The Drunkard's Walk, How Randomnes Rules Out Lives, by Leonard Mlodinow

[chin]

My daughter read a book about math for kids, kind of like a funny version of the Story of Math & Mathematicians, and it offers an even more simple solution of the birthday problem.

Let's say in the case of 30 people, you line them up on a straight line, and start counting from the left.

The 1st one would have no chance of having the same birthday as the person on the left, since there is no body on the left; The 2nd person would have 1/365 chance having the same birthday as the one single person on the left; then the 3rd person would have 2/365 since there are 2 person on the left; and so on.

At the end, you add up from 0/365 ... to 29/265, or 29*15/365.

[chin]

Let's say,

you play Sik-Bo in Casino.

After four SMALLs in a roll, engineering/scientists mind-set will always say the next game is 1/2 Big, 1/2 Small.  While some other people will start to believe the dice are biased and the chance of SMALL is higher on the next game.  So everything may come to an art.  I believe Chin will agree also.  No matter how accurate you model your wagering system, the assumptions behind are all set by human.

Let's get back to the three doors dillemma.  I will look into the host's eyes and say "Yes..............Maybe" and see if any clue from his eyes or his facial expression can tell.  Besides, I may also knock on the door and see if the goat bleats.  (Probability tells no difference)

The probability actually changed in the 3-door problem after the host open an empty door.

What you suggested is a but similar to in horse racing, some gamblers would guest the trainer intention, etc...

[kido]

Let's say,

you play Sik-Bo in Casino.

After four SMALLs in a roll, engineering/scientists mind-set will always say the next game is 1/2 Big, 1/2 Small.  While some other people will start to believe the dice are biased and the chance of SMALL is higher on the next game.  So everything may come to an art.  I believe Chin will agree also.  No matter how accurate you model your wagering system, the assumptions behind are all set by human.

Let's get back to the three doors dillemma.  I will look into the host's eyes and say "Yes..............Maybe" and see if any clue from his eyes or his facial expression can tell.  Besides, I may also knock on the door and see if the goat bleats.  (Probability tells no difference)

I also agreed with wongyan, without further thinking, 50-50 was my answer and switching doesn't make any difference. But it was so tempted to wiki the term and the result was so surprised.   (I should have know this problem should have odd results because it shouldn't be that trivial since Chin specifically posted it here.) 

[hangchoi]

Me too. Trapped by the Monty Hall problem.........

[hangchoi]

Yan,

If you want to visualise the Monty Hall Problem, here you go:

http://www.youtube.com/watch?v=mhlc7peGlGg

[wongyan]

that reminds me a very interesting quiz.

5个海盗抢到了100颗质地完全一样的钻石，但又不愿意平分，最后5个人同意抓阄解决：按照1，2，3，4，5的抓阄结果，将5个人编号。抓到1的是1号，抓到2的是2号，依次类推。现由抓到“1”的1号海盗提出分配方案，为了防止他分配不公，海盗们达成一致：他的方案必须有所有人（包括1号自己）的半数以上（注意，必须大于50％）通过才可执行。否则，他将被杀死，再由2号海盗提出分配方案，2号的方案也要所有剩下的人（包括他自己）的半数以上通过。否则他也将被杀死，依次类推。假设这5个海盗都贪婪成性、残忍无比、绝顶聪明而又一诺千金，都想自己得到最多，都想看到别人死去而自己活。请问，1号要怎样分配才能使自己得到的钻石最多？

another version is >=50%

[chin]

that reminds me a very interesting quiz.

5个海盗抢到了100颗质地完全一样的钻石，但又不愿意平分，最后5个人同意抓阄解决：按照1，2，3，4，5的抓阄结果，将5个人编号。抓到1的是1号，抓到2的是2号，依次类推。现由抓到“1”的1号海盗提出分配方案，为了防止他分配不公，海盗们达成一致：他的方案必须有所有人（包括1号自己）的半数以上（注意，必须大于50％）通过才可执行。否则，他将被杀死，再由2号海盗提出分配方案，2号的方案也要所有剩下的人（包括他自己）的半数以上通过。否则他也将被杀死，依次类推。假设这5个海盗都贪婪成性、残忍无比、绝顶聪明而又一诺千金，都想自己得到最多，都想看到别人死去而自己活。请问，1号要怎样分配才能使自己得到的钻石最多？

another version is >=50%

Game theory?!

So the first guy would divide the loot into pile 1 to 5?

I imagine 1 would need to keep 2 & 3 very happy and no incentive to vote against him. 1 would not care much about the vote from 4 & 5.

[kido]

that reminds me a very interesting quiz.

5个海盗抢到了100颗质地完全一样的钻石，但又不愿意平分，最后5个人同意抓阄解决：按照1，2，3，4，5的抓阄结果，将5个人编号。抓到1的是1号，抓到2的是2号，依次类推。现由抓到“1”的1号海盗提出分配方案，为了防止他分配不公，海盗们达成一致：他的方案必须有所有人（包括1号自己）的半数以上（注意，必须大于50％）通过才可执行。否则，他将被杀死，再由2号海盗提出分配方案，2号的方案也要所有剩下的人（包括他自己）的半数以上通过。否则他也将被杀死，依次类推。假设这5个海盗都贪婪成性、残忍无比、绝顶聪明而又一诺千金，都想自己得到最多，都想看到别人死去而自己活。请问，1号要怎样分配才能使自己得到的钻石最多？

another version is >=50%

I remembered I've seen this problem before.  However I didn't understand and thus didn't remember anything about the answer.  So I'll try my best here.

Let them be called A, B, C, D, E.

For anyone who has the chance to divide, he himself must agree on his own proposal, he just please the rest such that the agreed proportion > 50%.   E will not die in any case, so he probably will accept whenever the offer is >0, D will die hard if he's going to make a proposal. (because E will surely banned him).  And I think this is important, because he will try to make A-C's proposal successful.  So, D will always vote a YES to A,B,C's proposal to avoid dying.


If C is going to make a proposal: He just need to have most for himself, 0 to D, 0 to E. because D will vote YES to him to avoid D proposes, C doesn't needed to please D.  i.e. C:100 D:0 E:0
[in this case C vote YES, D vote YES, E not important]

If B is going to make a proposal: He just needed to proposed a better-than-C proposal. Because D will vote YES for sure. And he just needed to issue some small amt to E and E will vote a YES. He will probably propose this : B:(100-y-x) C:0 D:x>0 E:y>0
[in this case B vote YES, C vote NO, D vote YES(better than nothing), E vote YES(this is better than nothing)]

Seeing this, A needed to proposes a better-than-B proposal.  A's proposal should be :
A: 96
B: 0
C: 0
D: z=2 > x
E: u=2 > y

Is that the case?

[I modified it 4 times already...hehe ]

[wongyan]

"E will not die in any case, so he probably will accept whenever the offer is >0"

Are you sure, Kido?

[kido]

"E will not die in any case, so he probably will accept whenever the offer is >0"

Are you sure, Kido?

Errrr. Not sure. But it was my assumption.  Maybe it should read "E will not die in any case, so he probably needed to maximize his profit only"