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Can anyone get this tricky math problem?
- Thread starter Frisco
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Doctor Q said:
I'd like to help but I'm bad with word problems.ShaunPriest said:
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.
I solved a problem like that a while ago, but I can't remember how to do it now.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.
Here's a problem I like:
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B C D EThree 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.
Monty Hall problem
The key is the bold word:
The example posed in this thread is contrived, and eliminates 1/3 of the cases before you get to decide whether to change doors:
In the Monty Hall problem, the rules are slightly different, because all three cases are considered:
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 50-50. 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.
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 50-50 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 50-50. If door #1 is not a winner, it's still 50-50 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 50-50. 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.
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.
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.
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.
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.
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.
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.
Only one.
So, That-Is-Bull, is the answer one?
If so, I feel cheated because you used the words "and another."
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?
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?
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.
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.
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.
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.
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.
This problem comes up regularly on sci.math. Unfortunately, we don't have enough information about the rules to be able to answer. I'll assume that your choices as the player are to either take what is behind door #2, or to take what is behind door #3.
The first question is: What possible choices did the guard have when he opened the door? Did he have the choice to open a door or not open a door? Did he have the choice to open a door that showed wealth (and you are stuck with death
Someone posted that you should always switch. Now assume the guard really doesn't like you. Here is what he does: If your door holds wealth, he opens a door showing death. If your door holds death, he opens a door showing wealth. If he is allowed to do this, and you follow the advice you were given if possible, you will certainly die.
Another possibility: If your door holds wealth, he opens a door showing death. If your door holds death, he doesn't open a second door. So if he opens a door and you follow the advice, you die. But if you know that this is the strategy he follows, then you can beat him: If he opens a door sowing death, you stay with your choice. Otherwise you switch. Now your chances are one in three to win.
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.
I don't feel like writing it all out, but the answer involves something that looks like this...
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HI'll leave the rest up to you guys