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Monty Hall Paradox



make a deal (The Monty Hall Paradox)

Suppose that in a TV show you are participating and animator gives choice of three doors: what's behind the chosen is yours. Behind one of them is a new car, behind the other two a goat. You choose one of doors, say 1, and at that time (before opening) the driver, who knows what's behind each door, opens one of the other two, say 3, and shows a goat. And I asked "do you want to change your choice (opening the door 2)?" Do you should change?

This problem is paradoxical because common sense tells us that since we know nothing about what's behind the other two doors are the same chances of winning with which they choose with the other. But it can be shown that since the driver does know that behind the doors, it should change our decision. For a discussion of this point see http://math.ucsd.edu/ ~ crypto / Monty / montybg.html .

What implications does the driver know what's behind the doors? The most important is that the two two-thirds of the time (with probability 2 / 3) has no choice: the door that opens is the prize! Why, with probability 2 / 3? Well, eventually, for example of 100 players, about 67% of players choose a losing door (behind it is a goat) and in those cases the driver will have to show the contestant losing the second door (can not show the car). That is why the participants have more chances to win by changing your election to the third option as a strategy (ie, always, as a rule). Of course, if I had chosen the car door is going to lose, but that will happen with probability 1 / 3.

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This configuration of shapes and colors for the previous post. I decided to add it to the posts that try to be hoaxes. Moreover, it is a test to put pictures on the blog and try to make it more friendly (in the sense of friendly software). In the case of a post with a problem and see which configuration you get. Sincerely jmd

in vl

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Guide Hatter happy (and yet, tragic) Troubleshooting

Note: it is understood that the guide is in the air (not that, but present) to solve mathematical problems in contemporary education - no one should be attributed (in particular swear the author was not me!)

Rule 1: Whenever possible, avoid reading the problem statement. Reading takes time and causes confusion. (Especially when it includes words not commonly used in youth culture - which is almost always ... ... what little kindness.)

Rule 2: Extract the numbers stated in the order they appear. Be careful with the numbers written in words.
Rule 3: If rule 2 gives you three or more numbers, the best chance is to add them all.
Rule 4: If two about the same size, then the rest is a good bet to achieve a good result.
Rule 5: If two numbers but of very different sizes, then divided, but if the division is more accurate then multiplies.
Rule 6: If you beat the problem requires using a formula then choose one with enough literals (letters) to use all the numbers given in the problem.
Rule 7: If rules 1 to 6 seem not work, make a desperate attempt: take the set of numbers found in regulation 2 random filling operations run about two pages. Note: do not forget to enclose a circle or box with five or 6 answers in each page ... chance and one of them is the answer, remember that you can get some points for trying, and are such attempts.

final note: the method is particularly effective when the teacher is among those who like to put "real" problems (the classic example - called "the age of the captain" -: on a boat there are 25 goats, 10 and two bastards kids, how old is the captain?). Under Rule 3, the answer is 37 - and a captain of 37 years is typical, as can be seen in old pirate movies. (I think it is best to avoid giving this answer explanations when answering "undecidable response given the irrelevance of the data.")