = (1/2)[1 - (1/2)] = 9/36
C's chances of survival = (1 - a)(1 - b) = (2/3)(1/2) = 12/36
Finally, if we let b = 1 we get
aB = a
AxB = AxC = a(1 - a)
AxN = a
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| « Last Edit: May 14th, 2004, 5:12pm by ThudnBlunder » |
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rmsgrey
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Re: three-way pistol duel
« Reply #32 on: May 5th, 2004, 9:11am » |
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I notice King_T has a hidden assumption that the other two "duellists" pick their target randomly. If you instead assume that they play to maximise their own survival chances, then C wants to eliminate both A and B as quickly as possible - except in the case where he expects A and B to shoot at each other until one of them hits - at which point he kills the other. Given the choice of which to shoot, he will choose B over A in order to cut the chances of being killed by the survivor. A and B, knowing that if either of them hits the other while C survives, they both lose, won't shoot each other. If both deliberately miss, C has a choice between shooting B and deliberately missing - the latter meaning no-one wins. So C, if everyone survives the first round will kill B. B then has a strong incentive to kill C. So, if A kills either opponent, the other will get first shot at him, while if he misses, he gets first shot against the survivor (with a 1/2 chance of the survivor being B).
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ThudnBlunder
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Re: three-way pistol duel
« Reply #33 on: May 14th, 2004, 8:46pm » |
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Nice analysis, King_T.
However, for 'Shoot at C' you missed out
Miss C, B hits C
and
Miss C, B misses C, C hits you
Miss C, B misses C, C hits B
Adding these would increase A's probability of surviving the 1st round by shooting first at C to 7/14 = 50%
Quote:| 6, 7, 8 - (same as 4, 5, 6) |
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Typo - you meant 'same as 3, 4, and 5'.
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| « Last Edit: May 15th, 2004, 3:28am by ThudnBlunder » |
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TenaliRaman
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Re: three-way pistol duel
« Reply #34 on: May 15th, 2004, 5:29am » |
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Modifying the problem a bit,
There are 3 robots A,B and C.
A has a firing algorithm which has 33% accuracy
B has a firing algorithm which has 50% accuracy
C has a firing algorithm which has 100% accuracy
Each bot starts with a energy level of 100%. When a robot is hit, its the energy level reduces by about 5%.
(Assume that each bot knows every other bots firing accuracy,also every bot hits the bot which gives it the maximum survival chance).
What is the probability that A can survive such a duel?
(possible extension : whenever a robot B is hit by the bullet of robot A, robot A gains 5% energy, also allowing that energy level can go beyond 100%. what is the chance of A surviving now?)
(inspired from this post and robocodes)
(I haven't solved it and i don't know the answer.If this is tad bit too much for pen and paper then this question could be skipped and this post completely ignored)
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| « Last Edit: May 15th, 2004, 5:35am by TenaliRaman » |
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ThudnBlunder
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Re: three-way pistol duel
« Reply #35 on: May 15th, 2004, 11:00am » |
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Quote:| What is the probability that A can survive such a duel? |
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Assume that A shoots in the air at first ~ then his survival probability = 5/12
Assume that B always targets C whenever possible ~ then his survival probability = 1/4
Assume that C always targets B whenever possible ~ then his survival probability = 1/3
The number of rounds required will vary between 20 and 59.
So I make it 1 - [smiley=csigma.gif](7/12)20 * (5/12)k-20 * kCk-20
for k = 20 to 59
(This is a slight approximation as on the 59th and last round there will only be two players left.
However, the chance that 59 rounds are required is very small.)
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| « Last Edit: May 16th, 2004, 4:22pm by ThudnBlunder » |
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ThudnBlunder
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Re: three-way pistol duel
« Reply #36 on: May 15th, 2004, 2:21pm » |
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Quote:| [quote]whenever a robot B is hit by the bullet of robot A, robot A gains 5% energy, also allowing that energy level can go beyond 100%. what is the chance of A surviving now?) |
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Too loosely worded. 
Not sure about this, but...
Let the last man to lose 20 points be the survivor.
In 36 games A will win 15 of them, gaining 30 points.
