Maths Problem

[Dauphin]

Originally posted by Azazel

Let us expand this scenario a bit . Imagine there are a hundred cups , but only one with a prize . You pick one . 98 false options are eliminated . Now , does it make sense to switch ?

Again, it won't make any difference.

Answer me this: You have 3 cups. One of them has a prize under it. Then, an empty one is removed.
A) what's your chance of picking the right one?
B) how is that scenario different?

The point is you make the selection before an empty cup is removed.

[Az]

The point is you make the selection before an empty cup is removed.

No, not really. Since switching or not switching, is like making a choice all over again.

[Dauphin]

The second choice you make is not symmetrical though.

[Dauphin]

Ball is under Cup A

Case 1:

You choose the Cup A. Either Cup B or C is eliminated, it doesn't matter which. You switch and you lose. You stick and you win

Case 2:

You choose Cup B. Cup C is eliminated. You switch to A and you win. You stick and you lose

Case 3:

You choose Cup C. Cup B is eliminated. You switch to A and you win. You stick and you lose.

The fact that a cup is eliminated provides no useful information as Cup A will not be eliminated whatever happens.

[Az]

The fact that a cup is eliminated provides no useful information as Cup A will not be eliminated whatever happens.

Well, duh, of course the cup that has the ball won't be eliminated. This explains nothing. There is one certain thing: Before you choose to "stick or change", you'll have two cups. One of them will house the ball, the other will be empty. What cup you held to earlier is meaningless.

I think that an experiment is in order. Replace random choice with a roll of a dice

[RGBVideo]

EDIT: I just suck

[Az]

It seems that I have to concede defeat. I've repeated the experiment, and the results are quite discouraging, to say the least. (Btw, thanks mom for helping me with this one )

Good going, guys

[Mercator]

There's a 2/3 chance that you picked an empty cup.
The empty cup that is turned over changes nothing about the fact that there was a 2/3 chance that you picked an empty cup! It simply shows you that that other one is definitely empty.
In other words, there's still a 2/3 chance that your cup is empty.

If you're still not following...
If you were to repeat this experiment many times you would end up picking an empty cup in 2 out of every 3 cases.
In the case that you have picked an empty cup, with the other empty cup revealed, you should obviously switch to get the prize.
In other words, in 2 out of 3 cases, you have to switch to get the prize.

[Petek]

This puzzler is known as the Monty Hall problem, named after the host of the TV show Let's Make a Deal. Google for "Monty Hall Problem" for more info.

[General Ludd]

Originally posted by Mercator
There's a 2/3 chance that you picked an empty cup.
The empty cup that is turned over changes nothing about the fact that there was a 2/3 chance that you picked an empty cup! It simply shows you that that other one is definitely empty.
In other words, there's still a 2/3 chance that your cup is empty.

But only a 1/2 chance when you are given the second choice of switching your cup.



There are two choices here, each with a different chance, but the first choice does not really matter because the result is always the same.

You are given three choices.

You chose one.

One of the other choices, which is wrong, is removed.

You are then asked if you are sure of your original choice, or if you want to change it.

You now only have two choices.

[Solly]

Always switch.

A woman has two kids. At least one of them is a boy. What are the chances she has two boys?

[Worthingtons]

Ah, good stuff bods,

well, I will attempt to answer,
There are 4 permutations normally

Woman has 2 girls = 1/4
Woman has 1 each = 1/2 (boy+girl and girl+boy)
Woman has 2 boys = 1/4

With the information given only 3 can be possible, as two girls cannot be possible therefore we are left with 3/4, Woman having 2 boys now assumes the probability of 1 in 3, or 33.3%

Let me know how wrong I am

[Solly]

Very interesting Worthy's...

[KrazyHorse]

Originally posted by Azazel
Well, there obviously some trick here, but I am going to go at it straightforward. Correct me if I am wrong in any of my assumptions.

A person chooses a cup. No matter what that cup has under it, the number of choices he has will drop to 2, by the elimination of a wrong answer. and then he can choose to stay with the cup, or not, which is basically the same as choosing one cup out of two all over again. This way, it's 50/50. Thus, I say: no difference. His chances are the same.

Incorrect.Since you always turn over a cup with nothing under it (deliberately) the odds for the original cup do not improve. The odds for the other cup do.

[Mercator]

Originally posted by General Ludd
There are two choices here, each with a different chance, but the first choice does not really matter because the result is always the same.

The first choice does matter, because 2 out of 3 times you would have picked an empty cup, seeing as how 2 out of the 3 cups are empty. So obviously your first guess would be an empty cup 2 out of 3 times even when there are only two cups left to choose from. Consequently, 2 out of 3 times, you should switch.