Best...Puzzle...Ever....

ScotMart2000 said:

Label the chest A, B and C. We'll say the gold is in chest A. Here are the 3 possibilities if the characters switch their choice:

1. The adventurers pick chest A. The host reveals B is empty. The adventurers opt to switch their choice to C. They lose.
2. The adventurers pick chest B. The host reveals C is empty. The adventurers opt to switch their choice to A. They win.
3. The adventurers pick chest C. The host reveals B is empty. The adventurers opt to switch their choice to A. They win.

By switching, the adventurers win 2 out of 3 times. By keeping their original choice, they only win 1 out of 3 times.

This happens because the host always reveals an empty chest that the characters have not picked. So, if the characters pick an empty chest (2 out of 3 times they will), the host will have to remove the only other empty chest and leave the chest of treasure there for the characters to switch to.

I realize this is fairly anti-intuitive at first, but I think once the possibilities are laid out, you see why it works.


-Scot

Click to expand...

Ah, yes, BUT............

What if the "prize" in the applicable chest is worth only 90gp? Then paying 100gp in order to change the chest of choice merely results in the characters losing 10gp in the deal. Sorry I brought it up!

RingXero,

By your own list, if they narrow themselves down to options 1-3, which they DO in fact accomplish by switching, then they win in 2 of 3 occasions.

OTOH, if they don't they eliminate the possibilities 1-3 and are left with only a 1 in 3 chance.

But it is still only a fifty-fifty shot. Your logic is based upon only half the math.

Okay follow this:

Chest A and B are empty, C contains treasure. This holds true for the rest of my lecture.

Group chooses A, Host removes B. Group switches and Wins or stay and Loses.
Group chooses B, Host removes A. Group switches and Wins or stay and Loses.
Group chooses C, Host removes either A or B, doesn't matter. Group switches and Loses or stays and Wins.

Switching Wins twice, Loses once. Staying Wins once, Loses twice. So by your logic, switching provokes a 2/3 chance of winning.

However, the Group has no way of knowing if the chest they have is a winner or not. They have two chests to choose from, at this point either of which could be the winner.

Group stays with previous choice, A or B (the initial choice) and loses. Group Switches to C and Wins.
Group stays with C and wins. Group switches to either A or B (the remainder) and Loses.

Switching Wins once. Staying wins once.

Switching costs you 100¥.

Do More Than Just The Initial Math. Do All The Math.

TTFN

EvilE


Edit--
A further Twist:

If one chest contains the 'Golden Prize' (that which is worth dieing for) and the other two contain 'Certain Doom'. How do the Adventures choose?

BryonD said:

RingXero,

By your own list, if they narrow themselves down to options 1-3, which they DO in fact accomplish by switching, then they win in 2 of 3 occasions.

OTOH, if they don't they eliminate the possibilities 1-3 and are left with only a 1 in 3 chance.

Click to expand...

After making the list I ran a quick program, followed up by a quick search. and yep switching seems better.

aargh... and yet another search which shows it doesn't matter.

edit again: nope last search shows to switch, so strike that.

its all in the fact that Monty knows what the doors/chests are.



RX

*hugs google*

Well, two things, one - I knew I remembered the name properly, as it's sitting in my probability notes from college now.

and two - google is awesome. the best site so far that I have found is this:

http://math.ucsd.edu/~crypto/Monty/montybg.html

and to run the problem go here:
http://math.ucsd.edu/~anistat/chi-an/MonteHallParadox.html
or
http://www.shodor.org/interactivate/activities/monty/index.html
or
http://www.shodor.org/interactivate/activities/montynew/index.html

I would like to thank these boards for bringing this delightful 'paradox' back into my concious mind, I have forgotten way too much.


RX

Well how about this? (Regarding the three chests).

1. There are three chests, A, B & C.

2. One of them holds the treasure.

3. MONTY KNOWS WHICH ONE!!!

4. The party selects A (which, in fact, HOLDS the treasure)

5. Monty removes B and asks them if they wish to switch their answer for 100gp.

6. Monty only gives the players this option BECAUSE they've got it right the first time.

7. If the players fall for it, they pay 100gp for an empty chest (C).

8. All of this depends upon the DM's portrayal of "Monty" as either benevolent or malevolent. HOWEVER, ONLY THE DM KNOWS MONTY'S TRUE MOTIVES!!! This is a variable that no amount of math will be able to solve and I honestly believe that that is the point behind this particular "puzzle". It's somewhat connected to the false/true guardian puzzle in that the players have to figure out Monty's motives while considering his offer.

PennStud77 said:

Once the third chest is shown to have no gold in it, there is a 1 in 2 chance of WHICHEVER chest they pick to be correct, this is absolutely regardless of which one they have already picked. Switching their choice and keeping the chest they have both have the same probability of being correct.

Click to expand...

This came up in one of my college classes a long time ago, and one of my classmates explained it excellently.

If you had 100 chests, only one of which contains a prize, and you were to pick one, and the host opened up 98 other chests so to show that they were empty, and gave you the choice of switching the one you chose for the last remaining one, would you do it? Assuming a fair host who would offer you this choice regardless of whether you chose "correctly" or not, most people would, because you would only have a 1 in 100 chance of picking the correct chest initially, and so the chance that the other chest contains the prize is 99 in 100.

In the 3 chest problem, the odds that you chose the right chest initially are higher - 1 in 3 instead of 1 in 100. Still, probability-wise, it is in your interest to switch.

Ahh, I see the Monty Bug.

I was under the impression that an empty chest would be removed regardless of the initial pick.

That does change things if the puzzle is using the Monty Rules as a baseline.

Then as tuzenbach points out your screwed by a RatBastard DM and rewarded by a Kindly DM.

TTFN

EvilE

I've revised the dungeon room puzzle based on Merak's original dungeon.

This one seems to work fairly well. I put 3 secret doors to the final destination room, which improved the complexity significantly, though not to impossible levels.

It's rather easy, but the addition of the third wall meant that I could open up other choices and make the players really think about where they wanted to move to.

If a door is open, then the lever will close the door.

lever
1 opens G, E, I
2 opens F, I
3 opens D, B, J
4 opens C, F, A
5 opens B, E, J, H
6 opens C, E, H

Of course, it's a little easier with that information at your disposal, my players will have to solve that part first.

I will have a lever outside that will reset the whole contraption.

RingXero said:

After making the list I ran a quick program, followed up by a quick search. and yep switching seems better.

aargh... and yet another search which shows it doesn't matter.

edit again: nope last search shows to switch, so strike that.

its all in the fact that Monty knows what the doors/chests are.

RX

Click to expand...

Agreed, but the logic you presented did not get there. I didn't have it solved either, btw. I was just playing along. I knew you had to be right. But also new you had not proven it. And I was fairly frustrated/amused because it is a clever bit of math sleight of hand.


Obviously the trick has been revealed now. Very nice.