Now I was playing D&D the other night when our group was presented with this riddle: It's a basic cup and ball scenario. Three cups and a ball hidden under only one of the cups. We pick one cup that we think the ball is under. The npc lifts one of the cups and reveals that there is no ball under is under it and gives us the option to change our guess. What the npc wants to know is due to the odds would it be more logical to change your answer. The answer is apparently yes it is better to change your answer because it gives you a 2/3 chance of getting the right answer. ... Now that seems very very wrong to me you have three cups and each has a 1/3 chance of having the ball 0 0 0 1/3 1/3 1/3 Now when you are shown that one cup does not have a ball it doesn't give you any hint to where the ball actually is. I think the dm mistook 2/3 as the odds when 2/3 is the amount of cups you get to choose... and even if you stick with the same answer you have 2/3 of the cups. AND since you know 100% that one of the two cups you "chose" is incorrect that means that 1/3 of that 2/3 is null so you have two 1/3 chances (i.e 50% chance) I mean the cup and ball game is randomized so there is no way changing your answer would affect the outcome as there is just as much a chance of you changing your answer to the wrong answer as ther is of changing it to the right. Anyone else agree?

There was a whole thread on this a while back, but it has been statistically demonstrated that changing your answer does indeed give you better odds.

But how is that possible in a game that is completely random? You could change your answer and get it right 100% of the time or none of the time. Any demonstration could only be proven by pure coincidence.

This is my understanding of it (it may be dead wrong!) In the initial situation you have only a one in three chance of success. Once you have picked a cup, you are shown one of the others to be empty. Now if you stick with the original choice, the odds remain the same. You've picked one of three. It does not matter that you know that one of the other possibilities was false. With only one right answer and two wrong answers, at least one of the other possibilities had to be false. If you change your mind however, you are no longer picking one cup from three, you are picking one cup out of two. Thus the chance of success is now 50%, as opposed to the initial 33.33reacuring %. So whilst counter intuitive, it is more likely to win by changing your initial choice. More here:http://en.wikipedia.org/wiki/Monty_Hall_problem

one way to think about it is in reverse... there is a 66% chance you picked the wrong one to start off. then a wrong one is removed, but it's still more likely you chose wrong to start off, so take the other one.

Whether you change your guess or not after the first cup is removed you are still choosing between the two remaining. So you still are making the 50/50 guess whether you change or not. I never understood why people over think this.

That's wrong, as The Badger clearly explained. If you don't change your answer your chance of being correct stays the same as it was when you made the initial decision. You already made the choice with a one-in-three chance of being right. That doesn't change just because you got some extra information after the fact. Unless you change your decision it's still a one-in-three chance. Check out the link he provided. It explains the problem in great detail. Edit: removed attempt to further explain as you're probably better off following the link than my poor attempts at clarity.

I don't care how complex people try to make it. After the first cup is removed you are still making a choice between 2 remaining cups...a 50/50 chance. Choosing the same one and not changing does not invalidate that choice.

That's totally wrong. Switching will result in a win 2/3s of the time. http://en.wikipedia.org/wiki/Monty_Hall_problem http://people.hofstra.edu/Steven_R_Costenoble/MontyHall/MontyHallSim.html

That's a brilliant little site, Chaos Descending. I went through it once, sticking to my original answer every time, and got seven out of thirty wins. I then tried changing my selection, and won on twenty two out of thirty occasions. It's been a long time since I studied statistics so I'll not claim that the sample size was adequate, but whether we accept the explanation or not, there is empirical evidence that the effect occurs as described.

Isn't the point of the actual 3 cups trick is that the ball is under none of the cups. The correct answer in D&D is to shout "ENOUGH TALK" throw over the gaming table and smash a huge ham bone over the tricksters head.

I did it with 100 tries both. 35 wins if sticking with my original choice, 65 if switching when asked. Reality wins. (Or I'm a really dedicated scientist, or I'm a lazy bum: you decide!)

The Wikipedia article is extremely helpful in understanding the scenario. Both the playing cards and 1,000,000 doors example are good visual aids for the problem. Here's my way of explaining it now that I understand it. 1. There is a 1/3 chance of picking the right cup initially. 2. Since the host has to show one of the other cups to be empty, the chance to switch is like getting to pick both cups (a 2/3 chance).