as long as he doesnt select the one that you initially picked and eliminate/reveal it, youre right.

if he showed you the one you picked initially, you don't gain any new information.

Wrong.

If he *randomly* (!!!) selects which box to open of the boxes you didn't pick, and the one he selects happens to be empty, the ones that are left both have a 50% chance of containing the prize.

You see, in the original problem, you don't get any new information, because he will never reveal the box containing the prize, and there is always a box to open which doesn't contain a prize.

However, if a box is randomly selected, there is a 1/3 chance that it's the real one. If the box happens to be empty, that does not magically transfer 1/3 chance to the other box you didn't pick. It's just eliminated as a genuine possibility.


If a genuine possibility gets eliminated, chances get redistributed entirely.


The actual problem is:
X = A (1/3 chance) or Y (2/3 chance)
Y= (B or C)
Y will always contain at least 1 empty box, and an empty box will always be revealed. No genuine possibility will get eliminated by revealing a box.

The problem when used with randomness is:
X = A (1/3) or B (1/3) or C (1/3)
if X = B or C it might get revealed. A genuine possibility is confirmed or eliminated.


Non-random revealing: 1/3 vs 2/3
Random revealing: 1/2 vs 1/2

et tu, ytcracker? :(

wow
if you believe that you obviously failed math

not all math problems can be solved with math that is the lesson we should remember out of this


and now excuse me I am off to see wether my lotto numbers won or not ! :Graucho :Graucho

Wow you guys still on this crap?

You talkin' to me?

If the host randomly chooses which of the 2 boxes to open, then you get tons more information. Here's the possible outcomes, presuming for this example the cash is in C and you always select a different box (if the host selects C that doesn't eliminate it):

(1) You select A, host shows B, you choose C = WIN
(2) You select A, host shows C, you obviously choose C = WIN
(3) You select B, host shows A, you choose C = WIN
(4) You select B, host shows C, you obviously choose C = WIN
(5) You select C, host shows A, you choose B = LOSE
(6) You select C, host shows B, you choose A = LOSE

Your odds of winning by always switching, even though the host RANDOMLY chooses between the 2 remaining boxes = 66.7%
It's because you get the additional information when the host randomly chooses C that alters your odds.

Maybe you just didn't understand my original statement: You aren't randomly selecting the boxes...only the host is randomly selecting which of the 2 remaining boxes to show you in round 2. Sometimes, he will show you exactly which box to choose. Now if the host revealing a box eliminates it, then your odds of winning drop to 33.3% no matter what you do (because instances #2 and 4 will then be loses). So the end result is your odds are never 50% if the host randomly reveals one of the 2 remaining boxes.

what im saying

is lets say the host picks the boxes at random and picks the one you picked and opens it

and its empty

you havent gained any information because here was no guarantee he was picking a known empty

youre going to have to switch anyway

then it becomes a matter of 50/50

maybe i misunderstood

I see what the problem is.

You are assuming that when the host reveals a random box, after that you get to switch no matter if the box that was revealed was the right one or not, and the revealed box is not eliminated.

I am assuming that when the host reveals a random box, if it's the right box you just lose the game, and if it isn't you can choose whether to switch or not, with the revealed box being eliminated.
I am obviously counting your chance on switching only if the box revealed turns out to be the wrong one, because if it's the right one, the game is over and you don't get to switch anymore.
Switching, in this case, does not improve your chances.


Guess the problem here was just some differences in opinion on which definition of the rules of the alternative game to use :glugglug

Still????

Actually alex, the first time you made the pick, each box has a 1/3 chance of having the 10k.

In round 2, your initial pick still had a 33% of having the 10k. HOWEVER, switching DOES NOT increase your odds in round 2. Your odds of winning in round 2, is 50/50 , the second empty box had a 1/3 chance of having the 10k in round 1.

It's really simple and logical. Changing your choice will not alter the end result, and assuming because you are left with 2 options that the other empty box is a greater choice is FALSE. It had a 1/3 chance of having the 10k in round 1. In Round 2, with 2 options, each has a 50/50 split. Your chances of winning are 50/50 in round 2.

Your simple riddle is playing on your mind, it is switching the probability in round 2, but telling you that your initial choice had a 33.33% chance of being right. THAT IS ONLY true when no new information was presented. Both probabilities changed when the host showed the empty box, THEREFORE your initial probability changes as well.

THE END.

Not again :waaaaahh :waaaaahh :waaaaahh

just don't bother...

lol i had to chime in :) it was hard to resist

thing is...the odds are really 66/33...it has been proven like 100 times in this thread alone... :winkwink:

i'm a non believer , lol !!

well math is telling me I am wrong, I'm still going to start a movement with the rest of my believers on this board.

oh I don't really care anymore...we even ran a C program and tried it 500k times...

500k iterations:

$ ./a.out
Switch won 167024 (66%), lost 82927 (33%)
Stay won 83561 (33%), lost 166488 (66%)


I did like 2 graphics and countless explainations about it...if you don't bother to read'em, I really won't try to explain it anymore! ;)

like I said before, this is a known problem...not like we're the first one debating it....it HAS a solution...a fairly simple one if you take the time to try it out yourself.....

like I said earlier, it's just like someone poping in and saying gravity doesn't exist, or 2 + 2 is not = 4......

the odds ARE 2/3....it's been proven mathimatically....it's no philosophical debate...it's not one of thoes "Depends on how you look at it" problems...it's a statistical fact...