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 Post subject: Re: Brainteaser thread
PostPosted: Sun Oct 05, 2008 11:58 am 
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Just a mild one I got from an old puzzle book:

Once upon a time in the Old West, Alan, Bob and Chuck were 3 outlaws in town with a reputation of great shooting. All 3 of them never missed by more than an inch, and they're renowned for their accuracy when shooting at a vertical string 100 feet away, which they called "cutting with a bullet". Chuck was a respectable 50% shooter and could always cut a string once for every two shots. Bob was an even better 80% shooter who would only miss once for every five shots. However, Alan was the best among the three. Being a pure 100% shooter, he never missed with any shot he fired.

During one failed train-robbing attempt all 3 of them were captured. The Sheriff decided that for their crime two must die, with the remaining one exiled. To decide who was to survive he came up with this cruel plan:

The three were hanged up (tied around the torso, not the neck) very high above the ground by a long string, positioned in the corners of an equilateral triangle so that each was exactly 100 feet away from each of the other two. Each of them was equipped with a gun with only 2 bullets. Then they had to take turn shooting at the string hanging the others, in the following way:

The Sheriff would throw a fair die (dice) to determine the order for them to shoot. Considering there were only 6 possible different orders each of the 6 faces of the die represented a particular order. For example, if a 1 was throwed, the order would go like this: Alan -> Bob -> Chuck.

Then these were the rules:

1. During each turn a criminal must shoot only once. The goal was to "cut" the string hanging on the body of another criminal, so that the victim would immediately fall to the ground and die. If a criminal shot "out of turn", or shot twice instead of once within a turn, the string hanging himself would immediately be set loose and he would fall to the ground and die immediately.

2. The criminals must not shoot a body (of another criminal or any other people). If anybody got hit by a bullet from a gun, the shooter would immediately be dropped to the ground and die. But considering the exceptional shooting accuracy of these three, their bullets would only hit a body if they shot at that person intentionally.

3. If a criminal used up his 2 bullets (in 2 turns) and didn't hit a string at all, he would immediately be dropped to the ground and die.

4. Whenever one of them died (either by having his string cut by a bullet or by violating any of the rules above), the turn would be passed to the next criminal in the order.

5. If two of the criminals died, the remaining one (provided he didn't violate any of the rules above) would not be executed and would be exiled alive.

So these were all the rules. Also, we knew that all 3 of them were very cool-headed in calculations and would always act in ways which would maximise their own chance of survival. Also despite they were tied and hanged high above the ground their shooting abilities weren't affected by a single bit. So their respective accuracy of 100%, 80% and 50% of "cutting a string with a bullet" stayed intact. Lastly, all 3 of them were perfectly aware of each other's shooting accuracy and knew very well their common tendency to do whatever to maximise the chance of own survival.

With these conditions, could you work out who had the best chance to survive, and what was the survival probability for each of them? :pirate:

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 Post subject: Re: Brainteaser thread
PostPosted: Sun Oct 05, 2008 7:56 pm 
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Hi udosuk

Hidden Text:
How about Chuck 47/90 chance of survival, Alan, 3/10 and Bob 8/45


:drunk:


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 Post subject: Re: Brainteaser thread
PostPosted: Sun Oct 05, 2008 9:00 pm 
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I think this is the complete set up.

Hidden Text:
Whoops messed up the letters. And too lazy to correct it. So letter A is Chuck and letter C is Alan :P.

One rule: They always shoot the guy with the highest percentage of making the shot.

