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Public Zone 公開區 => Bookwyrm 書蟲天地 => Topic started by: chin on 18 September 2009, 06:30:33
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The Drunkard's Walk, How Randomnes Rules Out Lives, by Leonard Mlodinow
Another very interesting book. I got it this afternoon and read about half already.
The main line of the book follows the development of theory of randomness & probability (initially observed by gamblers, of course.) It's not dry history at all, but very captive narrative mixing with lots of interesting information, quiz and other bits and pieces.
(I am sure there are lots of reviews on the Internet for more details. I am just trying to write down part of the stories that interest me or touch my thought.)
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In the opening chapter, a good part was to illustrate how our life was affected by random events. I am quoting the following, hoping one day my kids will see the wisdom of it, and that the lesson make a lasting impression in their minds.
(page 10-11)
... That's why successful people in every field are almost universally members of a certain set - the set of people who don't give up.
A lots of what happens to us ... is as much the result of random factors as the result of skill, preparedness, and hardwork... That's not to say that ability doesn't matter - it is one of the factors that increase the chances of success - but the connection between actions and results is not as direct as we might like to believe.
Random events affecting one life are what some may call luck. As one grow older, the perception of what's luck changes too.
I once believed there is no such thing as lucky - that whatever good thing I haves, I deserve it because I worked hard and was smart enough. It take time to appreciate the random events that happen upon us, such that we can seize the opportunity and create good things out of it.
On the other hand, I have seen my share of people who are trying to read too much in random events, and becoming superstitious instead of taking events as they are.
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In Chapter 3 about sample space, and Gerolamo Cardano's attempt to solve a gambler's problem through relative simply probability concepts in 1500s, there is an interesting problem, supposedly called the Monty Hall problem.
(page 43)
Suppose the contestants on a game show are given the choice of three doors: Behind one door is a car; behind the others, goats. After a contestant picks a door, the host, who knows what's behind all the doors, opens one of the unchosen doors, which reveals a goat. He then says to the contestant, "Do you want to switch to the other unopened door?" Is it to the contestant's advantage to make the switch?
What's your answer? and why?
Please try without lookup reference on the Internet or book. :)
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In the next chapter dealing with paths and possible combinations, there is a short mentioning of the birthday problem.
My friends from high school would remember that in our F4 (?) class, we had about 40 students and 3 of us have the same birthday! What's the odds of that?
Our math teacher explained the solution, and the chance wasn't as remote as I thought. But now I have forgotten the solution. Perhaps someone can remind me...
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In the same chapter, the author talked about Blaise Pascal's life and his contribution to solve the combinatory problem by the now famous Pascal's Triangle.
Toward the end of Pascal's life, he completely transformed himself to a religious man, and "finally blasted off from planet Sanity." (Page 75) One of the products during this period was what's called Pascal's Wager, where he argued "the pros and cons of one's duty to God as if he were calculating mathematically the wisdom of a wager." (Page 76)
In another other book I read many years ago, the Pascal's Wager was said to be the "rational" arguement that has driven people en mass to Christianity. The Pascal Wager basically argued from the Expected Value angle. The key is that 1/2 of infinity is still infinity.
I never believed in his arguement but was not able to counter the idea in terms of expected value. Come to think of it now, the problem of his arguement is that not only the probability of God (as definited in the religious context) exists or not is not 50/50, the payoff is not infinity and the downside of believing not as trivial as "the sacrifices of piety." Or expressed as
EV = Prob * Positive Payoff + ( 1 - Prob) * Negative Payoff
Consider when Positive Payoff is not infinite, Negative Payoff is not trivial, & Prob is small enough...
[Some wise man said never to discuss religion or politics with your friends. In any case, the above can only serve as theoritical discussion. I learned many years ago that religion is emotional (taking a leap of faith), not rational (as in proving one way or the other.)]
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Before I forget, Benford's Law & Zeno's Paradox are all very interesting. Maybe I will comment more later.
(I heard of the Zeno's Paradox or something similar before, but did not realize it's usefulness in explaining calculus. :o)
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In the next chapter dealing with paths and possible combinations, there is a short mentioning of the birthday problem.
My friends from high school would remember that in our F4 (?) class, we had about 40 students and 3 of us have the same birthday! What's the odds of that?
Our math teacher explained the solution, and the chance wasn't as remote as I thought. But now I have forgotten the solution. Perhaps someone can remind me...
Really? Who are the 3 guys?
