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The riddle of the 4 trolls

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NIK11 said:
Sorry but this is not exactly the same puzzle. The card puzzle makes no mention of the dealer having 'inside' knowledge about which card is the K, so all the choices are pure chance. In the case of The Monty Hall thing (which I have never heard of before) it involves intervention by the host who makes informed choices knowing where the winning door is - so that changes all the odds.

If you are convinced you are right, then hand out 3 cards to your kids and record the results. Proof of the pudding ...
Click to expand...
Hi Nick,

You are technically correct that no specific mention was made of the dealer having knowledge of the cards identities. Under this assumption, your solution is correct and your simulation is also correct.

However this line of reasoning leads to an uninteresting and kinda moot puzzle rather than the famous one that I think was intended but not stated carefully enough. That's not your fault, however and I apologize for jumping to the conclusion that you were incorrect.

Your sticking to your guns forced me to look again more carefully at this puzzle and I thank you for it.

It's not a question of right or wrong here, but being careful of one's assumptions.

--
Don
http://www.pbase.com/dond
 
Don_D said:
It's not a question of right or wrong here, but being careful of one's assumptions.
Click to expand...
Thanks for your generous response. And yes, I too made incorrect assumptions with some of the earlier puzzles. It takes a while to 'read' the unwritten game rules.

Good luck with the next one.

Nick
 
Hiya NIK11,
NIK11 said:
Don_D said:
For me here's an insight into why it makes sense to switch.

If at the start, you pick the wrong card of three, then the host in turn reveals a card but this is not done at random....he KNOWS which cards are which and reveals the card that is NOT the winning card....this makes the remaining card more likely to be the winning card and you should switch to it.
Click to expand...
Sorry, disagree with your insight in this particular puzzle. The puzzle says the players know only that one of the cards is a King - that does not imply the dealer knows which one it is before he turns them over. Give this exercise to a couple of 10 year olds to play out and record the results.
Click to expand...
Nik, i have to disagree with you now.

Given the parameters, the player bets on one card, be it the king or not, the dealer has to turn one card, the player then gets the option to switch. In the wikepedia article it clearly states that the dealer has to know where the king is and thus has to turn the non king card.

It is stated in the puzzle that the dealler knows which is the king. otherwise how can he choose the box that does not contain the king.

After all, the player has to choose a card. The dealer has to know what that card is because the dealer also has to turn a card that is not the king. Then giving the player a chance to switch or not.

The wikepedia link explained it at very well.

The fact that the dealer can/could forget which card is which is irrelevant to the puzzle, that's just a goofup. Which probably means his salary will be cut for the next six months.
NIK11 said:
NIK11 said:
As the Wikipedia explanation states, if the host is clueless and forgets which cards are which, THEN it doesn't matter whether you switch or not as then the chances are 50/50 under these conditions.
--
Click to expand...
That is totally out of context. Of course if you are playing a game like bridge you weigh up the odds on what has gone before. But this is a 3 card trick, there is nothing to remember. It is pure luck, no skill.

Nick
Click to expand...
Greets.
Fotonut.

--
Snap snap - click click.

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NIK11 said:
Sunshine_boy said:
Look at it this way Nik: The winning card is in the lot that remains after you picked your original card. ALWAYS, except for the case that you picked the right card.
Click to expand...
OK. So it is a 1 in 3 chance.
NIK11 said:
Therefore, the winning chances of the remaining lot = 1 minus the chance of your card.
Click to expand...
Yes, until the point at which the dealer turns over one of his cards and finds it is not a K. Then the odds change from being 2 to 1 in favour of the dealer to 1 to 1.
NIK11 said:
Since all the wrong cards are turned over by the dealer except one, that last one represents the chances of the whole group.
Click to expand...
I don't follow this conclusion.
Click to expand...
The whole idea is that the player keeps playing. If the dealer turned over the king, (not knowing what the cards where at the start), there would be no further point in playing, (if he did know at the start and goofed) he'd be on half pay.

Greets.
Fotonut.
--
Snap snap - click click.

FujiLinks - http://fujilinks.110mb.com/index.html (Main Site)
FujiLinks - http://fujilinks.fizz.me.uk/index.html (Mirror Site)
 
Fotonut said:
Nik, i have to disagree with you now.

