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Just remembered something I encountered the other day. For all those who think 50% is the right answer, here's a master's thesis from a guy from my faculty proving you wrong:
http://www.philos.rug.nl/~barteld/monty.ps It's in postscript though, so get a viewer for that (e.g. ghostscript). |
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arghhhhhhhh no its nothing to do with 50% there are three boxes ahghghgihsiughdfiugiudfhiguhdfiughihwrihehhfdsoghf dghodfh shit |
Fuck that I can't believe people don't understand it...
Here we go again: Case 1: THE PRIZE IS BEHING DOOR B Let's say you pick door A.....The host opens door C... you STAY with A...YOU LOSE Case 2: THE PRIZE IS BEHIND DOOR C Let's say you pick door A.....The host opens door B... you STAY with A...YOU LOSE The only chance you would win is if the prize is behind door A (1/3) BUT IN BOTH CASE IF YOU STARTED WITH DOOR A AND SWITCHED, YOU WOULD WIN (2/3 chances of winning) come on it's not that hard to understand |
Originally posted by ytcracker
yes i think this will help **GFY VERSION OF THE RIDDLE** there are three hot looking women in front of you and two of them are trannies but they are all wearing baggy pants their pimp says choose the woman and you can have sex with her so you choose the one with the least defined jawline the pimp then removes one of the trannies from the lineup you notice the one you chose has a boner case closed ^- brilliance at work believe that |
RECAP!
In all this back and forth shit, everything has been kind of spread out. I won the ebay auction i was waiting for so i'm going to make one last post before i fuck off to bed for you guys to read and think about. 3 boxes. 1 has a prize in it. Initially, you have a 1 in 3 chance of picking the correct box. An incorrect box is then removed, leaving you with 2 boxes. One is the prize. The other is not. Two boxes. 50 percent chance. That makes it 1 in 2. You're then offered the ability to switch. If you choose not to switch, you're left with a 1 in 2 chance. If you elect to switch, you now have a 2 in 4 chance, which is the same as a 1 in 2 chance. Let me explain. At first, there are 3 boxes and a total of 3 outcomes each with a 33.3333% chance of occuring. A box is then removed. The 3 becomes 100% irrelevant. You now have two boxes and a total of 2 possible outcomes. These outcomes each now have a 50% chance of occurring. You are then offered a switch. This gives each of the 2 existing possible outcomes another possible outcome each, leaving you with a total of 4 possible outcomes. 2 of them will win if you elect to stay. 2 of them will win if you elect to switch giving you a 2 out of 4 chance, i.e., 1 in 2 chance of choosing the correct box. Illustrated below: Outcome Number 1: If A is right then the host reveals B and switching does NOT pay off OR Outcome Number 2: the host reveals C and switching does NOT pay off Outcome Number 3: If B is right then the host reveals C and switching does pay off Outcome Number 4: If C is right then the host reveals B and switching does pay off 2 out of 4 = 1 out of 2. That should explain it enough. But to further prove just how irrelevant the third box is, ponder this. You're given 2 boxes. You get to choose one. Then after making your choice, you get told you can change your mind. There are 4 possible outcomes. The third box is absolutely irrelevant. It serves no purpose whatsoever. This is further illustrated by the fact that NO MATTER WHAT YOU DO, an incorrect box will always be taken away and you'll be left with one right, and one wrong. No matter fucking what! Can't you see how that completely negates the third box? Anyways. I'm off. Bedtime for me. |
Let's take a look at the 3 cases again LOL
1 = prize, 0 = no prize CASE 1: A B C 1 0 0 You pick A, the host open eighter B or C...you stay with your choice, you win....you switch, you lose... ONLY CASE WHERE YOU WILL WIN CASE 2: A B C 0 1 0 You pick A, the host open C....you switch, YOU WIN...you stay with your choice YOU LOSE CASE 3: A B C 0 0 1 You pick A, the host open B....you switch, YOU WIN...you stay with your choice YOU LOSE Fuck....it's easy to understand! there's no 50/50 here damnit! |
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I'll blame myself, though, and assume that I haven't explained it clearly enough yet. So, to quote myself: Quote:
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Let's say you pick A.....
