Maths Problem

[Worthingtons]

I have three non see through tea cups
I place them upside down on a table
Under one of the I place a prize
Under the other two cups I place nothing.
I know which cup the prize is under.
I invite somebody to pick a cup.
If it has the prize under it then they get the prize.
My first player picks a cup
At this point the player has 1 chance in 3 of wining the prize.
But before showing what is under the cup I tell him
I will show you what is under one of the unselected cups.
I then always turn over a cup with nothing under it.
I then tell the player that he can stay with the cup he chose earlier
or switch to the remaining cup.

Some eminient mathematics professors thought that it did not matter
which cup he chose from the two. Other eminent maths professors said it did matter.

What is his chance of winning if he sticks to his original choice?

What is his chance of winning if he switches.

[DRoseDARs]

Flip the table, grab the prize, run like hell.

[Az]

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.

[Dauphin]

Switch.

If you don't pick the right cup, a 2 in 3 chance, then you guarantee that the remaining cup will be the one with the ball. Hence you have at least a 2 in 3 chance of winning if you switch.

Or put another way, if you don't switch, your chances of winning remain 1 in 3 regardless. Hence there must be a 2 in 3 chance of winning if you switch.

[Az]

Or put another way, if you don't switch, your chances of winning remain 1 in 3 regardless.

How so?

[aneeshm]

I know the answer , so I'll put it in spoiler tags .

Spoiler:

Draw a table :

Choice one : You pick junk . If you switch , you get the prize .

Choice two : You pick junk . If you switch , you get the prize .

Choice three : You pick the prize . If you switch , you get junk .

[Worthingtons]

I'm thinking the same as Dauphin but struggling to prove this, here is my attempt :


Three cups, name them A,B & C

Each probability of it having the prize is 1/3

A+B+C=1

A=B=C

Player picks A,
at this point

A= 1/3
B+C= 2/3

B&C are assesed with the wrong option removed, lets say this is C

B now equals 2/3 as C = 0

A= 1/3
B= 2/3

therefore he is better to switch??

I may do some actual tests

[Az]

3 choices are an illusion. One of them is certain to be eliminated through the process.

[Kuciwalker]

Az: if you switch always, you have a 2/3 chance of winning each time. If you don't ever switch, you have a 1/3 chance.

[Az]

Az: if you switch always, you have a 2/3 chance of winning each time. If you don't ever switch, you have a 1/3 chance.

Once again, how come?
whatever you choose at first, you're left with two cups: one of them has a prize, the other has no prize. "switching or not switching" is just like saying "choosing this cup, or choosing that cup". The fact that you chose some cup before that is irrelevant.

[aneeshm]

Azazel - ckeck my spoiler for the table .

[aneeshm]

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 ?

[Az]

сhecked it. It completely ignores the removal of a necessarily empty cup.

[Az]

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?

[Dauphin]

Originally posted by Azazel

Or put another way, if you don't switch, your chances of winning remain 1 in 3 regardless.

How so?

You have a one in three chance at stage 1, yes? If you refuse to change your choice your chances do not increase as you are told before hand that selection of non-balled cups will be made. ONLY if there was random selection would your odds increase as ONLY then will the extra information be provided and so ONLY then will conditional probabilities apply.