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alcibiades
Honorable
Undefeatable Hero
of Gold Dragons
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posted July 25, 2009 10:31 PM |
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Quote:
Easy puzzle
There are three boxes, one containing nails, one containing screws, and one containing both nails & screws. But the labels have been switched up; they are all labeled incorrectly!
Can you determine which box is which by pulling a single item out of a single box? No looking inside or shaking or weighing the boxes. You just observe a single item to see if it is a nail or a screw.
I know the clue to the second one (our math teacher gave us that problem, half the class were never convinced it was true  ), but I figured out the first one like this:
Pick item from box labeled "mixed". If item is nail, you know box contains nails only. Which means that box with screws only has to be the one labeled nails, leaving the box labeled screws to be mixed.
Similarly, if you draw screw, you know box labeled screws much be nails only, and box labeles nails must be mixed.
You know box labeled mixed has to contain only one or the other, as it is per default labeled wrong.
EDIT > Btw. you do know that the second one actually is taken from real life, right? The quizshow had a car behind one door and goats between the two others, and to spice it up, it was changed so that once you chose your door, one of the others were opened to show a goat, and you had the opportunity to repick if you bothered.
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What will happen now?
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TheDeath
Responsible
Undefeatable Hero
with serious business
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posted July 25, 2009 10:51 PM |
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For the second problem, no. It's irrelevant.
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Binabik
Responsible
Legendary Hero
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posted July 26, 2009 12:10 AM |
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We've done the second one at HC before, but I won't say anything about it since I already know the answer. It's an interesting riddle though.
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Corribus
Hero of Order
The Abyss Staring Back at You
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posted July 26, 2009 05:33 AM |
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Quote:
We've done the second one at HC before, but I won't say anything about it since I already know the answer. It's an interesting riddle though.
It's also one of the most famous statistics problems there is. You can find a whole lengthy page dedicated to it at Wikipedia if you're really interested in it. Even after being told the answer to it, most people don't believe the answer.
The first puzzle requires that you read the whole thing very carefully. There's a vital piece of information written there that most people would probably overlook as being insignificant. I won't say more lest I spoil it for others.
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I'm sick of following my dreams. I'm just going to ask them where they're goin', and hook up with them later. -Mitch Hedberg
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ihor
Supreme Hero
Accidental Hero
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posted July 27, 2009 08:10 AM |
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Edited by ihor at 08:32, 27 Jul 2009.
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About the second problem:
It's very strange but it does matter whether to change your mind.
As I remember it was mentioned in 21 movie about Black Jack game. If you haven't seen it I'll recommend to see it. It is worth.
In such a game if you want to win you should allways change your selection.
Let's think. You chose some door - door A. Now there are 3 cases.
1)The door A is needed. The host chose door B or door C - doesn't matter - you change your selection and lose.
2)The door B is needed. The host will open door C for sure - by rules.
You change your selection and chose B - voila you win.
3)The door C is needed. The host will open door B for sure - by rules.
You change your selection and chose C - voila you win.
Conclusion:
If you change your selection, you will have 2/3 chances. If not - only 1/3.
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alcibiades
Honorable
Undefeatable Hero
of Gold Dragons
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posted July 27, 2009 11:02 AM |
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Quote:
The first puzzle requires that you read the whole thing very carefully. There's a vital piece of information written there that most people would probably overlook as being insignificant. I won't say more lest I spoil it for others.
But: Isn't my solution above correct?
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JollyJoker
Honorable
Undefeatable Hero
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posted July 27, 2009 11:28 AM |
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Ihor, while the solution is right it doesn't explain why so many go wrong here.
The problem is finding the right "level" of how many possibilities are actually there.
If you take door A and it's the winning door, then there are TWO possibilities: The gamemaster chooses door B and you switch to C (wrong) or the gamemaster chooses door C and you switch to B (wrong)
In this case there are not THREE possibilities but 4:
A => B => C => wrong
A => C => B => wrong
B => C => A => right
C => B => A => right
So the problem is to realize that the decisive level of probability is the one where the player picks the door, because that's the probabilities that count. Depending on the player choice the picks of the game master are either restricted or not which is one level of probability beyond that.
You can check it easily by tackling it this way: Let's say you make a game with 100 doors, 99 are wrong, 1 is right. You pick one, let's say door 1. Now the gamemaster opens alle remaining doors exceptone, say door 53. Switch or not? Of COURSE you must switch, but you can make the same table as above. with 99 times switching would be right while once it were wrong, but for that wrong case the gamemaster had 99 possibilities which door to leave closed (the right one is picked) while in all others he MUST keep the right one closed.
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TheDeath
Responsible
Undefeatable Hero
with serious business
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posted July 27, 2009 11:27 PM |
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Quote:
The first puzzle requires that you read the whole thing very carefully. There's a vital piece of information written there that most people would probably overlook as being insignificant. I won't say more lest I spoil it for others.
Yeah, that they are all labelled wrong, not random. But alc already covered that.