Given the parameters, the player bets on one card, be it the king or not, the dealer has to turn one card, the player then gets the option to switch. In the wikepedia article it clearly states that the dealer has to know where the king is and thus has to turn the non king card.
Click to expand...
Please read this particular(DPRE forum) puzzle again, carefully. This puzzle makes no reference to Wikepedia or any U.S. TV gameshows.

Nick
 
Two old steam trains link 2 towns (A & B) in a perfectly straight line.
Train A starts from town A heading for town B.
At the same moment, train B starts from town B heading for town A
Train A travels at 30 Kmph, train B travels at 50 Kmph
The distance between the two towns is 180 Km

A pigeon is sitting on train A and can sense the danger of the two trains colliding if they stay on the same line. As soon as the two trains start to move towards each other, the pigeon also starts flying towards train B, trying to warn/indicate to the opposite train that it is on a collision course. The pigeon flies at a constant 100 Kmph.

As soon as the pigeon meets train B, it immediately turns back towards train A.

As soon as it meets train A, the pigeon turns back again towards train B and so on (flying from the one train to the other) until it is unfortunately crushed by the colliding trains.

The question is: How much distance (to the nearest meter) did the poor pigeon cover in its desperate attempt to warn the two trains about the danger ?

You can make the following assumptions:
  • The speed of the trains and pigeon does not change throughout the exercise
  • There is no loss of speed by the pigeon when turning back.
  • The length of the body of the two trains and the pigeon is nil, ie consider them all as single points.
No funny tricks, facts as told
Good luck... :)

--
Best Regards
Sunshine

Fuji F30, F31, S6500, OLY C4000Z, Canon Film EOS,
Nikon D60, 18-55VR, 18-105VR, 55-200VR, 35/1.8, SB-400
 
The trains combined speed is 50+30=80 km/h. Therefore in 180/80=2,25 hours they will meet each other.

This gives the pidgeon 2,25 hours * 100 km/h = 225 km.

Bas
Sunshine_boy said:
Two old steam trains link 2 towns (A & B) in a perfectly straight line.
Train A starts from town A heading for town B.
At the same moment, train B starts from town B heading for town A
Train A travels at 30 Kmph, train B travels at 50 Kmph
The distance between the two towns is 180 Km

A pigeon is sitting on train A and can sense the danger of the two trains colliding if they stay on the same line. As soon as the two trains start to move towards each other, the pigeon also starts flying towards train B, trying to warn/indicate to the opposite train that it is on a collision course. The pigeon flies at a constant 100 Kmph.

As soon as the pigeon meets train B, it immediately turns back towards train A.

As soon as it meets train A, the pigeon turns back again towards train B and so on (flying from the one train to the other) until it is unfortunately crushed by the colliding trains.

The question is: How much distance (to the nearest meter) did the poor pigeon cover in its desperate attempt to warn the two trains about the danger ?

You can make the following assumptions:
  • The speed of the trains and pigeon does not change throughout the exercise
  • There is no loss of speed by the pigeon when turning back.
  • The length of the body of the two trains and the pigeon is nil, ie consider them all as single points.
No funny tricks, facts as told
Good luck... :)

--
Best Regards
Sunshine

Fuji F30, F31, S6500, OLY C4000Z, Canon Film EOS,
Nikon D60, 18-55VR, 18-105VR, 55-200VR, 35/1.8, SB-400
Click to expand...
 
Hiya Nik,

But it does state the parameters. Which as it happens are the same as the wikepedia article.

Greets.
Fotonut.
NIK11 said:
Fotonut said:
Nik, i have to disagree with you now.

Given the parameters, the player bets on one card, be it the king or not, the dealer has to turn one card, the player then gets the option to switch. In the wikepedia article it clearly states that the dealer has to know where the king is and thus has to turn the non king card.
Click to expand...
Please read this particular(DPRE forum) puzzle again, carefully. This puzzle makes no reference to Wikepedia or any U.S. TV gameshows.

Nick
Click to expand...
--
Snap snap - click click.

FujiLinks - http://fujilinks.110mb.com/index.html (Main Site)
FujiLinks - http://fujilinks.fizz.me.uk/index.html (Mirror Site)
 
Fotonut said:
Hiya Nik,

But it does state the parameters. Which as it happens are the same as the wikepedia article.
Click to expand...
Before today I had never heard of the TV prgramme referred to in Wikipedia. When it comes to these puzzles of reasoning/mathematics/logic the critical parameters must be set out with the question and not assume special prior knowledge.