The only case you will win is if it's behind door A....but if it's behind door B or C...you will have it by switching!...as simple as that! Let's say you pick B..... The only case you will win is if it's behind door B....but if it's behind door A or C...you will have it by switching!...as simple as that! Let's say you pick C..... The only case you will win is if it's behind door C....but if it's behind door A or B...you will have it by switching!...as simple as that! Daaaaaaaaaaaaaaaaaamn |
chodadog
your analysis is incorrect the third box is incredibly significant it is the key to this whole thing if the host chose any box at random, then it wouldnt matter he doesnt though he chooses an empty box 100% of the time when you made your initial guess you chose the box based on a 1/3 assumption not knowing anything about what your box or any other box contains now when an empty is taken out of the picture, you are now switching based on a 66% chance because before you didnt know the box the host removed was empty by switching to the other box, you are now basing your opinion because of the presense of the known empty, not 50/50 chance because you picked that box based on 1/3 im having a hard time explaining this...fuckfukcukcukf |
silvertab your explanation is fucking easy thanks
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Let's have an invisible man be at box #2... ok? just standing there. Now, the person takes a box... invisible man get box #2 and stands there and the announcer opens box #3 to reveal that its' empty. Ok, no the person is standing there thinking "ok, that was opened, so the other box must have a higher chance of being the one with the prize." BUT WAIT. What would the invisible man be thinking?? Huh?? Would he be standing there thinking "ok, I have box #2, and the box #3 was empty, that must mean that the box infront of the other guy has a higher chance of having the prize!" Wait a sec.... does that mean that BOTH boxes have a 66% chance of having the prize? Can it possibly be... that just because you're standing with one box in your hands, that it DOESN'T AUTOMATICALLY MEAN THAT THE OTHER BOX HAS TO GET THE EXTRA 33%??? Who is person #1 to decide WHICH box has the higher probabillity?? huh?? What about the poor invisible man behind box #2?? His is automatically the lower probability because person #1 says so?? I think not. |
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I think I will give up! :( although the proof has been made...(I even think it's quite easy to figure out) some will just never admit it LOL |
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there is a 66% chance to win if he switches, not 50%... if you think it's 50-50, you're wrong... it's pretty simple |
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your choice is not 50/50 your initial choice was based on 3 boxes the host will always show you that one box is empty Quote:
do not think of this in terms of only two boxes |
SilverTab, ytcracker, I applaud your hard work but I fear that it will yield no results :(
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Try not to think of it in terms of one person... think of it being two boxes. Read my post... if there's someone at box #1 and someone at box #2... which one has the higher probability? You assume that the chances are higher so long as one person is making a choice, but when two people are, you can easily see... the odds are even. |
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you have to think of it in terms of one person one guess the reason why is because if the one person has INITIALLY chosen an empty box the host cannot open HIS box to show him his guess was wrong he will open the empty box that the guy hasnt selected in which case if he switches he will win however lets say he picks the prize box first the host has discretion over either empty box to open if he switches away from the prize he loses this is the only condition where he loses otherwise the host will always be taking away an empty box in favor of the contestant im trying to think of how i can illustrate this point in an easy fashion |
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One person selects two boxes (by actually selecting 1 box) while the other person is stuck with only 1 box. Look at it like this: You are trying to select 1 winning item out of 3 possible choices. If I give you one unknown box and keep two for myself, who has the better odds of winning? I have two chances of winning and you only one. Adding a fictional third person completely changes the equation. |
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because say the invisible man was sitting at the other empty box when the host takes that box from the invisible man he cant just switch boxes thats not the fucking point |
sheesh.. you guys go on picking that other box then. Good luck on that one.
I say one of you guys puts it to a test. Have someone know where the prize is so that he can remove a box every time.... and you go on picking the other box every single time for about 500 times. See whether or not you get it right 250 times, or more. |
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scroll down about a quarter of the way and play the simulation just hit the radio buttons to guess where the prize is try this yourself it will become clear to you |
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http://www.cut-the-knot.org/hall.shtml http://www.cut-the-knot.org/hall.shtml http://www.cut-the-knot.org/hall.shtml http://www.cut-the-knot.org/hall.shtml http://www.cut-the-knot.org/hall.shtml Just try it damn it!...the numbers are there....can't see why people won't understand it.... if you stay, you have 1/3 chances....(if you picked it correctly at your first pick) IF YOU DIDN'T PICKED CORRECTLY ON YOUR FIRST PICK, YOU WILL ALWAYS HAVE IT BY SWITCHING! |
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if you have selected EITHER one of the TWO empty boxes, he will remove the other empty box and you will switch to the prize you have a 2/3 chance of selecting an empty box the first time you have a 1/3 chance to select the prize this is not fucking hard |
Wow, there's gotta be a better way to write this "Riddle". I need some Tylonol
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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. |
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:thumbsup |
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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 |
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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. |
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Why wasn't the riddle written like this? 99.9% of us would get it right this way. |
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i have no faith in humanity anymore |
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