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27 replies to this topic

### #11 bushindo

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Posted 05 June 2009 - 01:21 AM

Spoiler for Damn i keep flip floping

Shouldn't the survival probabilities for A-E add up to 1? The game requires that only 1 person survives, so survival probabilities are mutually exclusive.

Spoiler for part 1

I think you're missing a 0. E's chance of killing anyone is 16/20 = 80%. A's chance of killing anyone is only 4/20 = 20%. A can't be the best place to be because eventually the game comes down to a 2-people duel. Even if A is among the last two, his low chance of killing makes him lose way too often.

Edited by bushindo, 05 June 2009 - 01:28 AM.

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### #12 final

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Posted 05 June 2009 - 04:21 AM

Spoiler for ...

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### #13 Era

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Posted 05 June 2009 - 04:38 AM

Spoiler for what I think. ^^

Edited by Era, 05 June 2009 - 04:40 AM.

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### #14 bushindo

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Posted 05 June 2009 - 05:52 AM

Spoiler for ...

Spoiler for HINT

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### #15 bushindo

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Posted 05 June 2009 - 05:55 AM

Spoiler for what I think. ^^

This is counter-balanced by the fact that C has a 1/2 chance of killing D on everyshot. Most of the time, C survives B's shot, so he gets to shoot D. If A kills B, that's even better for C. In this game, killing someone above you is not as important, or urgent, as having someone with low accuracy below you.
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### #16 final

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Posted 05 June 2009 - 09:22 AM

so i found the probability that each player is the first to die (easy) and the probability of who wins each pair with each first and i thought i understood what you meant by recursive untill i thought about it and are u actually suggesting that for example for a shoot at b i look at b shot c,d,e there likelyhoods of reducing to ab and continue as such because that means each will have 3 forks(at most 3 i think or exactly three havent really thought that much and then for those each has at most two and then well ive done that), but anyway i can just work forward if thats the case right? find the probabilities that each person dies second within the first already died then third then fourth is done and multiply and add. anyway thoughts? advice?
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### #17 Glycereine

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Posted 05 June 2009 - 10:37 AM

This is counter-balanced by the fact that C has a 1/2 chance of killing D on everyshot. Most of the time, C survives B's shot, so he gets to shoot D. If A kills B, that's even better for C. In this game, killing someone above you is not as important, or urgent, as having someone with low accuracy below you.

Additionally you said that likely A and E will die in the first round which is not possible. If E dies in the first round, no one can shoot at A because D has already fired his shot (killing E).

Spoiler for Not really the solution

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### #18 CaptainEd

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Posted 05 June 2009 - 04:21 PM

Assuming I've done my math AND coding correctly,
Spoiler for I'd choose to be

Spoiler for The math, including recursive functions:

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### #19 CaptainEd

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Posted 05 June 2009 - 05:05 PM

Assuming I've done my math AND coding correctly, (edit: bad assumption, typos corrected below)

Spoiler for I'd choose to be

Spoiler for The math, including recursive functions:

Edited by CaptainEd, 05 June 2009 - 05:10 PM.

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### #20 bushindo

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Posted 05 June 2009 - 05:13 PM

so i found the probability that each player is the first to die (easy) and the probability of who wins each pair with each first and i thought i understood what you meant by recursive untill i thought about it and are u actually suggesting that for example for a shoot at b i look at b shot c,d,e there likelyhoods of reducing to ab and continue as such because that means each will have 3 forks(at most 3 i think or exactly three havent really thought that much and then for those each has at most two and then well ive done that), but anyway i can just work forward if thats the case right? find the probabilities that each person dies second within the first already died then third then fourth is done and multiply and add. anyway thoughts? advice?

Spoiler for for final

Assuming I've done my math AND coding correctly,

Spoiler for I'd choose to be

Spoiler for The math, including recursive functions:

Good work, CaptainEd. I didn't actually expect you to actually implement the method, so your solution makes it all the more impressive. Well done.
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