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The Drunkard's Walk, How Randomnes Rules Out Lives, by Leonard Mlodinow
chin:
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)
kido:
--- Quote from: chin on 18 September 2009, 06:31:29 ---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...
--- End quote ---
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?
hangchoi:
They are Chin, Yan and Lam Wing Kei....all 29/10
chin:
--- Quote from: kido on 26 September 2009, 10:48:48 ---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?
--- End quote ---
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... ;))
wongyan:
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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