I don't know if 'sunshine boy' forgot or whether it was intentional, but at no time did he state the 'dealer' has knowledge of where the K is sitting - if the dealer knows, this makes quite a difference to the odds, so you would think it would get a mention if it is the case.

A few people seem to have rushed through reading the question and thought this is the same as that TV gameshow - wrong.

Let's move on. These games are fun if we don't get too bogged down in serious post mortems.

Try the next one. Good luck.

Nick
 
NIK11 said:
A few people seem to have rushed through reading the question and thought this is the same as that TV gameshow - wrong.
Click to expand...
Nike, Not to beat a dead horse, but your response above makes me refer to my earlier post:

"You are technically correct that no specific mention was made of the dealer having knowledge of the cards identities. However this line of reasoning leads to an uninteresting and kinda moot puzzle rather than the famous one that I think was intended but not stated carefully enough."

I give you credit for being technically right, but imho there was no way that this was not intended to be the same puzzle as the game show puzzle. If it was not, it was trivial.

--
Don
http://www.pbase.com/dond
 
Don_D said:
NIK11 said:
A few people seem to have rushed through reading the question and thought this is the same as that TV gameshow - wrong.
Click to expand...
Nike, Not to beat a dead horse, but your response above makes me refer to my earlier post:

"You are technically correct that no specific mention was made of the dealer > having knowledge of the cards identities.
Click to expand...
I am happy with that, all of the previous solutions have been 'technically correct'.
Don_D said:
However this line of reasoning leads to an uninteresting and kinda moot puzzle rather than the famous one that I think was intended but not stated carefully enough."
Click to expand...
Famous where? Remember this is an international forum.
Don_D said:
I give you credit for being technically right, but imho there was no way that this was not intended to be the same puzzle as the game show puzzle. If it was not, it was trivial.
Click to expand...
Well a few of the earlier puzzles have been trivial too and no one has suggested we change them half way through into something more interesting, or perhaps something that resembles a TV game.

Let's not get serious, this is beginning to sound like a 'clique'. Win or lose this should be fun, not half hearted apologies.

Nick
 
Of course your answer is correct. Normally you expect the Maths geeks to start forming mathematical series of the short trips of the pigeon and their sums instead of taking an overall view as you did...
Well done :)
--
Best Regards
Sunshine

Fuji F30, F31, S6500, OLY C4000Z, Canon Film EOS,
Nikon D60, 18-55VR, 18-105VR, 55-200VR, 35/1.8, SB-400
 
Yes well done Bas. Agree, it could be easy to get consumed with the arithmetic and miss the overall strategy.

Thanks, and keep 'em coming. This thread seems to be solving more problems than the guy's comparing IQ of F70/Wx1/ZX1/S90.

Nick
 
Hiya Nik,

just at th moment i cant be bothered to go back to the original question.
So i will leave it at my explanation.
NIK11 said:
Fotonut said:
Hiya Nik,

But it does state the parameters. Which as it happens are the same as the wikepedia article.
Click to expand...
Before today I had never heard of the TV prgramme referred to in Wikipedia. When it comes to these puzzles of reasoning/mathematics/logic the critical parameters must be set out with the question and not assume special prior knowledge.

I don't know if 'sunshine boy' forgot or whether it was intentional, but at no time did he state the 'dealer' has knowledge of where the K is sitting - if the dealer knows, this makes quite a difference to the odds, so you would think it would get a mention if it is the case.

A few people seem to have rushed through reading the question and thought this is the same as that TV gameshow - wrong.

Let's move on. These games are fun if we don't get too bogged down in serious post mortems.

Try the next one. Good luck.

Nick
Click to expand...
Sorry if i'm ducking the question , but i have other things to check/do,

Greets.
Fotonut.
--
Snap snap - click click.

FujiLinks - http://fujilinks.110mb.com/index.html (Main Site)
FujiLinks - http://fujilinks.fizz.me.uk/index.html (Mirror Site)
 
NIK11 said:
Famous where? Remember this is an international forum.
Click to expand...
Wikipedia states:

"Over 40 papers have been published about this problem in academic journals and the popular press (Mueser and Granberg 1999).

