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Yes, when Ben was first asked to choose a door, he had a 33.3% chance of picking the car. When the professor re-offers the question, Ben has a 66.7% chance of finding the car behind door #2. My question is how, when anyone in the world knows that there is a goat and a car behind 2 doors, there should be a 50% chance of choosing right? The answer is solved by accounting for "variable change", but unfortunately, this guy has know idea what that means and how it could be applied in a problem that makes more sense to me? | Yes, when Ben was first asked to choose a door, he had a 33.3% chance of picking the car. When the professor re-offers the question, Ben has a 66.7% chance of finding the car behind door #2. My question is how, when anyone in the world knows that there is a goat and a car behind 2 doors, there should be a 50% chance of choosing right? The answer is solved by accounting for "variable change", but unfortunately, this guy has know idea what that means and how it could be applied in a problem that makes more sense to me? | ||
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| + | Think of it this way. Let's look at the possible cases in which Ben picks door #1 and switches his choice to door #2 after the goat is revealed behind door #3. An initially correct guess (which happens 33.3% of the time) will result in losing because he switches from the winning door to door #2. However, a wrong guess (which happens 66.7% of the time) always ends up successful because he switches from the losing door to door #2 after being shown the goat behind door #3. In this case, since #1 and #3 are losers, door #2 must be the winner. | ||
Latest revision as of 11:30, 24 September 2008
Hey guys, the midterm is on October 23rd and is worth 30% of the grade!!! And the Final is on 12/17/08 from 10:20-12:20, in University 303! Just thought I'd give the heads up!
i did not like this homework. that is all.
I was not a fan of this either, because it gets confusing on some of the combinatorial proofs. But if you search on wikipedia for them, they give a couple good examples that help.Combinatorial proof
I thought some of the parts were pretty hard, but I think I got most of them right fortunately.
THE GAME SHOW HOST PROBLEM!
Has anyone seen the movie 21? If you have, you know what I'm talking about... If you haven't, I am having a little trouble understanding the game show host problem! The teacher asks Ben:
Ben, suppose you are on a game show. You are given the chance to choose between 3 doors. Behind one of the doors is a new car... behind the other 2 are goats
Ben chooses door #1.
The game show host opens door #3 to reveal one of the goats. He then proceeds to ask Ben, "Ben, do you want to stay with door #1 or go with door #2"?
Here is the punchline. Is it in Ben's interest to change his answer?
Answer
Yes, when Ben was first asked to choose a door, he had a 33.3% chance of picking the car. When the professor re-offers the question, Ben has a 66.7% chance of finding the car behind door #2. My question is how, when anyone in the world knows that there is a goat and a car behind 2 doors, there should be a 50% chance of choosing right? The answer is solved by accounting for "variable change", but unfortunately, this guy has know idea what that means and how it could be applied in a problem that makes more sense to me?
Think of it this way. Let's look at the possible cases in which Ben picks door #1 and switches his choice to door #2 after the goat is revealed behind door #3. An initially correct guess (which happens 33.3% of the time) will result in losing because he switches from the winning door to door #2. However, a wrong guess (which happens 66.7% of the time) always ends up successful because he switches from the losing door to door #2 after being shown the goat behind door #3. In this case, since #1 and #3 are losers, door #2 must be the winner.
