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Mar 13, 2009, 06:58 PM  #126 
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Mar 13, 2009, 07:12 PM  #127  
10990
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It took me a while to make sense of the solution to the Monty Hall problem, now when I think about it, it's logical. Quote:
Here's a problem I like: Code:
_________________A         _______________ B C D E
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Mar 13, 2009, 07:38 PM  #128 
Monty Hall problem
The key is the bold word:
Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat. He then says to you, "Do you want to pick door No. 2?" Is it to your advantage to switch your choice?The game show host is using his/her knowledge of the answers to pick which door to show you. That imparts to you additional information, which you can use to boost your chances. The example posed in this thread is contrived, and eliminates 1/3 of the cases before you get to decide whether to change doors: You pick door #2. You are then shown door #1, whether it's good or bad. If it's good (that's 1/3 of the time), you lose. If it's bad (that's 2/3 of the time), you can switch if you want, but it's 5050 either way. Overall, your chances of winning are therefore ( 1/3 x 0 ) + ( 2/3 x 1/2 ) = 1/3 with either strategy.In the Monty Hall problem, the rules are slightly different, because all three cases are considered: You pick any door, but without loss of generality we can assume it's door #2. You are then shown another door that is known to have a goat behind it. (This is always possible, because the host gets to peek.) In the case where door #1 is a winner, the host will show you door #3 instead of door #1, so you will win by switching doors. It's no longer 5050. If door #1 is not a winner, it's still 5050 no matter which strategy you pick (the first case above). Overall, your chances of winning by staying with door #2 are therefore ( 1/3 x 0 ) + ( 1/3 x 1 ) + ( 1/3 x 0 ) = 1/3 while your chances if you change doors are ( 1/3 x 1 ) + ( 1/3 x 0 ) + ( 1/3 x 1 ) = 2/3, i.e., changing doors is better.In a way, it's like this game: I think of a color (red, green, or blue). You guess what I'm thinking. If you get it right, you win. If you guess the color wrong, we play a second round where I think of an animal (dog or cat) and you guess which one. If you get it right, you win. If you get it wrong, I win. In this thread we're talking only about your chance in the second round, clearly 5050. In the Monty Hall problem, we're talking about the overall game, where you win 2/3 of the time, either by guessing the color or by guessing the animal.
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Mar 13, 2009, 07:41 PM  #129  
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Sorry, still not true. The odds are the same whether you switch doors or not, in the problem stated above. The odds might change if the other party gave you new information, e.g., by picking which door to show you based on already knowing what's behind them. That's not the case here, where what's revealed to you is predetermined to be door #1.I've only quickly read through these problems, but I believe that is the distinction. With the Monty Hall paradox, Monty Hall knows the location of the car, and purposefully choses a door hiding a goat. Dr. Q based his assumption on the fact that one door is opened at random. EDIT: And I'm too slow. 

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Mar 13, 2009, 07:45 PM  #130 
Ahh! As an undergraduate maths student you can see why I was curious!
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Mar 13, 2009, 07:46 PM  #131 
Does CalBoy's riddle address the guard's knowledge of what's behind each door? Can one form an accurate answer without that information?


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Mar 13, 2009, 08:01 PM  #132  
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Without that info, it's a very different problem. 

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Mar 13, 2009, 08:01 PM  #133 
Monty Hall problem
There's even another possibility that's fun to consider. What if the game show host gets to decide whether or not to show you what's behind another door based on which door you pick, and he/she wants you to lose?
Supposed you are a trusting contestant. You start by picking door #2. If that's a goat (which is 2/3 of the time), he/she doesn't offer you a chance to change doors, so you lose. If it's a winner, he/she shows you a goat behind another door and asks if you want to change. Since you don't know any better, you remember the Monty Hall problem, say yes, change doors, and lose. Your chances are zilch in this rigged game! However, if you know (or suspect) that the game show host is dishonest, then you'll know to stick with your initial guess, and you'll win 1/3 of the time, which is the most you can hope for. Once again, it's the amount of information you have that determines the odds. * * * Now let's assume the host wants you to win (to boost the show's ratings, since sponsors pay for the prizes anyway). If you pick a winner (1/3 of the time), he/she won't offer you another door, so you win. If you pick a goat (2/3 of the time), he/she shows you the other goat and asks if you want to change. If you think he/she is against you, you never switch in this case, and always lose. Total odds: 1/3. If you don't know that he/she is trying to fix the game, you might decide at random in this case and win 50% of the time. Total odds: 2/3. If you know that he/she is on your side, you always switch in this case and win. Total odds: 3/3. If you think it's the Monty Hall case (even though it's not), you always switch, so you always win. Total odds: 3/3.
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Mar 13, 2009, 08:56 PM  #134 
You have one bucket filled with three gallons of water, and another bucket filled with eight gallons of water. How many buckets do you have?
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Mar 13, 2009, 08:59 PM  #135 
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Mar 13, 2009, 09:00 PM  #136 
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Mar 14, 2009, 12:02 AM  #137  
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Err, doesn't that not work, because Angle ABE= 90 Deg. and Angle ACE= 90 deg. Together they equal 180, and in order for that to be a square, each angle has to be 90 deg. so Angle ADE doesn't equal that. 

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Mar 14, 2009, 12:51 AM  #138  
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If so, I feel cheated because you used the words "and another." 

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Mar 14, 2009, 01:40 AM  #139 
The answer's two. You have one bucket, and another bucket. That makes two.
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Mar 14, 2009, 01:57 AM  #140 
Another Fun Brain Teaser
Here's an old one I learned a long time ago:
A farmer has a chicken, a bag of chicken feed, and a fox. He needs to get across the river with all three items. However, the small boat he has only allows him to carry one item across at a time. He can't leave the chicken with the feed because he will eat it. Likewise, he can't leave the fox with the chicken because he will eat him. How can the perplexed farmer get all three across successfully?
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Mar 14, 2009, 02:01 AM  #141  
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Second trip: take the feed, swap it with the chicken. Third trip: take the fox. Fourth trip: take the chicken. 

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Mar 14, 2009, 03:35 AM  #142 
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Mar 14, 2009, 11:22 AM  #143  
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Angle ADE is 45º and angles ABE and ACE are both less than that.
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Mar 14, 2009, 11:33 AM  #144 
Oh, I figured that the line was Line A not just Point A.
Point A, and Point B (If the line is drawn between them), is two lines cut by a transversal, so the angle A(Top side), is congruent to Angle ABE. The same way, Segment AC is a transversal to the two parallel lines, and the new angle of Angle A (Top side), is congruent to Angle ACE. And also, Segment AD is two lines cut by a transversal, and the top angle of A is congruent to Angle ADE. Thats all I got, I suck with proofs. Mostly because our Geometry teacher doesn't believe that they exist . And we don't have to ever do it. 

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Mar 15, 2009, 12:31 AM  #145 
Good job! It's one of those simple "think outside of the box" puzzles. Most don't think about bringing an item back across after they have taken it. Like I said, one of my favorites.
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Mar 25, 2009, 05:55 PM  #146  
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My granddad told this one back before who flung the chunk.
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Mar 25, 2009, 06:05 PM  #147 
Now that's just not a statement you read everyday...
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Mar 25, 2009, 06:08 PM  #148  
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Mar 25, 2009, 07:12 PM  #149  
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Be good 

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Mar 25, 2009, 07:51 PM  #150  
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_____F     G________________         _______________ H I'll leave the rest up to you guys 

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