MacRumors Forums Can anyone get this tricky math problem?

 Mar 13, 2009, 05:58 PM #126 Shaun.P macrumors 68000     Join Date: Jul 2003 Location: Omicron Persei 8 Can you explain this then? http://en.wikipedia.org/wiki/Monty_Hall_problem I'm confused! __________________ I support the May 2013 MacRumors Blood Drive! 0
Mar 13, 2009, 06:12 PM   #127
iMacmatician
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Join Date: Jul 2008
Quote:
 Originally Posted by Frisco Question: How many legs are there in the bus?
10990

Quote:
 Originally Posted by Doctor Q …
Quote:
 Originally Posted by ShaunPriest …
I'd like to help but I'm bad with word problems.

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:
 Originally Posted by CalBoy I still like the 12 unmarked coins with 1 counterfeit and only a basic "heavy" "light" scale problem. That one required good old fashioned thinking power.
I solved a problem like that a while ago, but I can't remember how to do it now.

Here's a problem I like:

Code:
``` _________________A
|     |     |     |
|     |     |     |
|_____|_____|_____|
B     C     D     E```
Three squares are put together like this, with lines AB, AC, and AD drawn. Prove that angle ABE + angle ACE = angle ADE by using simple geometry only.
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Last edited by iMacmatician; Mar 13, 2009 at 06:27 PM.
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Mar 13, 2009, 06:41 PM   #129
EricNau
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Join Date: Apr 2005
Location: San Francisco, CA
Quote:
 Originally Posted by ShaunPriest Can you explain this then? http://en.wikipedia.org/wiki/Monty_Hall_problem I'm confused!
Q's response (emphasis mine):
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, 06:45 PM #130 Shaun.P macrumors 68000     Join Date: Jul 2003 Location: Omicron Persei 8 Ahh! As an undergraduate maths student you can see why I was curious! __________________ I support the May 2013 MacRumors Blood Drive! 0
 Mar 13, 2009, 06:46 PM #131 EricNau Demi-God (Moderator emeritus)     Join Date: Apr 2005 Location: San Francisco, CA Does CalBoy's riddle address the guard's knowledge of what's behind each door? Can one form an accurate answer without that information? 0
Mar 13, 2009, 07:01 PM   #132
CalBoy
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Join Date: May 2007
Quote:
 Originally Posted by EricNau Does CalBoy's riddle address the guard's knowledge of what's behind each door? Can one form an accurate answer without that information?
Well the riddle is originally from Monty Hall, where it was known that the guard is fully aware of what is behind each door.

Without that info, it's a very different problem.
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 Mar 13, 2009, 07:01 PM #133 Doctor Q Administrator     Join Date: Sep 2002 Location: Los Angeles 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. __________________ Oh do pay attention 007. In the wrong hands, this cylindrical 12-core Mac Pro with three 4K displays, FirePro graphics, and Thunderbolt 2 could be very dangerous. 0
 Mar 13, 2009, 07:56 PM #134 That-Is-Bull macrumors 6502   Join Date: Sep 2007 Location: Edmond, Oklahoma 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? __________________ my head feels like a frisbee 0
Mar 13, 2009, 07:59 PM   #135
CalBoy
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Join Date: May 2007
Quote:
 Originally Posted by That-Is-Bull 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?
A minimum of 2, a maximum of infinite.

But then I gather this is a riddle of some kind.
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Mar 13, 2009, 08:00 PM   #136
EricNau
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Quote:
 Originally Posted by That-Is-Bull 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?
Only one.
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Mar 13, 2009, 11:02 PM   #137
fireshot91
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Join Date: Aug 2008
Location: Northern VA
Quote:
 Originally Posted by iMacmatician 10990 Code: ``` _________________A | | | | | | | | |_____|_____|_____| B C D E``` Three squares are put together like this, with lines AB, AC, and AD drawn. Prove that angle ABE + angle ACE = angle ADE by using simple geometry only.

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 13, 2009, 11:51 PM   #138
CalBoy
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Join Date: May 2007
Quote:
 Originally Posted by That-Is-Bull 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?
Quote:
 Originally Posted by EricNau Only one.
So, That-Is-Bull, is the answer one?

If so, I feel cheated because you used the words "and another."
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 Mar 14, 2009, 12:40 AM #139 That-Is-Bull macrumors 6502   Join Date: Sep 2007 Location: Edmond, Oklahoma The answer's two. You have one bucket, and another bucket. That makes two. __________________ my head feels like a frisbee 0
