A math question

[Insectoid]

My family had a huge argument about this, so I'm coming to you to see who comes up with the same answer I did, and if nobody does, I'll explain my reasoning in an upcoming post.

The story-

There are 3 cups set out , upside down, on a table. One has some token inside(hereafter referred to as 'the right one'), and it has not been revealed which one. After one has been selected by Jeremy, one cup is removed and identified as a cup without the token. Now there are 2 cups on the table, one the right one, one a wrong one. Jeremy still doesn't know if his is the right one. Jeremy is given a choice: Change his selection to the other cup, or keep his current selection.

The question-

Should Jeremy keep the cup he chose when there were 3 on the table, or change it now that there are only two?

possible answers-

1. Either; it doesn't make a difference
2. Change it, the other cup has better odds!
3. Don't change it, the one he chose at first is likely the right one!

My answer- 2. I wil explain after some other people answer, unless someone else does first.

Submit your answer in the poll, explain yourself in a post. PLEASE, explain yourself.


+100 bits to the first one to have the same reason as me.

[r691175002]

The odds are better if you change.

I recall writing a program that did a few hundred thousand iterations of this question a while back to "prove" it. You could also show it by identifying all the possibilities since there isn't that many.

[Saad]

Like r691175002 said, the odds are better if you change. This is known as the Monty Hall Problem

[apomb]

I still dont think it matters, he should keep his hand on the one at the beginning... how does the probability change for either cup left on the table now? His has just as likely a probability to have the token than the other one.

However, after reading that Wikipedia article, i am completely wrong. lol

[Nick]

every cup has a 33% chance... it doesn't matter

sure the odds are the cup he has isn't the right one, but the odds are just the same for each cup

[Mackie]

I saw this on Daily Planet. They told me that you would have better odds if you changed it. So, I'll go with that. Blindly following people I don't know.

[apomb]

Ok, well bevause the game show host KNOWS that there is none behind the door he didnt open, i get it ... your question is a bit ambiguous

[Insectoid]

I believe I used cups, though the original did use doors I believe. Well, I have proof now, what with Daily Planet (great show) AND a wikepedia article. My reasoning:

The percent chance of the cups removed * % chance per cup is divided by and added to the odds of number of cups remaining - 1

I.E 3 cups with 33% each. 1 cup removed
(1 * 33) / ( 2 - 1 ) = 33.
33 + 33 = 66
Thus, the one cup not chosen has a 66% chance of being the right one, whereas the other cup has 33%. Switch cups man, better odds.

A bigger example.

50 cups. choose 1. likely the wrong one, right?
remove 48 cup, does your selection magically turn to the right one? no, it's still (likely) the wrong one.

Math:

50 cups with 2% chance each, - 48
(48 * 2) / (2-1) = 96
96+2=98%

The unselected cup has 98% chance of being the right one. The originally selected cup still has only 2% chance of being the right one.

I made this up myself! If this was mentioned in the wiki article or Daily Planet, well, I haven't read/watched those yet.


I'm only getting a 71% in math right now.
but if I take out all the other students...

(31*71) + 71 = 2272%! Woo!

Now how to eliminate them...


oh yeah, +100 bits to r691175002

EDIT: Don't vote in the poll unless you know (or think you know) what you're talking about. No 'I'll just blindly follow people I don't know' *couph*macki*couph*

[tenniscrazy]

they also did this in the beggining of the movie 21. Thats the one with the smart guy who counts cards in vegas.

Edit: right?

[Nick]

your logic is seriously flawed

[tenniscrazy]

man, i keep coming up with different reasonings...i'll have to put some serious thought into this then get back...tho i know it has been proven many times, i still want to prove it to myself. not just follow blindly

[apomb]

Nick @ Wed May 28, 2008 8:39 pm wrote:

your logic is seriously flawed

I know it seems flawed, but the actual reasoning, if you read the Wiki is solid... if you make sure you read the question about the host of the show knowing which door to open, it might just click. thats how it happened with me, i thought the same way you did at first.

[rdrake]

Like Saad said above, this is the classic Monty Python Question. It came from the show "Let's Make A Deal" or something like that.

[r691175002]

Here is a potentially clearer explanation.

When you choose your first case, the chance that you chose the right case is 1/3.

Here is the key point of the matter. When one case is removed, the probability you chose the right case does not increase.
Even after one case is removed, the chance you picked the right case is still 1/3.

You made a choice that had a 1/3 chance of being correct. Nothing can change that anymore.

Now, since between the two cases, one of them must have the prize, therefore the remaining 2/3 probability goes into the second case.

The hinge to this question is whether or not the probability you chose the right case increases as other options are eliminated.

[apomb]

rdrake @ Wed May 28, 2008 10:16 pm wrote:

Like Saad said above, this is the classic Monty Python Question. It came from the show "Let's Make A Deal" or something like that.

I believe the show was actually called "Faulty Towers" and yes, it did have john cleese, but it wasnt about Monty Python at all!