*the* smart question

[Bozos of Bones]

Actually, Q'alooaith is quite right on this one.

You have three boxes. In one there is a stone, in the other two there is no stone. If you take one, and the seller opens one of the two remaining and shows it empty, you now have two boxes and one stone in the game. There is a 50:50 chance. The reasoning that the seller aimed to open the empty one holds no water; you say that it is in his interest to show the one without the price, and that would increase the chance of the remaining box having the stone, price, whatever. One can also argue that it is in the seller's, host's or whatever best interest to take out the stone, price, gold or whatever and thus keep it. The minute the first box is opened, wether it has the price or stone or whatever it is out of the equation, it only matters if there is to be a new equation(if the box is empty) or that that's it.
And now, ladies and gentlemen, prepare for something of a higher level!
You have a deck of 52 cards. There are 4 aces, 4 kings, 4 queens, 4 jacks and yadda yadda, a standard poker deck. What are the chances of drawing an ace, any ace out of that deck? It's 4 against 52, or 1 against 13, meaning that one in every thirteen cards should be an ace. Now, there comes a person, takes one card and hides it. You don't know what card he has. Now, what are the chances of getting an ace from that deck with 51 cards?
Please, no stupid answers.

[Timber Loftis]

What if I don't have to tell monty? What if I could write it on a secret ballot?

Then Monty wouldn't get to eat up air time creating suspense by showing you the doors you didn't pick, he'd run through twice as many contestants, and give away twice as many prizes each show. And, it'd be really boring. So... it won't happen.

It's there for the same reason reality shows use language like "The next name I am going to call from my list is...." They don't just read the names, do they? Why? Because they've got 15 mins. to kill at the end of the show, and need to stretch it out.

[Illumina Drathiran'ar]

Originally posted by ZFR: nah... DONT THINK CREATIVE... ok another hint: poly****

He marries the maiden, then marries the princess?

[ZFR]

Originally posted by Illumina Drathiran'ar: quote: Originally posted by ZFR: nah... DONT THINK CREATIVE... ok another hint: poly****

He marries the maiden, then marries the princess? [/QUOTE]yup. simple as that.

[mad=dog]

The doors: It reminds me of the ancien story about the gladiator who gets a hint from his beloved.

The deck: Well there are two posibilities from the first draw. 51 card with 3 aces with a probability of 1/13 and 51 cards with 4 aces and a probability of 12/13.
So 3/51*1/13+4/51*12/13 = 1/13 the chance remains the same as the first pick is completely random.

[Aragorn1]

Originally posted by ZFR: quote: Originally posted by Illumina Drathiran'ar: quote: Originally posted by ZFR: nah... DONT THINK CREATIVE... ok another hint: poly****

He marries the maiden, then marries the princess? [/QUOTE]yup. simple as that. [/QUOTE]THis isn't really a logic problem, as it requires an assumption that cannot be infered from the problem, you cannot know from the problem that the Prince can practice polygamy, or that he is in a country where this practice is legal.

[ZFR]

Originally posted by Bozos of Bones: Actually, Q'alooaith is quite right on this one. You have three boxes. In one there is a stone, in the other two there is no stone. If you take one, and the seller opens one of the two remaining and shows it empty, you now have two boxes and one stone in the game. There is a 50:50 chance. The reasoning that the seller aimed to open the empty one holds no water; you say that it is in his interest to show the one without the price, and that would increase the chance of the remaining box having the stone, price, whatever. One can also argue that it is in the seller's, host's or whatever best interest to take out the stone, price, gold or whatever and thus keep it. The minute the first box is opened, wether it has the price or stone or whatever it is out of the equation, it only matters if there is to be a new equation(if the box is empty) or that that's it.

I hate digging this up
You are wrong

Look at it this way.
Two guys always picks the first box. But guy A doesnt change his decision after empty is revealed, while guy B changes his choice.

Guy A will win only if the prize is in first box (33% chance)
Guy B will win if the prize is in either the second or the third box (67% chance)

[Morgeruat]

Originally posted by ZFR: Guy A will win only if the prize is in first box (33% chance) Guy B will win if the prize is in either the second or the third box (67% chance)

Chances are 33.33~% for A and 66.66~% for B

[VulcanRider]

Originally posted by mad=dog: The doors: It reminds me of the ancien story about the gladiator who gets a hint from his beloved. The deck: Well there are two posibilities from the first draw. 51 card with 3 aces with a probability of 1/13 and 51 cards with 4 aces and a probability of 12/13. So 3/51*1/13+4/51*12/13 = 1/13 the chance remains the same as the first pick is completely random.

51 cards with 4 aces has a probability of 12/13? Are you using a marked deck?

[Aragorn1]

Originally posted by ZFR: quote: Originally posted by Bozos of Bones: Actually, Q'alooaith is quite right on this one. You have three boxes. In one there is a stone, in the other two there is no stone. If you take one, and the seller opens one of the two remaining and shows it empty, you now have two boxes and one stone in the game. There is a 50:50 chance. The reasoning that the seller aimed to open the empty one holds no water; you say that it is in his interest to show the one without the price, and that would increase the chance of the remaining box having the stone, price, whatever. One can also argue that it is in the seller's, host's or whatever best interest to take out the stone, price, gold or whatever and thus keep it. The minute the first box is opened, wether it has the price or stone or whatever it is out of the equation, it only matters if there is to be a new equation(if the box is empty) or that that's it.

I hate digging this up
You are wrong

Look at it this way.
Two guys always picks the first box. But guy A doesnt change his decision after empty is revealed, while guy B changes his choice.


Guy A will win only if the prize is in first box (33% chance)
Guy B will win if the prize is in either the second or the third box (67% chance) [/QUOTE]OK, do it this way...

Draw three doors on piece of paper. pick one door. THe probability if this containing the prize is 1/3, put this above the door. the other doors each have 1/3 chance of containing the prize. Therefore, together there must be a 2/3 probablity the prize is behind one of them. Draw a box round the other two doors and write 2/3 above this box. OK, now one of the doors in the box must be empty, this is reveal to you, therefore, put a cross through that door. You are now left with one door with the probability of 1/3 and a box with only one possible door in it with the probability of 2/3.

THe point is, the ordinary rules of probability do not aply, as there are not two separate events, so the normal probability rules do not apply.

[ 06-23-2005, 02:45 PM: Message edited by: Aragorn1 ]