The earliest of several probability puzzles related to the Monty Hall problem is Bertrand's box paradox, posed by Joseph Bertrand in 1889 in his Calcul des probabilités (Barbeau 1993). In this puzzle there are three boxes: a box containing two gold coins, a box with two silver coins, and a box with one of each. After choosing a box at random and withdrawing one coin at random that happens to be a gold coin, the question is what is the probability that the other coin is gold. As in the Monty Hall problem the intuitive answer is 1/2, but the probability is actually 2/3.

The problem continues to resurface outside of academia. The syndicated NPR program Car Talk featured it as one of their weekly "Puzzlers," and the answer they featured was quite clearly explained as the correct one (Magliozzi and Magliozzi, 1998). An account of the Hungarian mathematician Paul Erdos's first encounter of the problem can be found in The Man Who Loved Only Numbers—like many others, he initially got it wrong. The problem is discussed, from the perspective of a boy with Asperger syndrome, in The Curious Incident of the Dog in the Night-time, a 2003 novel by Mark Haddon. The problem is also addressed in a lecture by the character Charlie Eppes in an episode of the CBS drama NUMB3RS (Episode 1.13) and in Derren Brown's 2006 book Tricks Of The Mind. Penn Jillette explained the Monty Hall Problem on the "Luck" episode of Bob Dylan's Theme Time Radio Hour radio series. The Monty Hall problem appears in the film 21 (Bloch 2008). Economist M. Keith Chen identified a potential flaw in hundreds of experiments related to cognitive dissonance that use an analysis with issues similar to those involved in the Monty Hall problem (Tierney 2008)".

Nik,

You are an intelligent person, I find it very hard to believe with all the discussion and analysis that you still not see that the card problem posed here is the same problem as the game show problem? The only difference being the careless omission of one assumption, ie that the poser knows the id of the cards.

I'm done....it really was a fun discussion. Thanks :)

--
Don
http://www.pbase.com/dond
 
Don_D said:
You are an intelligent person, I find it very hard to believe with all the discussion and analysis that you still not see that the card problem posed here is the same problem as the game show problem? The only difference being the careless omission of one assumption, ie that the poser knows the id of the cards.

I'm done....it really was a fun discussion. Thanks :)
Click to expand...
Flattery will get you everywhere. Of course I can see the puzzle has close similarities. Since the originator here has chosen not to offer clarification of what he intended with this puzzle, we can only assume there is a 50% chance you are right and a 50% chance I am right.

But if you factor in peer pressure and pride, the probability is, that if pushed, he would likely agree with your line - let's say 66%? Deja vous.

Thanks for the interesting background info.

Nick
 
NIK11 said:
Yes well done Bas. Agree, it could be easy to get consumed with the arithmetic and miss the overall strategy.

Thanks, and keep 'em coming...
Click to expand...
OK, you asked for it!...

Here's a problem that in the past caused me the loss of many hours of deep maths, calculations and mental gymnastics to the point of near obsession, till I finally gave up.

Let's start off by saying that in a given situation, if all variables are known except for one, the unknown variable can be worket out, especially if it can only have a unique value. So, here goes the 'simple' problem:

You have a corridor 3m wide between two vertical walls and a horizontal floor

You have a ladder 5m long and you place it at the lower right corner of the corridor to lean against the opposite (left) wall.

You have a second ladder 8m long which you place at the lower left corner of the corridor to lean against the opposite (right) wall.

You make sure that the two ladders are on the same vertical plane, ie they are touching each other.

The point of contact of the ladders is unique. There is only one point at which they can touch. You are requested to calculate the height (vertical distance) of that point from the horizontal floor.

This problem is so tidy with clear, well defined parameters that not solving it makes it so aggravating but I am not going to start all over again. I'll leave that to any and all of you, but don't lose too much sleep over it!...

For consistency, please use the following symbols:

D = corridor width/distance between the walls = 3m
L1= first ladder = 5m
L2= second ladder = 8m
H = height of the point of ladder intersection from the floor = ???

Good luck :)

--
Best Regards
Sunshine

Fuji F30, F31, S6500, OLY C4000Z, Canon Film EOS,
Nikon D60, 18-55VR, 18-105VR, 55-200VR, 35/1.8, SB-400
 
Could the answer be 2,598482672 height?