Let's exercise our maths. Consider a case of pre-fixed 3 people. Given a day-of-year number[1..365], the prob. of another one has the same bday is 1/365, likewise, the prob. of a third guy having the same bday is 1/(365^2).
Assume our class size was 40. There were 40C3 ways to choose 3 people... i.e. 9880 ways. At the end 40C3 / (365^2) = 7.4%
Anyone can verify it for me?
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They are Chin, Yan and Lam Wing Kei....all 29/10
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Really? Who are the 3 guys?
Let's exercise our maths. Consider a case of pre-fixed 3 people. Given a day-of-year number[1..365], the prob. of another one has the same bday is 1/365, likewise, the prob. of a third guy having the same bday is 1/(365^2).
Assume our class size was 40. There were 40C3 ways to choose 3 people... i.e. 9880 ways. At the end 40C3 / (365^2) = 7.4%
Anyone can verify it for me?
Just happened that we had similar discussion in the office on the same problem. We figured that the logic goes like this:
a. pick some one who has a birthday, any birthday. The chance of this is 100%.
b. randomly pick another person, and the chance of having the same birthday would be 1/365.
c. so by now, we established that for any PARTICULAR pair, the chance of having same birthday is a * b or 1/365.
d. in a group of n people, the # of permutations of pairs is nC2, thus the chance of having two persons having same birthday is nC2/365.
Example 1: I often heard that 23 people is enough to have a GOOD chance that a pair will have the same birthday. Using d above, 23C2/365=253/365 or about 2/3 chance.
Example 2: We have about 40 classmates in F5, 3 having same birthday. Using the same logic about, the chance should be 40C3/(365*365)=7.4% which is exactly what Kido had! :o (How clever we were... ;))
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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)
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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.
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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...
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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.) ::)
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Me too. Trapped by the Monty Hall problem......... :)
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Yan,
If you want to visualise the Monty Hall Problem, here you go:
http://www.youtube.com/watch?v=mhlc7peGlGg
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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%
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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.
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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 ]
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"E will not die in any case, so he probably will accept whenever the offer is >0"
Are you sure, Kido?
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"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"
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I think E will say NO to every suggestion unless he got quite a lot of it......because he will not die anyway (Just opposite to Kido's assumption) but, as all A, B, C, D know that, they will not give him any. E will ban all D's proposal in order to get the most.
C will be killed when D says NO (in principle), but if D really says NO, C will be killed and D will get nothing, so C may seek for majority i.e. C ~ 100%, D>0, E=0%
B, knowing that he will die if any two of C, D and E say NO, so he will have to make a proposal that any two of C, D and E would accept, It is clear that C would say NO anyway and B has to get the vote from both D and E to survive. If B dies, E must get nothing, so B has to offer E a little bit to make him say YES. So B~99%, D > 0%, E > 0%
A needs two votes only and notes that both D and E will get a bit even if he is killed, can offer them a bit more and keep the majority. So A ~ 99%, D > 0% and E > 0%
After I have finished writing the above, I check that my answer is quite like Kido's......but I have different view from yours.
1) E will not die in any case....Yes. He will accept whenever the offer is > 0, I think it is right too. He will absolutely get nothing if C makes proposal. So whenever A or B makes a proposal that he will have some share, he will say Yes.
2) The votes from D and E matter much. These two votes are the votes that A or B has to get in order to survive.
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you guys never question how E execute D if vote is not >50%???
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What do you mean? If A, B and C are killed, E will ban D's proposal anyway because D will be killed after that and E will take all..
So D will alway let C's proposal pass no matter what C proposes.
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yeah, you are right, hang. E will reject any proposal from D. But how can he execute D? (隻抽？）
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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。 <----that's the assumption.
So.....If E ban D's proposal, D will be 一诺千金 and let E kill him.......
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I simply do not believe 一诺千金can override survival.
anyway, model answer listed.