ABC:
A survives: A hits, B misses, A hits; A hits, B misses, A misses, B misses; A misses, (doesn't matter), A hits = 0.5*0.2*0.5 + 0.5*0.2*0.5*0.2+0.5*0.5 = 31%
B survives: A hits, B hits; A hits, B misses, A misses, B hits; A misses, B hits, A misses = 0.5*0.8 + 0.5*0.2*0.5*0.8+ 0.5*0.8*0.5 = 64%
C survives: A misses, B misses, A misses = 0.5*0.2*0.5 = 5%

ACB:
A survives: A hits, B misses, A hits; A hits, B misses, A misses, B misses; A misses, A hits = 0.5*0.2*0.5 + 0.5*0.2*0.5*0.2 + 0.5*0.5 = 31%
B survives: A hits, B hits; A hits, B misses, A misses, B hits = 0.5*0.8 + 0.5*0.2*0.5*0.8 = 44%
C survives: A misses, A misses = 0.5*0.5 = 25%

BAC:
A survives: B hits, A hits; B hits, A misses, B misses, A hits; B misses, A hits, B misses; B misses, A misses, A hits 0.8*0.5 + 0.8*0.5*0.2*0.5 + 0.2*0.5*0.2 + 0.2 * 0.5 * 0.5 = 51%
B survives: B hits, A misses, B hits; B hits, A misses, B misses, A misses; B misses, A hits, B hits 0.8*0.5*0.8 + 0.8*0.5*0.2*0.5 + 0.2*0.5*0.8 = 44%
C survives: B misses, A misses, A misses = 0.2*0.5*0.5 = 5%

BCA:
A survives: B hits, A hits; B hits, A misses, B misses, A hits; B misses, A hits 0.8 * 0.5 + 0.8*0.5*0.2*0.5+0.2*0.5 = 54%
B survives: B hits, A misses, B hits; B hits, A misses, B misses, A misses= 0.8 * 0.5 * 0.8 + 0.8*0.5*0.2*0.5 = 0.32 + 0.04 = 36%
C survives: B misses, A misses = 0.2 * 0.5 = 10%

CAB:
A survives: A hits = 50%
B survives: not possible = 0%
C survives: A misses = 50%

CBA:
A survives: A hits = 50%
B survives: not possible = 0%
C survives: A misses = 50%

A(Chuck) = (31 + 31 + 51 + 54 + 50 + 50)/6 = 267/6 = 44.5%
B(Bob) = (64 + 44 + 44 + 36 + 0 + 0)/6 = 188/6 = 31.33%
C(Alan) = ( 5 + 25 + 5 + 10 + 50 + 50)/6 = 145/6 = 24.17%


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 Post subject: Re: Brainteaser thread
PostPosted: Sun Oct 05, 2008 10:26 pm 
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Para,

Great answer but you have made an assumption that it is everyone's best interest to take a shot at the most dangerous opponent.

In fact, it is in Chuck's (ie the 50% guy) best interest just to fire at the ground, leaving the other two to fight it out between them. That way, Chuck gets first stab at the survivor. I think that if you rework your analysis on this basis, Chuck gets a higher chance of surviving, ie more than 50%.

You can see the principle of the tactics most clearly if A and B were both 100% dead shots. Chuck would be out of his mind trying to kill one of then knowing that the survivor would certainly kill him. By firing into the ground, he has a 50% chance of killing the survivor. If in fact the survivor is the 80% guy, even better!

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 Post subject: Re: Brainteaser thread
PostPosted: Sun Oct 05, 2008 11:31 pm 
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Yeah, i thought about that too for a while. Just thought the rules kinda meant, they have to shoot to try and cut a string.

And yes, i had nothing to do :P.


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 Post subject: Re: Brainteaser thread
PostPosted: Mon Oct 06, 2008 7:09 am 
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Great answr Para & a good insight Bigtone53,

Chuck's survival increases as long as Alan & Bob stay alive ..

Because Alan is going to shoot 1st 2nd or 3rd

The 1st round will always result in somebody dead.