Here is what I did: use pythagoras to find the heights of the ladders against the wall:
  • Left = 7,416198487 meters (square root 55)
  • Right = 4 (square root 16)
You can then use these numbers to construct 2 lineair equasions, assuming bottom left is the point with x=0 and y=0. Then the equasions look like

y = 4/3 * x (5 meter ladder)
y = - 7,416198487/3 * x + 7,416198487 (8 meter ladder)

Solving it gives (4/3 * x) + (7,416198487/3 * x) = 7,416198487

So x = 1,948862004

Mulitplied with the 4/3 from the first equation gives the height (y) of 2,598482672 meters. The numbers are a bit odd though and maybe I got sloppy, it's late. But I'm sure this strategy must work.

Bas
Sunshine_boy said:
NIK11 said:
Yes well done Bas. Agree, it could be easy to get consumed with the arithmetic and miss the overall strategy.

Thanks, and keep 'em coming...
Click to expand...
OK, you asked for it!...

Here's a problem that in the past caused me the loss of many hours of deep maths, calculations and mental gymnastics to the point of near obsession, till I finally gave up.

Let's start off by saying that in a given situation, if all variables are known except for one, the unknown variable can be worket out, especially if it can only have a unique value. So, here goes the 'simple' problem:

You have a corridor 3m wide between two vertical walls and a horizontal floor

You have a ladder 5m long and you place it at the lower right corner of the corridor to lean against the opposite (left) wall.

You have a second ladder 8m long which you place at the lower left corner of the corridor to lean against the opposite (right) wall.

You make sure that the two ladders are on the same vertical plane, ie they are touching each other.

The point of contact of the ladders is unique. There is only one point at which they can touch. You are requested to calculate the height (vertical distance) of that point from the horizontal floor.

This problem is so tidy with clear, well defined parameters that not solving it makes it so aggravating but I am not going to start all over again. I'll leave that to any and all of you, but don't lose too much sleep over it!...

For consistency, please use the following symbols:

D = corridor width/distance between the walls = 3m
L1= first ladder = 5m
L2= second ladder = 8m
H = height of the point of ladder intersection from the floor = ???

Good luck :)

--
Best Regards
Sunshine

Fuji F30, F31, S6500, OLY C4000Z, Canon Film EOS,
Nikon D60, 18-55VR, 18-105VR, 55-200VR, 35/1.8, SB-400
Click to expand...
 
Sunshine_boy said:
NIK11 said:
Yes well done Bas. Agree, it could be easy to get consumed with the arithmetic and miss the overall strategy.

Thanks, and keep 'em coming...
Click to expand...
OK, you asked for it!...

Here's a problem that in the past caused me the loss of many hours of deep maths, calculations and mental gymnastics to the point of near obsession, till I finally gave up.

Let's start off by saying that in a given situation, if all variables are known except for one, the unknown variable can be worket out, especially if it can only have a unique value. So, here goes the 'simple' problem:

You have a corridor 3m wide between two vertical walls and a horizontal floor

You have a ladder 5m long and you place it at the lower right corner of the corridor to lean against the opposite (left) wall.

You have a second ladder 8m long which you place at the lower left corner of the corridor to lean against the opposite (right) wall.

You make sure that the two ladders are on the same vertical plane, ie they are touching each other.

The point of contact of the ladders is unique. There is only one point at which they can touch. You are requested to calculate the height (vertical distance) of that point from the horizontal floor.

This problem is so tidy with clear, well defined parameters that not solving it makes it so aggravating but I am not going to start all over again. I'll leave that to any and all of you, but don't lose too much sleep over it!...

For consistency, please use the following symbols:

D = corridor width/distance between the walls = 3m
L1= first ladder = 5m
L2= second ladder = 8m
H = height of the point of ladder intersection from the floor = ???

Good luck :)

--
Best Regards
Sunshine

Fuji F30, F31, S6500, OLY C4000Z, Canon Film EOS,
Nikon D60, 18-55VR, 18-105VR, 55-200VR, 35/1.8, SB-400
Click to expand...
2.59879 meters is the vertical height from the floor where the two ladders intersect - if I understand the description properly.

Let me know how I did. Have to run to Back to School night now.
 
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