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< Earlier Kibitzing · PAGE 117 OF 172 ·
Later Kibitzing> |
Jun-13-12
 | | OhioChessFan: Ali would have killed him. |
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| Jun-13-12 | | Jim Bartle: Has the NBA made a bulk purchse of black-rimmed glsses for postgame press conferences, or are the players passing around a single pair to wear? |
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Jun-13-12
 | | HeMateMe: What's next, a return to the Gazelles frames, of the 80s? In NYC people were getting killed for the expensive Gazelle frames. |
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Jun-17-12
 | | talisman: who is the only world chess champion born on Father's day? :) kinda easy huh? |
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| Jun-17-12 | | diceman: <OhioChessFan:
Fill in the blanks with the same seven letters in the same order:
The __________ doctor was ___________ to operate because she had __________.>The itching doctor was itching to operate because she had itching. (sounds a bit rash to me) |
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Jun-18-12
 | | Sneaky: You know what I find more annoying than the Monty Hall problem? (I am guessing most of you know what the Monty Hall Problem is already so I'll spare you my rendition of it, but just in case there are some noobs here, have a link: http://en.wikipedia.org/wiki/Monty_...) Here's what I find annoying, the puzzle that goes like this: <A woman says "I have two children, and one of them is a boy." What's the probability that the other child is also a boy?> If you don't immediately want to say 50% you are not a normal person. And yet, that's not the right answer. It's a little like the Monty Hall problem in that you're given some knowledge of the situation that changes the odds from the "instinctive" answer of 50%. With the Monty Hall problem, I see the wrinkle that confounds everybody, and I've watched the actual show Let's Make a Deal so many times I had the "switching trick" figured out as a kid. But with THIS puzzle.... oooooh, it drives me mad. I know the right answer but I want to deny its reality. |
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Jun-18-12
| | Sneaky: I should add for clarity: when the lady says "One of them is a boy" she means "at least one of them" ... perhaps she has two boys. She's a very precise speaker. |
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Jun-18-12
| | Sneaky: From Jeopardy the other day, "Sports quotations" was the category. Who said the following? <I just paid lip service to being the greatest. Joe, he was the greatest.> Easy question I know, but great quote. |
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| Jun-18-12 | | Jim Bartle: Who is the Louisville Lip? |
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Jun-18-12
| | Sneaky: (1) The noble SS doctor was nobless to operate because she had no bless. (2) The Hungary doctor was "hungary" to operate because she had hun, Gary. |
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Jun-18-12
| | Sneaky: I should clarify on the Jeopardy quote: Of course he wasn't speaking of Joe Frazier, he was speaking of Joe Louis. A rare moment of humility for Ali... must have been in his retirement years. |
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| Jun-18-12 | | diceman: <Sneaky: You know what I find more annoying than the Monty Hall problem?> The Wink Martindale problem? |
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Jun-18-12
 | | Marmot PFL: <<A woman says "I have two children, and one of them is a boy." What's the probability that the other child is also a boy?> Seems simple, with 2 children you have 4 possibilities: G G, B G, G B, and B B. If you know at least 1 child is a boy, than G G is eliminated, leaving B G, G B, and B B. Hence the chance that the other child is a boy is 1/3. |
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| Jun-18-12 | | Petrosianic: There's only two possibilities. Boy or girl. The first one is given in advance. So the odds are 50/50. (Or if you want to be picky, you could go by the population as a whole, which is often slightly higher female.) |
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Jun-18-12
| | Marmot PFL: <With the Monty Hall problem, I see the wrinkle that confounds everybody, and I've watched the actual show Let's Make a Deal so many times I had the "switching trick" figured out as a kid. > I watched that show too but it's tricky, sometimes they rotate the stage and the bad looking prize is actually the best. |
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Jun-19-12
| | OhioChessFan: Several have gotten it, but for future visitors:
The <notable> doctor was <not able> to operate because she had <no table> |
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Jun-19-12