If Chuck misses on purpose then his next shot is the last & carries with it a 50% chance

Chuck's turn will always be 1st when there is only 2 men

therefore his chances are always 50%

Except when he has 2 bullets & is aginst Bob. that happens if the order is:
BCA or BAC that is 1/3 * 80 % ... his chances are 55 %

so in 26.667 % of the time his chances are 55%
in 73.333% of the time his chances are 50%

His overall survival is 51.33335 %

tarek


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 Post subject: Re: Brainteaser thread
PostPosted: Mon Oct 06, 2008 3:19 pm 
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Tarek,

Your answer differs from mine. While I work through yours, here is my full story, assuming that C aims to miss.

A – the 100%er.

1/2 chance of first shot – shoots B overall 0.5
1/2 chance of second shot then 1/5 chance of surviving B in which case shoots B. overall 0.1
Combined chance of A surviving B 0.5 + 0.1 = 0.6

C then gets a shot, with a 0.5 chance of killing A. otherwise he is shot

Probability of A surviving 0.5 x 0.6 = 0.3

B – the 80%er

Chance of surviving A 1.0-0.6 (above) = 0.4. Will he survive C?
C get a shot 0.5 chance, then B fires 0.8 chance. B’s chance of killing C in this round is 0.5 x 0.8 = 0.4
But there is a 0.2 chance that B will miss, so C has a 0.5 x 0.2 = 0.1 chance of surviving to round 2
C’s next shot 0.5 chance, B fires 0.8 chance . B’s chance of killing C on the second round is 0.1 (ie getting through first round) x 0.5 X 0.8 = 0.04

Going on with B missing will give B’s chance of beating C = 0.44444444 … = 4/9, which together with the 0.4 chance of surviving A gives B’s overall chances 4/10 x 4/9 = 16/90 = 0.1777777…

C’s chance is therefore 1 – 0.3 – 0.1777777 = 0.522222222… = 47/90

Tony

:pray:

[Edit - OK, I have just noticed the bullet constraint. :oops: Let me rework]


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 Post subject: Re: Brainteaser thread
PostPosted: Mon Oct 06, 2008 4:26 pm 
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well i guess that I have to think it again ....

the main reason behind my answer is the fact that if you miss twice then you die ...

so essentially , if chuck decides to intentionally miss then his chance of surviving is 50% (if he hits then he's the only one left & therefore survives, if he misses then he dies by the virtue of his two misses) .... if chuck intentionally misses his first shot, he will not live to see anyone shoot him :idea:

the exception to this 50% rule happens 80% of 1/3 of the time ....

That is when Bob shoots 1st & doesn't miss. Chuck has 2 bullets now to shoot bill :idea:
Chuck hits ... that is 50%
if chuck misses ... then Bob shoots. By the virtue of his 1st kil, Bob will not die if he misses (20%) ... Leaving Chuck with the 50% dilemma again ... 50% of the 20% ... that is 50% of the 20% of the 50% which is 5% ....

so in 80 % of 1/3 of the time he has a 55% survival rate
in the rest (2/3 + 20% of 1/3) of the time he has 50%...

44/300 + 110/300 = 154/300 = .5133333

tarek


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 Post subject: Re: Brainteaser thread
PostPosted: Mon Oct 06, 2008 6:47 pm 
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Looks good, Tarek

Not sure though what you meant by

Quote:
Leaving Chuck with the 50% dilemma again ...


Chuck at this stage has one bullet against a guy who is out. No dilemma, but as you say a 50% chance of winning, as a miss would be his downfall.


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 Post subject: Re: Brainteaser thread
PostPosted: Mon Oct 06, 2008 8:21 pm 
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Bigtone53 wrote:
Looks good, Tarek
Not sure though what you meant by
Quote:
Leaving Chuck with the 50% dilemma again ...

Chuck at this stage has one bullet against a guy who is out. No dilemma, but as you say a 50% chance of winning, as a miss would be his downfall.

Yes, No dilemma ... but remains a 50% situation ....

based on that scenario ....

The survival rates are:
Alan 30%
Bob 18.6667%
Chuck 51.3333%

tarek


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