In 36 games B will win 9 of them, gaining 18 points.
In 36 games C will win 12 of them, gaining 24 points.
In 36 games A will lose 21 of them, losing 21 points.
In 36 games B will lose 27 of them, losing 27 points.
In 36 games C will lose 24 of them, losing 24 points
Hence in 1 game
A can expect to gain 1/4 of a point.
B can expect to lose 1/4 of a point.
C can expect to break even.
Consider C.
Each game is a Bernoulli trial with probability 1/2 of winning more points than losing.
He starts with 20 points and must win another 40 points to survive.
Therefore his chance of not being ruined (surviving) is 20/60 = 1/3 
Now what?
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| « Last Edit: May 15th, 2004, 5:11pm by ThudnBlunder » |
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ThudnBlunder
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Re: three-way pistol duel
« Reply #37 on: May 15th, 2004, 3:24pm » |
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Considering King_T's variant wherein B and C choose a target randomly:
Let probability that A hits his target = a
Let probability that B hits his target = b
Let probability that C hits his target = 1
Let 0 < a < b < 1
Assume that there is only A and B left, and that it is A to shoot.
Let probability that A survives in this case = aB
Then aB = a / [1 - (1 - a)(1 - b)]
Shoots First at B:
Let probability that A survives in this case = AxB
1 - Hit B, C hits you
2 - Miss B, B hits you
3 - Miss B, B misses you, C hits you
4 - Miss B, B misses you, C hits B ~ probability = a(1 - a)(1 - b) / 4
5 - Miss B, B hits C ~ probability = b(1 - a)aB / 2
6 - Miss B, B misses C, C hits you
7 - Miss B, B misses C, C hits B ~ probability = a(1 - a)(1 - b) / 4
Total probability of surviving = (a/2)(1 - a)(1 - b) + (1/2)[b(1 - a)]aB
Shoots First at C:
Let probability that A survives in this case = AxC
1 - Hit C, B hits you
2 - Hit C, B misses you ~ probability = a(1 - b)aB / 2
3 - Miss C, B hits you
4 - Miss C, B misses you, C hits you
5 - Miss C, B misses you, C hits B ~ probability = a(1 - a)(1 - b) / 4
6 - Miss C, B hits C ~ probability = b(1 - a)ab / 2
7 - Miss C, B misses C, C hits you
8 - Miss C, B misses C, C hits B ~ probability = a(1 - a)(1 - b) / 4
Total probability of A surviving = (a/2)(1 - a)(1 - b) + (1/2)[a(1 - b) + b(1 - a)]aB
Shoots at Nobody:
Let probability that A survives in this case = AxN
As before, AxB = (1 - a)AxN
Hence AxN = (a/2)(1 - b) + (1/2)baB
And AxC = AxB + (a/2)(1 - b)aB
For example,
let
a = 1/3
b = 1/2
Then
aB = a / [1 - (1 - a)(1 - b)] = (1/3) / [1 - (2/3)(1/2)] = 1/2
AxB = (a/2)(1 - a)(1 - b) + (1/2)[b(1 - a)]aB
= (1/6)(2/3)(1/2) + (1/2)[(1/2)(2/3)(1/2) = 10/72
AxC = (a/2)(1 - a)(1 - b) + (1/2)[a(1 - b) + b(1 - a)]aB
= (1/6)(2/3)(1/2) + (1/2)[(1/3)(1/2) + (1/2)(2/3)](1/2) = 13/72
AxN = (a/2)(1 - b) + (1/2)baB
= (1/6)(1/2) + (1/2)(1/2)(1/2) = 15/72
In each case, A's chance of survival is exactly half of what it was previously, when B and C tried to maximize their chances by targeting each other. This is explained by the fact that they both now target A more often.
AxB = 10/72 < 10/36 previously.
AxC = 13/72 < 13/36 previously.
AxN = 15/72 < 15/36 previously.