| | Marmot PFL: A casino offers a game of chance for a single player in which a fair coin is tossed at each stage. The pot starts at 1 dollar and is doubled every time a head appears. The first time a tail appears, the game ends and the player wins whatever is in the pot. Thus the player wins 1 dollar if a tail appears on the first toss, 2 dollars if a head appears on the first toss and a tail on the second, 4 dollars if a head appears on the first two tosses and a tail on the third, 8 dollars if a head appears on the first three tosses and a tail on the fourth, and so on. What would be a fair price to pay the casino for entering the game? |
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Jun-19-12
 | | Shams: I brute-forced a solution to <Marmot>'s puzzle, but I've been feeling probability-stupid lately and I'd like to learn more. Can someone post the equation that solves it? |
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Jun-20-12
| | Sneaky: Marmot's puzzle is a famous problem that I've heard dubbed the "St. Petersburg Paradox". I had a huge discussion on the subject some years ago on my chessforum because, to my surprise, it's the cornerstone of a related paradox that was driving me positively up the wall (the Envelope-Switching paradox). <Shams> If you found a "brute force" solution I imagine you wrote a computer program to simulate the game and then ran it for a billion iterations, then divided by a billion, or something like that. If that was your method, your answer is not just wrong, but INFINITELY wrong. I'm not trying to be a jerk--I just honestly find it humorous how a sensible approach like that could fail not just by an order of magnitude, but by "an infinitude!" In my version you get rewarded for flipping heads, and another detail is different, but its effectively the same paradox. Here's a link that explains it --> Sneaky chessforum |
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Jun-20-12
| | Shams: <Sneaky> Well, I said I brute-forced "a" solution, not "the" solution. I'm not yet convinced there is an infinite EV for this game. (I ended up at around $3.) Still, I can see that I'll have to start my crash course in probability with a still-easier problem. :) |
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Jun-20-12
| | Marmot PFL: <Sneaky> The envelope problem says a lot about human nature, but the calculations seem easy. You are shown two envelopes, told that one has twice as much money as the other and asked to choose one. Then you are asked if you want to keep the money or switch envelopes. You open the envelope and see $100, so you think, I can gain $100, and the most I can lose is $50, and it's a 50-50 chance... 100 x 1/2 + (-50) x 1/2 = 25
so you stand to make $25 by switching.
Since this is true for any amount
2 X A x 1/2 + (-A)/2 x 1/2 = A/4
don't even bother opening the first envelope, just instantly trade. To save even more time, don't even take the 1st envelope, just go directly to the 2nd one... |
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Jun-20-12
| | Marmot PFL: <Shams> Here is the Wiki explanation- <What would be a fair price to pay the casino for entering the game? To answer this we need to consider what would be the average payout: With probability 1/2, the player wins 1 dollar; with probability 1/4 the player wins 2 dollars; with probability 1/8 the player wins 4 dollars, and so on. The expected value is thus> equation does not copy <Assuming the game can continue as long as the coin toss results in heads, in particular that the casino has unlimited resources, this sum diverges without bound, and so the expected win for the player, at least in this idealized form, is an infinite amount of money. According to the usual treatment of deciding when it is advantageous and therefore rational to play, one should therefore play the game at any price if offered the opportunity. Yet, in published descriptions of the game, e.g., (Martin 2004), many people expressed disbelief in the result. Martin quotes Ian Hacking as saying "few of us would pay even $25 to enter such a game" and says most commentators would agree.> http://en.wikipedia.org/wiki/St._Pe... |
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| Jun-20-12 | | RMKvdS: There is no fair price. The number of times heads can be thrown consecutively is not bounded from above. Though the probability itself is limited to zero, the probability of a number of n heads in a row will never be zero. Ergo, a fair price does not exist. |
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| Jun-20-12 | | izimbra: <one should therefore play the game at any price if offered the opportunity. > The expected value is infinite, the rational is a different matter. http://en.wikipedia.org/wiki/Utility
<Money:....The utility function is concave in the positive region, reflecting the phenomenon of diminishing marginal utility.> |
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Jun-20-12
| | Marmot PFL: <2 X A x 1/2 + (-A)/2 x 1/2 = A/4 > should be
A x 1/2 + (-A)/2 x 1/2 = A/4
Anyway, the paradox is the conviction that no matter what choice you make in a symmetric situation with no information, you are led to think you would better off making the other choice. |
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