As for the second, well, can't say I put much thought into it
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Rarensu
Known Hero
Formerly known as RTI
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posted July 28, 2009 10:18 AM |
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On the Gameshow problem and it's propensity to fool most people
I once showed this problem to my uncle, who is not a stupid man (he attended CalTech). My uncle fell for the trap, believing that the switch is irrelevant. I discussed it with him for awhile. Eventually I discovered the bad assumption in his logic.
We both agreed that the original guess's probability is 1/3. However, my uncle believed that when the host opens a door, the quality of your original guess improves to 1/2, to match the probability of being right when you switch. In most possibility-elimination problems, this is correct; if the host had chosen a door to reveal at random, then the problem would follow that pattern as well.
However, in this case, it does not. This is because the host cannot reveal the grand prize. That is to say, you original guess has 0 probability of being eliminated, so it's quality does not improve. The guess's probability of being right can only increase if there is a chance that you lose the game when information is revealed.
I believe that this bad assumption that my uncle fell into is also at the root of the reasons why other people fall for this problem. The host cannot reveal the Grand Prize, and therefore your original guess remains 1 in 3 quality, but this small fact gets overlooked by our intuition.
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Sincerely,
A Proponent of Spelling, Grammar, Punctuation, and Courtesy.
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Corribus
Hero of Order
The Abyss Staring Back at You
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posted July 28, 2009 04:34 PM |
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Here's the solution to the Monty Hall problem.. Contains spoilers.
Sorry Alc about earlier. I didn't read responses before I posted.
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ihor
Supreme Hero
Accidental Hero
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posted July 28, 2009 07:58 PM |
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Edited by ihor at 20:01, 28 Jul 2009.
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Quote:
Here's the solution to the Monty Hall problem.. Contains spoilers.
Sorry Alc about earlier. I didn't read responses before I posted.
Yes, probabilistic solution there seems to be quite accurate and all questions dissapear.
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Corribus
Hero of Order
The Abyss Staring Back at You
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posted September 22, 2009 06:41 PM |
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*bump*
Anybody out there consider themselves experts at matrices?
This is a quantum chemistry problem I'm working on. I don't think knowing the chemistry is important, as at this point it seems to be just a math problem. Plus it would take me too long to explain the chemistry.
The problem is as follows:
I need to find a general solution to the determinant of an n x n matrix (call it M) with elements M(ij) defined as:
M(ij) = x, if i = j
M(ij) = a, if |i-j| = 1 and i+j = 3, 7, 11, ....
M(ij) = b, if |i-j| = 1 and i+j = 5, 9, 13, ....
M(ij) = 0 otherwise.
For example, a 6 x 6 matrix would be:
x a 0 0 0 0
a x b 0 0 0
0 b x a 0 0
0 0 a x b 0
0 0 0 b x a
0 0 0 0 a x
Basically, the "a's" and "b's" alternate.
Express the (general) determinant as a function of x, a, b and n.
Thoughts?
____________
I'm sick of following my dreams. I'm just going to ask them where they're goin', and hook up with them later. -Mitch Hedberg
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ihor
Supreme Hero
Accidental Hero
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posted September 22, 2009 07:48 PM |
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I think I got a recurrent formula if it suitable.
Just use formula of determinant decomposition by rows and columns.
Here's image.
Hope, this will help.
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Ecoris
Promising
Supreme Hero
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posted September 23, 2009 12:05 AM |
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ihor is correct. I don't think it gets better than that.
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Corribus
Hero of Order
The Abyss Staring Back at You
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posted September 23, 2009 03:07 PM |
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Edited by Corribus at 15:13, 23 Sep 2009.
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Right.. that's about as far as I got, but I need to go a little farther. Probably will require that I elaborate on the problem a bit, which I'll do when I get a chance.
EDIT:
Ok, elaboration provided below:
What I really need is a general formula for the roots (values of x, as a function of a, b, and n) when the determinant of M is set equal to zero.
A simpler version of the problem involves replacing all values of a with b, so that all off-diagonal elements of M are b.
i.e.
M(ij) = x if i = j
M(ij) = b if |i-j| = 1
M(ij) = 0 otherwise
A solution for the kth value of x when the determinant is set equal to zero is
x(k) = -2b*cos(R)
where
R = k * pi / (n + 1)
And k = 0, 1....n.
In case anyone is interested (to give the problem some relevance), the values of x in these matrices represent the energy states of some model conjugated conducting polymers. Unfortunately, the simpler problem, which appears in many introductory p-chem texts, treats all carbon-carbon bonds as equal in length, which isn't really true. A more accurate representation of most true polymers is that the bonds alternate long and short (hence the alternating a and b terms in the more complex matrix in my original post). I do have the derivation of the solution to the simpler problem, but trying to apply the strategy involved there toward the more complex problem has ran me into a wall; the solution involves breaking down the matrix into its constituent linear equations.
I do believe the solution to the complex problem is something of the form
x = +/- SQRT(a^2 + b^2 + 2ab*cos(R))
But in this case R is a different function of n and I'm not sure of any coefficients.