When A shoots at nothing
B's chances of survival = (b/2)[(1 - b) + (1 - ab)]
= (1/4)[(1/2) + (1/2)] = 18/72
and
C's chances of survival = (1/2)(1 - b)[(1 - a) + (1 - b)] + b/2
= (1/2)((1/2)[(2/3) + (1/2)] + 1/4 = 39/72
Hence A's chances of survival have decreased from 5/12 to 5/24
Hence B's chances of survival have remained the same at 1/4
Hence C's chances of survival have increased from 1/3 to 13/24
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| « Last Edit: Jul 14th, 2007, 11:05am by ThudnBlunder » |
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Three Hands
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Re: three-way pistol duel
« Reply #38 on: May 15th, 2004, 5:44pm » |
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Seems like a rather interesting conclusion - even though A is now more often targetted, he is more likely to survive. Of course, the only possibilities A has of walking out of the duel alive is by shooting C after B has been shot, shooting B after C has been shot, or killing C, and then surviving B's attempt to kill him, before shooting B - so shooting at B is inherently stupid, since killing B also (indirectly) kills A and so, given the two-target option (B or C), you'd prefer to shoot at C - as makes more sense.
However, one interesting factor is if you miss what you target, then we're assuming you'd hit nothing. However, if you target nothing, then you'd presumably hit something. Given there are two other targets given in this example, either you hit B or you hit C - and there is no stated probability of hitting one or the other if you miss "nothing". Assuming a 50% chance to hit either of them if you miss nothing, then shooting at nothing would lead to a (2/3 * 1/2) 1/3 chance of hitting either randomly. Given you don't want to hit B (as you then automatically get killed by C), I would prefer to shoot at C in the first place - same chance of killing C, but without the 1/3 chance of killing B, and so being killed by C subsequently.
I am, of course, assuming that missing either B or C doesn't lead to similar problems, but automatically hit "nothing" (something about targetting being good enough to be close, and so miss the other), and that missing nothing will not potentially cause you to hit yourself (since you could be considered a valid target, but I felt that such a death (killing yourself by "missing nothing") would be too embarassing for any robot to consider worth the risk...). However, if you bring about some kind of random "scatter" effect for missed shots, then this would probably change the dynamics of the puzzle quite a bit...
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ThudnBlunder
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Re: three-way pistol duel
« Reply #39 on: May 15th, 2004, 6:07pm » |
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Quote:| However, one interesting factor is if you miss what you target, then we're assuming you'd hit nothing. However, if you target nothing, then you'd presumably hit something. |
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Are you suggesting that we consider 'nothing' as a 4th target and A's probability of hitting B or C while trying not to hit them as 2/3?
But if we are being realistic we would set the probability of hitting something while trying not to hit it at, for example, 0.00001; this is about as close to zero as would make very little difference to the previous calculations.
Quote:| ...and that missing nothing will not potentially cause you to hit yourself |
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Oh, now I see! While I have been lazily adopting the usual stereotyped way of looking at things, you on the other hand have quickly hacked and probed your way to the essential core of the problem!
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| « Last Edit: May 18th, 2004, 4:44pm by ThudnBlunder » |
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TenaliRaman
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Re: three-way pistol duel
« Reply #40 on: May 16th, 2004, 1:42am » |
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wow! nice work
and as rmsgrey says, it is quite an interesting conclusion considering the two cases.
i think it also explains why sometimes in robocodes,with around 10 robots in the battle, the weaker ones sometimes survive till the end.
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Three Hands
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Re: three-way pistol duel
« Reply #41 on: May 16th, 2004, 5:31am » |
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True, I was going with a rather silly view of what would happen if A aimed at nothing and missed The obvious solution is for A to aim instead at a passing fly which is nowhere near B or C - or any other target which is not "nothing" "B" or "C" (or indeed himself - that could end up being embarassing...)
It is an interesting question, though, of what would happen if A targetted nothing, and managed to miss. Surely he would then have to hit something, otherwise he would fail to miss nothing. If the only other things he could hit were himself, B or C, then aiming at nothing could be rather hazardous...
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ThudnBlunder
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Re: three-way pistol duel
« Reply #42 on: May 16th, 2004, 10:10am » |
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Quote:| it is quite an interesting conclusion considering the two cases. |
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Alas, I found a problem with my numerical calculations. It is now fixed. Although the original conclusion is not true, perhaps more interestingly, B's chances of survival remain unchanged.
