发信人: hero080 (APM=080), 信区: BrainTeaser
标 题: zz三人决斗
发信站: BBS 未名空间站 (Tue Jul 1 14:04:28 2008), 转信

发信人: GGGGDDDDK (和某些人生气不值得), 信区: IQDoor
标 题: 古老的三人决斗问题的一般形式
发信站: 水木社区 (Tue Jul 1 09:03:27 2008), 站内

三个人的射中概率分别为0<a<b<c<=1，当然是按照a, b, c的顺序射击。
请问：

1. 是否真的无论a, b, c等于多少，第一个人的最佳策略都是对天开枪？
2. 是否存在这样的a, b, c，使得三个人的胜出概率都是1/3？

--
╱╲
╠┃║﹍﹍﹍﹍﹍﹍﹍﹍﹍﹍﹍﹍﹍﹍﹍
瞳凝秋水剑流星，裁诗为骨玉为神? ╱╤╤╝┃ _____________________________╲
翩翩白衣云端客，生死为谁一掷轻? ╲╧╧╗┃ ╱
╠┃║﹉﹉﹉﹉﹉﹉﹉﹉﹉﹉﹉﹉﹉﹉﹉
╲╱



※ 来源:·BBS 未名空间站 海外: mitbbs.com 中国: mitbbs.cn·[FROM: 136.152.]

发信人: siss (爱还在 义无赖 曲未终 戏台谢), 信区: BrainTeaser
标 题: Re: zz三人决斗
发信站: BBS 未名空间站 (Tue Jul 22 20:39:31 2008), 转信

没看懂 -_-||| 我google去... 太受打击了，hero来点简单的热热身

【 在 hero080 (APM=080) 的大作中提到: 】
: 发信人: GGGGDDDDK (和某些人生气不值得), 信区: IQDoor
: 标 题: 古老的三人决斗问题的一般形式
: 发信站: 水木社区 (Tue Jul 1 09:03:27 2008), 站内
: 三个人的射中概率分别为0<a<b<c<=1，当然是按照a, b, c的顺序射击。
: 请问：
: 1. 是否真的无论a, b, c等于多少，第一个人的最佳策略都是对天开枪？
: 2. 是否存在这样的a, b, c，使得三个人的胜出概率都是1/3？



--

※ 来源:·BBS 未名空间站 海外: mitbbs.com 中国: mitbbs.cn·[FROM: 75.153.]

发信人: OhRe (Re), 信区: BrainTeaser
标 题: Re: zz三人决斗
发信站: BBS 未名空间站 (Wed Jul 23 00:17:27 2008), 转信

下面那个题题目就很简单，而且保证你做出来时热得不行。
【 在 siss (爱还在 义无赖 曲未终 戏台谢) 的大作中提到: 】
: 没看懂 -_-||| 我google去... 太受打击了，hero来点简单的热热身



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※ 来源:·BBS 未名空间站 海外: mitbbs.com 中国: mitbbs.cn·[FROM: 76.199.]

发信人: siss (爱还在 义无赖 曲未终 戏台谢), 信区: BrainTeaser
标 题: Re: zz三人决斗
发信站: BBS 未名空间站 (Wed Jul 23 20:08:48 2008), 转信

你不是黑肉哥哥（哥哥是替midt mm加的~）

【 在 OhRe (Re) 的大作中提到: 】
: 下面那个题题目就很简单，而且保证你做出来时热得不行。



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※ 来源:·BBS 未名空间站 海外: mitbbs.com 中国: mitbbs.cn·[FROM: 75.153.]

发信人: hero080 (APM=080), 信区: BrainTeaser
标 题: Re: zz三人决斗
发信站: BBS 未名空间站 (Wed Jul 23 21:46:13 2008), 转信

siss妹妹好，黑肉哥哥在此 ^_^
【 在 siss (爱还在 义无赖 曲未终 戏台谢) 的大作中提到: 】
: 你不是黑肉哥哥（哥哥是替midt mm加的~）


--
╱╲
╠┃║﹍﹍﹍﹍﹍﹍﹍﹍﹍﹍﹍﹍﹍﹍﹍
瞳凝秋水剑流星，裁诗为骨玉为神? ╱╤╤╝┃ _____________________________╲
翩翩白衣云端客，生死为谁一掷轻? ╲╧╧╗┃ ╱
╠┃║﹉﹉﹉﹉﹉﹉﹉﹉﹉﹉﹉﹉﹉﹉﹉
╲╱



※ 来源:·BBS 未名空间站 海外: mitbbs.com 中国: mitbbs.cn·[FROM: 76.199.]