答案
要回答这个问题，一般人肯定会想到，1号必须先让另外两个人同意，所以，他可以自己得到32颗，而给2号3号各34颗。但只要仔细想想，就会发现不可能，
2号和3号有积极性让1号死，以便自己得到更多。所以，1号无奈之下，可能只有自己得0，而给2和3各50颗。但事实证明，这种做法依然不可行。为什么呢？
因为我们要先看4号和5号的反应才行。很显然，如果最后只剩下4和5，这无论4提出怎样的方案，5号都会坚决反对。即使4号提出自己要0，而把100颗钻石都给5，5也不会答应――因为5号愿意看到4号死掉。这样，5号最后顺利得到100颗钻石——因此，4的方案绝对无法获得半数以上通过，如果轮到4号分配，4号只有死，只有死！
由此可见，4号绝对不会允许自己来分。他注定是一个弱者中的弱者，他必须同意3号的任何方案！或者1号2号的合理方案。可见，如果1号2号死掉了，轮到3 号分，3号可以说：我自己100颗，4号5号0颗，同意的请举手！这时候，4号为了不死，只好举手，而5号暴跳如雷地反对，但是没有用。因为3个人里面有 2个人同意啊，通过率66.7％，大于50％！
由此可见，当轮到3号分配的时候，他自己100颗，4和5都是0。因此，4和5不会允许轮到3来分。如果2号能够给4和5一些利益，他们是会同意的。
比如2的分配方案是：98，0，1，1，那么，3的反对无效。4和5都能得到1，比3号来分配的时候只能得到0要好得多，所以他们不得不同意。
由此看来，2号的最大利益是98。1号要收买2号，是不可能的。在这种情况下，1号可以给4号和5号每人2颗，自己收买他们。这样，2号和3号反对是无效的。因此，1号的一种分配方案是：96，0，0，2，2。
这是不是最佳方案呢？再想一想，1号也可以不给4号和5号各2个，而只需要1个就搞定了3号，因为如果轮到2号来分配，2号是可以不给3号的，3号的得益只有0。所以，能得到1个，3号也该很满意了。所以，最后的解应该是：97，0，1，2，0。
好，再倒推。假设1号提出了97，0，1，0，2的方案，1号自己赞成。2和4反对。3∶2，关键就在于3号和5号会不会反对。假设3号反对，杀掉1 号，2号来分配，3自己只能得到0。显然，3号不划算，他不会反对。如果5号反对，轮到2号、3号、4号来分配，5号自己最多只能得到1。
所以，3号和5号与其各得到0和1，还不如现在的1和2。
正确的答案应该是：1号分配，依次是：97，0，1，0，2; 或者是：97，0，1，2，0。
人类如强盗，一直分钻石
出现这样的结果，相信是很多人没有想到的。
1、谁负责分配，谁的收益就最大！
2、看起来最强势的2号居然什么都得不到！他的反对是无效的！
3、次弱势的人如果位置恰当，稍微可以得到一些，如3号、4号、5号，但如果位置不当，也无所得，他们的反对票是根本无用的！如2号。
4、要使决策获得通过，只要收买个别次弱势群体就可以了。如果愿意多给他们一点，比如，给2号和4号各1个，而给3号和5号都增加1个，结果是：94，1，2，1，2。那么，他们4人也许都会感激涕零、高呼万岁、欢呼“青天大老爷”甚至于死心塌地地为决策者效力。
5、尽管1号的位置风险大，因为大家都会争先恐后去当1号。
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For those like counting, here is another interesting passage:
http://sciencenews.org/view/generic/id/60598/title/When_intuition_and_math_probably_look_wrong (http://sciencenews.org/view/generic/id/60598/title/When_intuition_and_math_probably_look_wrong)
I have two children, one of whom is a son born on a Tuesday. What is the probability that I have two boys?
Gary Foshee, a puzzle designer from Issaquah, Wash., posed this puzzle during his talk this past March at Gathering 4 Gardner, a convention of mathematicians, magicians and puzzle enthusiasts held biannually in Atlanta. The convention is inspired by Martin Gardner, the recreational mathematician, expositor and philosopher who died May 22 at age 95. Foshee’s riddle is a beautiful example of the kind of simple, surprising and sometimes controversial bits of mathematics that Gardner prized and shared with others.
“The first thing you think is ‘What has Tuesday got to do with it?’” said Foshee after posing his problem during his talk. “Well, it has everything to do with it.”
Even in that mathematician-filled audience, people laughed and shook their heads in astonishment.
When mathematician Keith Devlin of Stanford University later heard about the puzzle, he too initially thought the information about Tuesday should be irrelevant. But hearing that its provenance was the Gathering 4 Gardner conference, he studied it more carefully. He started first by recalling a simpler version of the question called the Two Children Problem, which Gardner himself posed in a Scientific American column in 1959. It leaves out the information about Tuesday entirely: Suppose that Mr. Smith has two children, at least one of whom is a son. What is the probability both children are boys?