 Mar 14, 2009, 12:57 AM #140 mscriv macrumors 68040     Join Date: Aug 2008 Location: Dallas, Texas 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? __________________ I'm a professional therapist. If I deem our forum interaction to be professional in nature then I will bill you. Prompt payment is expected. 0
Mar 14, 2009, 01:01 AM   #141
CalBoy
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Join Date: May 2007
Quote:
 Originally Posted by mscriv 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?
First trip: take the chicken.

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, 02:35 AM   #142
UWSpindoctor
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Join Date: Oct 2007
Location: Sunnyvale, CA
Quote:
 Originally Posted by fireshot91 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.
AEB and AEC would be 90 degrees, not ABE and ACE
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Mar 14, 2009, 10:22 AM   #143
iMacmatician
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Join Date: Jul 2008
Quote:
 Originally Posted by fireshot91 Err, doesn't that not work, because Angle ABE= 90 Deg. and Angle ACE= 90 deg.

Angle ADE is 45º and angles ABE and ACE are both less than that.
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 Mar 14, 2009, 10:33 AM #144 fireshot91 macrumors 601     Join Date: Aug 2008 Location: Northern VA 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. 0
Mar 14, 2009, 11:31 PM   #145
mscriv
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Join Date: Aug 2008
Location: Dallas, Texas
Quote:
 Originally Posted by CalBoy First trip: take the chicken. Second trip: take the feed, swap it with the chicken. Third trip: take the fox. Fourth trip: take the chicken.
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, 04:55 PM   #146
spork183
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Join Date: Jul 2006
Quote:
 Originally Posted by EricNau There is no missing dollar. You shouldn't be adding the \$2, you need to subtract it. Remember, the total is \$25, not \$30. Therefore, each guest paid \$9 x 3 = \$27 - the stolen \$2 = \$25. ...
No, you aren't trying to get to \$25, you're trying to get to \$30. Based on that, the 9x3=27 simply adds up what is downstairs (\$25) with what the dishonest bell hop stole (\$2). Therefore, each individual paid \$9 for a total of \$27. If you then look at what each has left over (\$1x3 people), your totals add up to the original \$30.

My granddad told this one back before who flung the chunk.
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Mar 25, 2009, 05:05 PM   #147
spork183
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Quote:
 Originally Posted by Doctor Q It could be that there are two types of certain death.
Now that's just not a statement you read everyday...
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Mar 25, 2009, 05:08 PM   #148
EricNau
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Quote:
 Originally Posted by spork183 No, you aren't trying to get to \$25, you're trying to get to \$30. Based on that, the 9x3=27 simply adds up what is downstairs (\$25) with what the dishonest bell hop stole (\$2). Therefore, each individual paid \$9 for a total of \$27. If you then look at what each has left over (\$1x3 people), your totals add up to the original \$30.
I do believe that's the same explanation, just described differently.
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Mar 25, 2009, 06:12 PM   #149
MOFS
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Location: Liverpool, UK
Quote:
Like you I disagree with the answer, but for different reasons. You initially have a 1/3 chance of getting the door with certain death, due to 3 options. However, when you're shown the door with certain death, your options are immediately reduced to 2; of which one is certain death and the other wealth. Therefore why change door? 1/2 is certain death, but 1/2 is wealth.
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Mar 25, 2009, 06:51 PM   #150
benbondu
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Join Date: Jul 2004
Quote:
 Originally Posted by iMacmatician Here's a problem I like: Code: ``` _________________A | | | | | | | | |_____|_____|_____| B C D E``` Three squares are put together like this, with lines AB, AC, and AD drawn. Prove that angle ABE + angle ACE = angle ADE by using simple geometry only.
I don't feel like writing it all out, but the answer involves something that looks like this...
Code:
```             _____F
|     |
|     |
G___________|_____|
|     |     |     |
|     |     |     |
|_____|_____|_____|
H```
...and recognizing that triangle FGH is isosceles right.

I'll leave the rest up to you guys
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