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| « Last Edit: Oct 27th, 2006, 10:21am by ThudnBlunder » |
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ThudnBlunder
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Re: three-way pistol duel
« Reply #43 on: May 16th, 2004, 4:32pm » |
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on May 16th, 2004, 5:31am, Three Hands wrote:| It is an interesting question, though, of what would happen if A targetted nothing, and managed to miss. Surely he would then have to hit something, otherwise he would fail to miss nothing. If the only other things he could hit were himself, B or C, then aiming at nothing could be rather hazardous... |
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...but evidently much less hazardous than studying too much philosophy!
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| « Last Edit: May 16th, 2004, 7:03pm by ThudnBlunder » |
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Three Hands
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Re: three-way pistol duel
« Reply #44 on: May 17th, 2004, 2:24am » |
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Depends on which you prefer - insanity or a possibility of death...
I chose insanity 
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towr
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Re: three-way pistol duel
« Reply #45 on: May 17th, 2004, 5:09am » |
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He might hit himself regardless of what he targets.
If he targets C and misses, he'll either hit nothing, B or himself (assuming the universe is otherwise empty)
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rmsgrey
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Re: three-way pistol duel
« Reply #46 on: May 17th, 2004, 5:24am » |
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A variation I came across many years ago had the three robots all perfect shots, but with doctored ammunition, so that instead of always firing live ammo, A fires harmless blanks 2/3 and B 1/2 the time. This avoids the problem of what happens when A fires at "the set of things which aren't C" - thereby invalidating some of the previous analysis as he is now a bigger threat to C than B is...
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Three Hands
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Re: three-way pistol duel
« Reply #47 on: May 17th, 2004, 6:05am » |
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on May 17th, 2004, 5:09am, towr wrote:He might hit himself regardless of what he targets.
If he targets C and misses, he'll either hit nothing, B or himself (assuming the universe is otherwise empty)
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Granted, I was assuming that a miss would result in hitting nothing (the generally assumed victim of a miss), but such a result would raise the question "If he aims at nothing and misses, then what does he hit?"
If it is a random effect on what anyone hits if they miss anything, then it is only slightly less likely to kill A if he aims at anything other than himself - and leads to his survival chances plummeting, I would imagine...
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Re: three-way pistol duel
« Reply #48 on: Dec 11th, 2004, 6:27pm » |
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Man you don't need all of those algorithms and whatever to prove it. If you shoot at B, you have a 33% chance of hitting. If you hit, C shoots you, and you die for sure. If you miss, B will shoot at C. Then, If B misses, C kills B. Then you have a shot at C. 33% chance you live, 66% chance you die. So, if you shoot at B, 1/2 you die for sure and 1/2 you have a 66% chance of dying. That evens out to a 17% chance of living. If you shoot at C and you hit(33% chance) then B shoots at you (50% chance). If he misses, then I think you will keep shooting back and forth and it will be a 16.5% chance of you living. I'm not sure. If you miss C (66% chance) then B shoots at C. If he hits (50% chance) then you have a 33% chance of hitting him, and I think(not sure) it's the same 16.5% chance of living. If B misses, then C kills B and You get a shot at B. Same 16.5% chance. So I think that if you shoot at C, you have a 16.5% chance of living. Not sure, but I know it's less that a 50% for sure. If you attack no one then B attacks C. If he hits, (50%) then you get a shot at B. 1/3 of hitting. If not, 1/2 of dying. If not 1/3 of hitting. Ect. I think that this is a 66% chance of living. I dono. If he misses, C kills B, and you have a 33% chance of living, if you hit C. The average of 66% and 33% is 50%, so you have the best chance of living by deliberately missing on your first shot, with a 50% chance! Pretty good for the worst shooter!
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Icarus
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Re: three-way pistol duel
« Reply #49 on: Dec 11th, 2004, 7:25pm » |
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Those "algorithms" are how you calculate probabilities correctly. The probabilities you have stated are not correct.
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"Pi goes on and on and on ...
And e is just as cursed.
I wonder: Which is larger
When their digits are reversed? " - Anonymous
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