ultimate probability question (1 Viewer)

[hainsay]

there are three cowboys, each with a different shooting ability. Cowboy 1 has a 1/3 chance of hitting on any one shot. Cowboy 2 has a 1/2 chance of hitting on any one shot. Cowboy 3 has a 2/3 chance of hitting on any one shot. (perhaps someone imaginative could come up with some names)

The three cowboys have an argument and it happens that this can only be settled over a nice shootout. Cowboy 1 shoots first, then 2, then 3, then 1,2,3... until only one cowboy is left standing. (unless one cowboy is dead, in which case his turn is skipped for some reason)

Each cowboy only shoots at the most accurate shooter left alive, unless this is he, in which case he shoots at the next best shooter . (Each successful shot is lethal)

calculate the probability of each being the last cowman standing.

(Hint: Use geometric series with limiting sums to aid in addition of probabilities)

Good Luck!

 

[Estel]

Infirst round
P(1) = 1/3, P(2) = 2/3*1/2 = 1/3, P(3) = 2/3*1/2*2/3 = 2/9
In 2nd round it is merely these numbers times the chance none of them wins... (1/9 chance nobody won that round)
i.e. P(1) = 1/27 P(2) = 1/27 P(3) = 2/81 and so on
P(1) is a GP a = 1/3 r = 1/9
s = a/(1-r) = (1/3)(8/9) = 3/8
P(2) = 3/8
and P(3) = 1/4

oops just erad i didn't answer the q
um after they have shot the highest person, repeat process.

 

[Archman]

there are three cowboys, each with a different shooting ability. Cowboy 1 has a 1/3 chance of hitting on any one shot. Cowboy 2 has a 1/2 chance of hitting on any one shot. Cowboy 3 has a 2/3 chance of hitting on any one shot. (perhaps someone imaginative could come up with some names)

The three cowboys have an argument and it happens that this can only be settled over a nice shootout. Cowboy 1 shoots first, then 2, then 3, then 1,2,3... until only one cowboy is left standing. (unless one cowboy is dead, in which case his turn is skipped for some reason)

Each cowboy only shoots at the most accurate shooter left alive, unless this is he, in which case he shoots at the next best shooter . (Each successful shot is lethal)

calculate the probability of each being the last cowman standing.

(Hint: Use geometric series with limiting sums to aid in addition of probabilities)

Good Luck!

Not answering the question, but just on a side note, its a pretty stupid strategy for cowboy1 to use. I mean what if he actually kills someone on the first go . now there are only 2 of them left and the cowboy with the better aim gets to go first.
A cleverer cowboy 1 would aim somewhere so that he wont harm any of the others, and let them have a shoot off between them because they are each other's greatest threat. so ye, theres a chance that cowboy 1 is down to a shoot off between another shooter BUT him having the first shot. Being a hopeless shooter, every extra shot is good

..yes im just ranting, this has nothing to do with the actual question.

 

[Estel]

Actually Archman, you answered the version of this question I'd seen before: What should cowboy 1 do..

 

[Xayma]

Actually Archman, you answered the version of this question I'd seen before: What should cowboy 1 do..

Run like all buggery? It shouldnt take too long to get out of range.

Or shoot for a part of the body where they wont die (at least not very quickly, eg torso) and then extort them before putting them out of their misery.

 

[Estel]

If you were on that vein...
screw honour and just keep shooting

 

[withoutaface]

/slowly backs away from thread

 

[Templar]

Actually Archman, you answered the version of this question I'd seen before: What should cowboy 1 do..

Ah, the classical truel from game theory...or at least a modification of it.

 

[mojako]

Infirst round
P(1) = 1/3, P(2) = 2/3*1/2 = 1/3, P(3) = 2/3*1/2*2/3 = 2/9
In 2nd round it is merely these numbers times the chance none of them wins... (1/9 chance nobody won that round)
i.e. P(1) = 1/27 P(2) = 1/27 P(3) = 2/81 and so on
P(1) is a GP a = 1/3 r = 1/9
s = a/(1-r) = (1/3)(8/9) = 3/8
P(2) = 3/8
and P(3) = 1/4

oops just erad i didn't answer the q
um after they have shot the highest person, repeat process.

so whats the final answer?

 

[ND]

Ok for cowboy 1 to win, there are basically 3 alternatives:

- C1 hits, C2 misses, C1 hits = 1/3*1/2*1/3 = 1/18
- C1 misses, C2 hits, C1 hits = 2/3*1/2*1/3 = 1/9
- C1 misses, C2 misses, C3 hits, C1 hits = 2/3*1/2*2/3*1/3 = 2/27

However, they can all miss (2/3*1/2*1/3=1/9), in which case the above scenarios are repeated. So:

P(C1 wins) = 1/18+1/9+2/27+1/9(1/18+1/9+2/27)+(1/9)^2(1/18+1/9+2/27)+...+(1/9)^n(1/18+1/9+2/27) as n-->infinity, which of course we know how to evaluate.


edit: Crap didn't work. Anyone know how to make the fake html tags work?

edit: cool thanks, it's now in white.

 

[withoutaface]

Just make the post in white [COLOR=WHITE ] [/COLOR ]

 

[mojako]

Just make the post in white [COLOR=WHITE ] [/COLOR ]

in which case you dont have to click on Quote.
Just highlight the hidden stuff..

eh, wait.. I think u need to make it light blue coz the background colour of this forum is light blue... probably print screen and use Paint to determine what exact colour it is.

 

[withoutaface]

in which case you dont have to click on Quote.
Just highlight the hidden stuff..

eh, wait.. I think u need to make it light blue coz the background colour of this forum is light blue... probably print screen and use Paint to determine what exact colour it is.

White is still pretty much impossible to read on this background anyway

see?

 

[mojako]

apparently so,
but if you have a long text then its notice-able
qwertyuiopasdfghjklzxcvbnm
qwertyuiopasdfghjklzxcvbnm
qwertyuiopasdfghjklzxcvbnm
qwertyuiopasdfghjklzxcvbnm
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qwertyuiopasdfghjklzxcvbnm
qwertyuiopasdfghjklzxcvbnm
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qwertyuiopasdfghjklzxcvbnm
qwertyuiopasdfghjklzxcvbnm
qwertyuiopasdfghjklzxcvbnm
qwertyuiopasdfghjklzxcvbnm

 

[Rorix]

lol i dont know what you forum noobs are on about this is how you do proper invisitext


woah how do you make those colours

 

[Li0n]

say if you come across a real bullshit question in 4u , and you write down the funniest joke as an answer, do you rekon the marker would give you a mark for cracking him/her up?

 

[withoutaface]

say if you come across a real bullshit question in 4u , and you write down the funniest joke as an answer, do you rekon the marker would give you a mark for cracking him/her up?

Do you seriously want an answer to that?

 

[Trev]

haha good point; but seirously that quesiton - would u get anything like that in 4unit? thats something to put me off doing it..........i fckn hate probability! urghhh

 

[Mill]

I started trying to solve this with Markov chains but got bored and forgot all my Marko stuff. Might come back to it later.

 

[Mill]

I didn't want to calculate it by hand so my answers are decimals not fractions.

Pr(Cowboy 1 wins) = 0.3884
Pr(Cowboy 2 wins) = 0.4687
Pr(Cowboy 3 wins) = 0.1429

Would be interested to know if I'm correct. Also 0.1429 is 1/7 to 4dp I notice.

No method here, so I don't imagine I spoilt anything for anyone.