发信人: siss (爱还在 义无赖 曲未终 戏台谢), 信区: BrainTeaser
标 题: Re: zz三人决斗
发信站: BBS 未名空间站 (Thu Jul 24 18:16:37 2008), 转信

^-^ 原来是博弈论啊
黑肉哥哥怎么不贴好玩的游戏了

【 在 hero080 (APM=080) 的大作中提到: 】
: siss妹妹好，黑肉哥哥在此 ^_^



--

※ 来源:·BBS 未名空间站 海外: mitbbs.com 中国: mitbbs.cn·[FROM: 75.153.]

发信人: hero080 (APM=080), 信区: BrainTeaser
标 题: Re: zz三人决斗
发信站: BBS 未名空间站 (Thu Jul 24 18:22:04 2008), 转信

因为每次我发贴了你都不顶！
【 在 siss (爱还在 义无赖 曲未终 戏台谢) 的大作中提到: 】
: ^-^ 原来是博弈论啊
: 黑肉哥哥怎么不贴好玩的游戏了


--
╱╲
╠┃║﹍﹍﹍﹍﹍﹍﹍﹍﹍﹍﹍﹍﹍﹍﹍
瞳凝秋水剑流星，裁诗为骨玉为神? ╱╤╤╝┃ _____________________________╲
翩翩白衣云端客，生死为谁一掷轻? ╲╧╧╗┃ ╱
╠┃║﹉﹉﹉﹉﹉﹉﹉﹉﹉﹉﹉﹉﹉﹉﹉
╲╱



※ 来源:·BBS 未名空间站 海外: mitbbs.com 中国: mitbbs.cn·[FROM: 136.152.]

发信人: siss (爱还在 义无赖 曲未终 戏台谢), 信区: BrainTeaser
标 题: Re: zz三人决斗
发信站: BBS 未名空间站 (Thu Jul 24 18:41:13 2008), 转信

我每次都认真地玩了！但都没有耐心地玩到底... 下回挪个鼎来顶一鼎

【 在 hero080 (APM=080) 的大作中提到: 】
: 因为每次我发贴了你都不顶！



--

※ 来源:·BBS 未名空间站 海外: mitbbs.com 中国: mitbbs.cn·[FROM: 75.153.]

发信人: hero080 (APM=080), 信区: BrainTeaser
标 题: Re: zz三人决斗
发信站: BBS 未名空间站 (Thu Jul 24 18:46:46 2008), 转信

赞，一定要鼎！
【 在 siss (爱还在 义无赖 曲未终 戏台谢) 的大作中提到: 】
: 我每次都认真地玩了！但都没有耐心地玩到底... 下回挪个鼎来顶一鼎


--
╱╲
╠┃║﹍﹍﹍﹍﹍﹍﹍﹍﹍﹍﹍﹍﹍﹍﹍
瞳凝秋水剑流星，裁诗为骨玉为神? ╱╤╤╝┃ _____________________________╲
翩翩白衣云端客，生死为谁一掷轻? ╲╧╧╗┃ ╱
╠┃║﹉﹉﹉﹉﹉﹉﹉﹉﹉﹉﹉﹉﹉﹉﹉
╲╱



※ 来源:·BBS 未名空间站 海外: mitbbs.com 中国: mitbbs.cn·[FROM: 136.152.]

发信人: pcasnik (pcasnik), 信区: BrainTeaser
标 题: Re: zz三人决斗
发信站: BBS 未名空间站 (Fri Oct 3 00:46:47 2008)

laptop was down for the past few weeks:(

let me try reviving this forum by discussing this old classic problem. i will slowly post my thoughts. feel free to join if you are interested.

: 三个人的射中概率分别为0<a<b<c<=1，当然是按照a, b, c的顺序射击。
: 请问：
: 1. 是否真的无论a, b, c等于多少，第一个人的最佳策略都是对天开枪？
: 2. 是否存在这样的a, b, c，使得三个人的胜出概率都是1/3？

here is my take of the problem:

assumptions:

(0) no one would be killed by a wild shot not intended for him, including the ones shot into the sky :)

(1) everyone is adopting the best strategy to maximize his probability of winning the duel.
note the difference between "winning" and "surviving". since they agree at the first place to participate the duel, i would say these three people tend to be more risky. this is to eliminate the possibility of the "coward equilibrium", i.e., everyone shoots into the sky, which we will see below.

