Stop being such retards and read my post on page two again. If the removed box is random, you don't gain any chance of winning by switching, because the removed box might be the "winner" and is as such still "in effect" (just not pickable).

If the removed box is always empty, you of course gain its probability by switching and so have a higher chance of winning.

:thumbsup

Uh no, I already said that. You're the second one, not the first.

The box is not random, that would defeat the purpose of the gameshow.

So assuming you always pick A as your first choice... (again)

A B C
1 0 0

eighter B or C is removed...you switch, you lose, you stay you win...only case....


now if the prize isn't behind your original pick:

PRIZE IS BEHIND DOOR B:
A B C
0 1 0

you pick A....C is removed, you end up with:

A B
0 1

will you switch to B and WIN or stay with A and LOSE?

now, PRIZE IS BEHIND DOOR C
A B C
0 0 1

Your original pick is A...now B is removed, you end up with
A C
0 1

will you STAY WITH A and LOSE or switch and WIN...

fuck that LOL it's simple

JUST FUCKING SWITCH

No. Read your own post, and the mistake in it:

You are saying that because it's NOT random, it's 50/50. However, the opposite is the case... because it's not random, it's 33.3r/66.6r.

it doesn't remain 33.333% it changes to 50% because there are only 2 boxes. no need to get all complicated on it it's just trivial.

:1orglaugh

Why wasn't the riddle written like this? 99.9% of us would get it right this way.

sorry you are wrong

i have no faith in humanity anymore

NOOOOOOOOOOO

fuck i'm going crazy here...

you are WRONG... LOL


the host KNOWS where the prize is...

the only times when you will win by STAYING is if you got it right the FIRST TIME (1/3 chances of winning)


I did all those fancy graphics for nothing...:(

silvertab and punkworld are you guys going to vegas

we need to get some drinks

Here they are again...

IT IS SIMPLE DAMNIT....my 12y.o. cousin is understanding it...


http://www.americanthumbs.com/silvertab/fordumb.jpg

http://www.americanthumbs.com/silvertab/fordumb2.jpg

Why would anybody bother to read the explanation, look at the simple illustrations that have been made or even read what a Masters student has to say about it when one could spit ignorance with so much less trouble?

still not sure! if I go, we need to hook up for sure bro! :thumbsup

Your right, I flipped my logic.

well shit when you put it that way...

50% sounds pretty good!

http://www.ytcracker.com/images/smartchart.jpg

hoooooly shit! we're making progress!

all is not lost...

at beggining of contest you have 1 in 3 or 33.3% chance of winning.....once one box is opend and its empty your chances have increased for finding the right box to 1 in 2 or 50% ...but the intial 33.3 chance of winning still exists as far as the whole contest is concerned..chooosing the remaing box isnt going to increase your chances becuase you dont know whats in that one or the one you picked...wich means no matter what you have 2 choices or 50 /50 chance of getting the money:2 cents

The real riddle is whats the chances you get offered to choose 1 box out of 3 for 10k.? 1 in 500 million ?

wrrrrrrrrrong

Here's another way of looking at it.

There are a million doors. You choose one of them. Monty opens 999,998 empty doors leaving the one you chose and another one. Do you really think you picked the car on your first try and shouldn't switch.

A reasonable person on GFY. I'm shocked :helpme

I was not agruing the rule, meerly pointing out that the method used to select the box to open adjusts the outcome... sorry to burst your bubble, no progress has been made...

Ok. I'm not reading 4 pages of conflicting views, but let me state for the record your odds are the same whether or not you switch.

ROUND 1: Your odds are 33.3% that you'll choose the correct box. Every box has a 33.3% chance of being the right box

ROUND 2: Your original odds were indeed 33.3%, and that's what you selected from. But now each box has a 50% chance of being the right box: There's only 2 boxes and 1 of them has the cash. Providing everything is random, then no box can possess a higher probability of being right: one has it; the other doesn't. You're odds don't differ from this. If the original box has a 50% chance of being right, then your odds are the same, because it's the same box. It's only your odds that have changed.

Look at it this way. The original post stated that if you switched, the odds of choosing the correct box is 50%, but if you don't switch, it is 33.3%. A LAW of statistics is the sum of all probabilities must equal 100%. This reasoning only totals 83.3% and has no other choices. Therefore, that logic cannot be correct. If your odds of choosing the other box is 50%, then your odds of staying with the original box has to be 50% as well.

Another way to look at it: The choice in round 2 is completely independent of round 1. You're asked to make a choice between 2 boxes. Being that you previously selected one of the boxes has no bearing on this decision. It is just as if a random audience member, who was taking a leak during round 1 and didn't witness anything that happened before, was asked to choose between just those 2 boxes. Obviously, his chances (and yours) are 50% to select the right box.

Ok here's the deal....