Intuition would suggest that the answer should be 1/2, since the sex of one child is independent of the sex of the other. And indeed, had he been told which child was a boy (say, the younger one), this reasoning would be sufficient. But since the boy could be either the younger or the older child, the analysis is more subtle. Devlin started by listing the children’s sexes in the order of their birth:
Boy, girl
Boy, boy
Girl, boy
Since one child is a boy, we know that girl, girl isn’t a possibility. Of the three approximately equally likely possibilities, one has two boys and two have a girl and a boy — so the probability of two boys is 1/3, not 1/2, Devlin concluded.
He used this same method on the Tuesday birthday puzzle, enumerating the equally likely possibilities for the sex and birth day of each child and then counting them up.
If the older child is a boy born on Tuesday, there are 14 equally likely possibilities for the sex and birth day of his younger sibling: a girl born on any of the seven days of the week or a boy born on any of the seven days of the week. (This analysis ignores minor differences like the fact that slightly more babies are born on weekdays than on weekend days.)
Now suppose that the older child isn’t a boy born on Tuesday. The younger child then must be, of course. Now we count up the possibilities for the sex and birth day of the older child. If she’s a girl, she might have been born on any day of the week, generating seven more possibilities. If he’s a boy, he could have been born any day except Tuesday. (Otherwise this case would already have been counted in the first scenario: the older child a boy born on Tuesday). This second scenario generates just six, rather than seven, more possibilities.
Since each of these cases is (approximately) equally likely, we can compute the probability by dividing the number of cases in which there are two boys by the total number of cases. The total number of cases is 27: 14 if the older child is a boy born on Tuesday and 13 if the older child isn’t. In 13 of those cases both children are boys (7 if the older child is a boy born on Tuesday and 6 if he isn’t), yielding a probability of 13/27.
Devlin was astonished by this answer. As a mathematician, he had long been familiar with the Two Children Problem and its answer of 1/3. “Knowing the birth day is a Tuesday may (and does) make a difference, but it surely cannot make much of a difference, right?” he wrote in his blog, Devlin’s Angle. “Wrong.” After all, 13/27 is far closer to 1/2 than 1/3.
So why does intuition seem to lead us so astray? Both the intuitive and the mathematically informed guesses are wrong. Are human brains just badly wired for computing probabilities?
Not so fast, says probabilist Yuval Peres of Microsoft Research. That naïve answer of 1/2? In real life, he says, that will usually be the most reasonable one.
Everything depends, he points out, on why I decided to tell you about the Tuesday-birthday-boy. If I specifically selected him because he was a boy born on Tuesday (and if I would have kept quiet had neither of my children qualified), then the 13/27 probability is correct. But if I randomly chose one of my two children to describe and then reported the child’s sex and birthday, and he just happened to be a boy born on Tuesday, then intuition prevails: The probability that the other child will be a boy will indeed be 1/2. The child’s sex and birthday are just information offered after the selection is made, which doesn’t affect the probability in the slightest.
Gardner himself tripped up on his simpler Two Children Problem. Initially, he gave the answer as 1/3, but he later realized that the problem is ambiguous in the same way that Peres argues that the Tuesday Birthday Problem is. Suppose that you already knew that Mr. Smith had two children, and then you meet him on the street with a boy he introduces as his son. In that case, the probability the other child is a son would be 1/2, just as intuition suggests. On the other hand, suppose that you are looking for a male beagle puppy. You want a puppy that has been raised with a sibling for good socialization but you are afraid it will be hard to select just a single puppy from a large litter. So you find a breeder who has exactly two pups and call to confirm that at least one is male. Then the probability that the other is male is 1/3.
In the scenario of Mr. Smith, you’re randomly selecting a child from his two children and then noticing his sex. In the puppy scenario, you’re randomly selecting a two-puppy family with at least one male.
The remarkable thing that Foshee’s variation points out is that any piece of information that affects the selection will also affect the probability. If, for example, you selected a family at random among those with two kids, one of whom is a boy who plays the ukulele and wants to become a dancer, the ukulele-playing and dancing ambitions would affect the probabilities about the sex of his sibling.
Peres says that we shouldn’t despair about our probabilistic intuition, as long as we apply it to familiar situations. The difficulty of these problems is rooted in their artificiality: In real life, we almost always know why the information was selected, whereas these problems have been devised to eliminate that knowledge. “The intuition develops,” he points out, “to handle situations that actually occur.”
Still, Gardner’s initial overly narrow interpretation warns of the dangers of over-hasty analysis of probability questions — and shows the wonder that can come from them.