(2) no one is going to change his strategy until someone is eliminated from the duel.
this is, surprisingly, not true at all generally (i'm assuming this just for simplicity; hopefully to relax it later)
let me give some examples: there could be several strategies with same probability of winning for a particular person. things become complicated in these cases - one needs to specify other criteria to determine his strategy (e.g., minimizing another player's probability of winning, because he is not very likable). i will ignore these.
another possibility is that let's say A decided to shoot into the sky, because he found his chance of surviving is better if he converts the situation into the one with the order of B, C then A. But in this case, if B, C somehow survives and it's A's turn again, his best strategy is not necessarily shooting into the sky again.

----- ----- ----- ----- ----- ----- ----- ----- ----- -----

i will start with only two people in the duel, let's say between A and B, with probability of a<b, so A shoots first. since we need the results for later. suppose B survives after A has acted. following our assumptions, B has to shoot or otherwise he will never win (even in the case that A shot into the sky - he could survive by doing the same thing but the winning probability is 0!). knowing B will definitely shoot, A has to shoot as well, otherwise he loses his priority advantage, which we will see below.

So there are no strategies for two-person-duel: A and B shoot in turns, and A has the probability of winning: p(a,b)=a/(a+b-ab), while for B is q(a,b)=(b-ab)/(a+b-ab). note p+q=1, and the notation here is that the first variable in the parenthesis indicates who shoots first. the priority advantage is ab/(a+b-ab).
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※ 修改:·pcasnik 於 Oct 3 00:51:04 2008 修改本文·[FROM: 75.34.]

发信人: pcasnik (pcasnik), 信区: BrainTeaser
标 题: Re: zz三人决斗
发信站: BBS 未名空间站 (Fri Oct 3 01:40:36 2008)

now for triangular duel.

let's consider the first part of the problem:

: 三个人的射中概率分别为0<a<b<c<=1，当然是按照a, b, c的顺序射击。
: 请问：
: 1. 是否真的无论a, b, c等于多少，第一个人的最佳策略都是对天开枪？

i will start with the case that both A and B shoot into the sky. by assumption (1), C will shoot, and he will shoot B. this is because by (2), he will be the only one shooting till he kills someone. of course he would like to face A in the two-person-duel instead of B, as summarized in the following table:
(columns are A, B, C's action, their probability of winning, X means not preferable)

A B C P(A) P(B) P(C)
1 - - B: p(a,c) 0 q(a,c) [X by B]
2 - - A: 0 p(b,c) q(b,c) [X by C]

as q(b,c)<q(a,c) if a<b. then we can conclude that B should at least shoot someone, otherwise he would never win.

so for the case B shoots C:

3 - C -: p(a,b) q(a,b) 0 [X by C]
4 - C B: p(b,c)p(a,b)+q(b,c)p(a,c) p(b,c)q(a,b) q(b,c)q(a,c)
5 - C A: p(b,c)p(a,b) p(b,c)q(a,b)+q(b,c)p(b,c) q(b,c)q(b,c) [X by C]

apparently C has to shoot or he would never win; he is going to shoot B, because he would like to face A when he gets to eliminate someone before B.

mathematically, for example, case 5, the winning probability for B is either in the case when he gets C first, which has a probability of p(b,c), then facing A with a winning probability of q(a,b) as A shoots first; or C gets A first, which has a probability of q(b,c), then B facing C with a winning probability of p(c,b) as B shoots first. easy to see 0<q(b,c)<q(a,c) if a<b.

now for the case B shoots A:

6 - A -: 0 q(c,b) p(c,b)
7 - A B: q(b,c)p(a,c) p(b,c)q(c,b) p(b,c)p(c,b)+q(b,c)q(a,c)
8 - A A: 0 p(b,c)q(c,b)+q(b,c)p(b,c) p(b,c)p(c,b)+q(b,c)q(b,c) [X by C]

same argument as above, if C shoots, he has to shoot B.
now consider case 6 vs 7 for C:
when p(c,b)>p(b,c)p(c,b)+q(b,c)q(a,c), i.e., p(c,b)>q(a,c), or a/(1-a)>b(1-c), then C would choose not to shoot.

to summarize, when A shoots into the sky, C would always shoot B, unless B shoots A, and a/(1-a)>b(1-c).

now consider B. for all the cases left, which are 4, 6 and 7. note 7 is always worse than 4 for him, as q(c,b)<q(a,b), so only in the case that q(c,b)>p(b,c)q(a,b), i.e., a/(1-a)>bc/(1-c) and a/(1-a)>b(1-c), he would shoot A, otherwise he would always shoot C.

thus one conclusion we can draw so far is that if a/(1-a)>bc/(1-c) && a/(1-a)>b(1-c), A's best strategy is not shooting in the sky - he would not win in this case!



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※ 修改:·pcasnik 於 Oct 3 02:15:48 2008 修改本文·[FROM: 75.34.]