NOW EVERYONE WHO DOUBT IT JUST READ THIS AND TRY TO UNDERSTAND

We will assume that there is a winning door and that the two remaining doors, A and B, both have goats behind them. There are three options:

1. The contestant first chooses the door with the car behind it. She is then shown either door A or door B, which reveals a goat. If she changes her choice of doors, she loses. If she stays with her original choice, she wins.

2. The contestant first chooses door A. She is then shown door B, which has a goat behind it. If she switches to the remaining door, she wins the car. Otherwise, she loses.

3. The contestant first chooses door B. She is then is shown door A, which has a goat behind it. If she switches to the remaining door, she wins the car. Otherwise, she loses.



Just fucking try it on your OWN.....

If your first pick is A...
The only time you will by NOT SWITCHING is if it's really behind door A...

Now if the the car is behind eighter B or C...you WILL WIN BY SWITCHING...IN BOTH FUCKING CASE

FUUUUUUUUUUUUUUUUUUUUUUUUCK

Silver Tab, man it's amazing how you're willing to keep reading all the crap some people here are saying and keep trying to explain the problem to them... :thumbsup

I've given up a long time ago

Maybe just read the Thesis paper that is included as a link on those four pages. You'd find out you're in fact wrong.

Another way of explaining it is to break it down as this:

Three boxes of which two are empty and one has a prize. If I hand you one box and keep two for myself would you be satisfied that you have the best chance of finding the prize or would you rather switch with me and have the two boxes to my one?

That's the fucking best explanation yet. Even if someone else worded a similar earlier.

and ? ? ?what the fuck happened ? ? did that guy end up winning that $10k?? :Graucho

I will explain it to the dumb fucks who don't understand it yet


A /

B /

C /


>>> I chose C


so its not A / its not B /


they open B/ and its empty


so now we have A / and C / left closed


why would I have more chance betting on A / now ?


the 33% chance now turned into 50% chance wether I change or not it just doenst matter

Exactly.... thanks, you did a better job than me...

The host KNOW where the prize is....he will open an empty box for SURE.....

how does that matter ??


you dont know where its at ,

your original 33% turned into 50% wether you change your choice or not

No.

The choices, on the simplest level, are these: Open one box of three or open two boxes of three. Which has the better odds of finding the one box containing the prize? Is one choice of three bettern than two choices of three?

ok let's keep your example....
You chose C.....now if you decide to stay with your choice...the only time you will have it right is if it was REALLY in C to begin with...now, let's say it's in A...the host knows it...he will open B....if you switch, you will HAVE IT....

Now if, again, you chose C.....the prize is in B...the host KNOWS IT...he will open A...you switch to B...you will HAVE IT....

so the only chances to get it by sticking with you first choice (C) is if it's really in C....if it is in eighter A or B, you will NOT have it if you stay with C...but you will HAVE IT if you SWITCH.... :Graucho

thats irrelevant


your odds after opening one box will be 50% wehter you change or not

they cant grasp this

they completely eliminate the third box like it doesnt even exist anymore

its the most important part of the equation

no

that is not true

you need to read seriously

read my reply to your post...!

I understand where you are getting at and a certain 50% change is always better than a 33% chance thus you should change

but your original 33% changed into 50% wether you change or not

you are saying if you choose now again you will have a 50% chance of getting it right , this means that I stay with my choice and dont change I still have 50% chance , thats why I am saying it doesnt matter if you change or not

I read it and fully understand it but 50% is 50% either way

no no no no no

ok look

the host
will ONLY pick an empty box

he will not pick the box with the prize

http://www.americanthumbs.com/silvertab/fordumb2.jpg

Na, Silvertab is correct, Think of it this way and it should be more clear.

"wrrrrrrrrrong

Here's another way of looking at it.

There are a million doors. You choose one of them. Monty opens 999,998 empty doors leaving the one you chose and another one. Do you really think you picked the car on your first try and shouldn't switch."

no ^^^^^

there's no 50/50!


it's 66/33 if you SWITCH....


if you take a million box, you have 1 chances of winning out of 1 million if you stick with your choice, and 999 999 chances of winning if you switch LOL

EVEN IF YOU END UP WITH 2 BOX YOUR FIRST GUESS WAS OUT OF A MILLION BOXES

Same rule apply with 3 boxes!

thats where you are wrong , this case is about 3 not about 100 or any other number in this case it doesnt matter if you switch , I havent looked at the graph yet , but I do know it wont be correct if doesn't support my claim :winkwink:

nick do you have aim
seriously just IM me
and i can break it down

i know youll get it once i explain it

aim: brycec

just give me a chance to explain it

yeah I just look at the graph and tha graph called me retarded :1orglaugh

I read it, don't understand it, but running the numbers of "always switching" to "never switching" don't lie -- always switching gets you more wins.

I grasp it now how they spin it mathematically but in a real life situation it wouldn't matter if you switched







you know that they always pick the one that is empty first , so you know you start out with a 50% chance