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TheDeath
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posted September 23, 2009 04:00 PM |
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If this is a serious problem (that is, practical) rather than just an exercise, can I use software to do it if you need the answer?
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Corribus
Hero of Order
The Abyss Staring Back at You
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posted September 23, 2009 04:08 PM |
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Quote:
If this is a serious problem (that is, practical) rather than just an exercise, can I use software to do it if you need the answer?
Yes and no.
It is a "serious" problem, and the answer would be useful to me regardless of how it is derived.
On the other hand, to incorporate the problem into a manuscript, I'll have to be able to explain the derivation.
But sometimes knowing the answer in advance can help you figure out how to solve the problem. So, by all means...
I've tried to use Maple to solve the problem, but it's been several years since I used the program and I'm finding myself spending more time trying to relearn the program than I am actually working on the problem.
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I'm sick of following my dreams. I'm just going to ask them where they're goin', and hook up with them later. -Mitch Hedberg
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TheDeath
Responsible
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with serious business
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posted September 23, 2009 04:48 PM |
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Well I haven't used matrices in Maxima so I had to learn the functions and operators a bit. Turns out I had to manually devise a pattern as 'n' changes (so it's no rigurous mathematical proof, sorry)
I came up with this polynomial (that has to be solved for 'x'). This is for the simplified problem (with b only, no a):
sum(i=0,n/2) [x^(n-i*2) * b^(i*2) * C(n-i,i) * (-1)^i]
C(n,k) is combinations of n taken k at a time.
Solving the roots of this is, well, more algebrically (is it even possible with very high-order polys?)
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dimis
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posted September 24, 2009 05:47 PM |
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Edited by dimis at 17:55, 24 Sep 2009.
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just a thought
Ok Corribus,
I didn't spend time to try to figure the determinant because others seem to have tried that but it didn't work.
So, if splitting the cases to "odd x odd" and "even x even" doesn't help with induction (or figuring out the formula is not easy so that you can perform induction), then my second thought was, try to find the Jordan canonical form. May be it works for you. Here are some simple examples I generated with Maple (but already after 6x6 it takes too long):
> restart;
> A6 := array([[x,a,0,0,0,0],[a,x,b,0,0,0],[0,b,x,a,0,0],[0,0,a,x,b,0],[0,0,0,b,x,a],[0,0,0,0,a,x]]):
> A2 := array([[x,a],[a,x]]);
> A4 := array([[x,a,0,0],[a,x,b,0],[0,b,x,a],[0,0,a,x]]);
> linalg[jordan](A2,P2);
[-a + x 0 ]
[ ]
[ 0 a + x]
> linalg[jordan](A4,P4);
[ 2 2 1/2 ]
[ (b + 4 a ) ]
[b/2 + x - -------------- , 0 , 0 , 0]
[ 2 ]
[ ]
[ 2 2 1/2 ]
[ (b + 4 a ) ]
[0 , - b/2 + x - -------------- , 0 , 0]
[ 2 ]
[ ]
[ 2 2 1/2 ]
[ (b + 4 a ) ]
[0 , 0 , b/2 + x + -------------- , 0]
[ 2 ]
[ ]
[ 2 2 1/2]
[ (b + 4 a ) ]
[0 , 0 , 0 , - b/2 + x + --------------]
[ 2 ]
> B3 := array([[x,a,0],[a,x,b],[0,b,x]]);
> J3 := linalg[jordan](B3,P3);
[x 0 0 ]
[ ]
[ 2 2 1/2 ]
J3 := [0 x + (b + a ) 0 ]
[ ]
[ 2 2 1/2]
[0 0 x - (b + a ) ]
> B5 := array([[x,a,0,0,0],[a,x,b,0,0],[0,b,x,a,0],[0,0,a,x,b],[0,0,0,b,x]]);
> J5 := linalg[jordan](B5,P5);
[x , 0 , 0 , 0 , 0]
[ ]
[ 2 2 1/2 ]
[0 , x + (a - a b + b ) , 0 , 0 , 0]
[ ]
[ 2 2 1/2 ]
J5 := [0 , 0 , x - (a - a b + b ) , 0 , 0]
[ ]
[ 2 2 1/2 ]
[0 , 0 , 0 , x + (a + a b + b ) , 0]
[ ]
[ 2 2 1/2]
[0 , 0 , 0 , 0 , x - (a + a b + b ) ]
So, hopefully, you have a perfectly diagonal similar matrix to your initial, and perhaps you can perform induction on "n" and prove this argument. Once you do that, end of story, because the determinant of your similar matrix is the product along the diagonal, which is the same as that of your original matrix.
This is just a thought. I might look into the problem again during the weekend, but I thought I could let you know in the meanwhile.
Good luck!
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Corribus
Hero of Order
The Abyss Staring Back at You
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posted September 24, 2009 06:36 PM |
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Thanks for the thoughts; I've been distracted from the problem for the last day, but I'm going back at it this afternoon. I'll take a look at what you guys wrote.
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