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See: http://tierneylab.blogs.nytimes.com/200 … -problems/
Quick Summary from the article:
1. Mr. Smith has two children, at least one of whom is a boy. What is the probability that the other is a boy?
2. A 10-door version of the Monty Hall Problem. You pick one door, but before its opened Monty will always open seven other doors to reveal goats. Should you switch? And what are the odds?
3. Monty Hall: suppose, when youre confronted with the final choice to stick or switch doors, you flipped a coin heads, you stick with your original door; tails, you switch. What would be your odds of winning the car?
Last edited by MathsIsFun (2008-04-14 00:51:17)
"The physicists defer only to mathematicians, and the mathematicians defer only to God ..." - Leon M. Lederman
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The articles from adjacent days are on the Monty Hall problem as well, containing variations to try to convince people that the solution really does work.
I was fooled by one of them:
"Suppose that Andy (A), Ben (B), and Chris (C) are three men selected at random from the telephone book. If you learn that Andy is taller than Chris, then what is the probability that Andy is the tallest of the three?"
Why did the vector cross the road?
It wanted to be normal.
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"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."
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Wrap it in bacon
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"The physicists defer only to mathematicians, and the mathematicians defer only to God ..." - Leon M. Lederman
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"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."
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Try this version:
Monty Hall with 4 doors - hiding 2 cars and 2 goats.
You pick a door.
The host knows what's behind the doors and tosses a coin.
If it's heads, he will eliminate a car from the remaining doors.
If it's tails, he will eliminate a goat.
As it happens, the coin comes up heads and a car is eliminated.
There are three doors left (including the one you chose). Should you stick or switch?
Or are the odds 1/3 regardless?
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You should stick with the door you picked originally. If you keep your first choice you have a 50% chance of winning, while if you switch doors you only have a 25% chance of winning. To picture why this is, below are all the possible configurations:
CCGG
CGCG
CGGC
GCCG
GCGC
GGCC
G is for goat, C is for car. Assume that you chose door #1, the choice on the far left (it doesn't matter which you choose, #1 is just used for an example). In the first 3 cases you'll win if you keep your choice, while you'll lose in the second 3 cases. This gives you a 50% chance of winning.
Now, if you decide to switch, you'll lose every time in the first 3 cases. In the second 3 cases you have a 50% chance of winning. Those 3 cases only make up half of the possibilities, so your total chance of winning is 50% * 50% = 25% chance of winning, so you're better off keeping your first choice.
Wrap it in bacon
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