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Post by TDR on Jun 17, 2009 15:07:24 GMT -5
let me put it this way. Without Monty opening any door, the odds of you winning are always one in three. We can all agree to that. So lets set up three doors. For the purposes of this explanation we will call these doors CAR, DONKEYA, DONKEYB There are three possible initial setups. You have picked either car OR donkeyA OR donkeyB. With me so far? IF you pick the car, Monty comes along and opens either DonkeyA or Donkeyb and offers you the choice and you take it. this moves you from a car to a donkey. This sucks. So if you start on door CAR, you LOSE. IF you pick donkeyA, Monty comes along and has to show you DonkeyB (the only other non winning door on the floor). You swith and get CAR. You WIN. IF you pick donkeyB, Monty comes along and has to show you DonkeyA (the only other non winning door on the floor). You swith and get CAR. You WIN. So from any starting point where your odds were intially one in three, the odds of winning after a switch become, let's count (lose, win, win), two in three. Those odds are better. That's the most persuasive explanation of the switch strategy so far. |
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Post by billhammond on Jun 17, 2009 15:10:29 GMT -5
Yeh, I posted my last note before reading Tramp's post.
I am ALMOST there.
Us right-brain folk don't do math, but we sure can improvise melodic lines!
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Post by millring on Jun 17, 2009 15:10:38 GMT -5
Here's another way to frame it:
The geniuses are telling me that when Monte took the 2 out of 3 doors with the 2/3 chance of having the car, and then revealed one of the doors and proved that it was not a car.... ....the geniuses are telling me that this reveal confers upon Monte's remaining door the full 2/3 chance that both doors had before the reveal.
That's nonsense. The reveal PROVED that Monte did not have 2/3 chance -- it cancelled out one of the possibilities. It did not confer upon Monte's remaining door 2/3 chance. Hell, after the reveal there is no longer 3/3 for there to be 2/3 of.
The reveal left Monte with the 50/50 choice that the contestant had all along.
And, again, there is no advantage to changing choices in a 50/50 proposition.
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Post by Fingerplucked on Jun 17, 2009 15:10:48 GMT -5
let me put it this way. Without Monty opening any door, the odds of you winning are always one in three. We can all agree to that. So lets set up three doors. For the purposes of this explanation we will call these doors CAR, DONKEYA, DONKEYB There are three possible initial setups. You have picked either car OR donkeyA OR donkeyB. With me so far? IF you pick the car, Monty comes along and opens either DonkeyA or Donkeyb and offers you the choice and you take it. this moves you from a car to a donkey. This sucks. So if you start on door CAR, you LOSE. IF you pick donkeyA, Monty comes along and has to show you DonkeyB (the only other non winning door on the floor). You swith and get CAR. You WIN. IF you pick donkeyB, Monty comes along and has to show you DonkeyA (the only other non winning door on the floor). You swith and get CAR. You WIN. So from any starting point where your odds were intially one in three, the odds of winning after a switch become, let's count (lose, win, win), two in three. Those odds are better. You have 3 choices. You pick one. Your odds are one in three. Using your scenario with the car behind door#1, and donkeys A & B behind doors #2 & #3: STRATEGY: DON'T SWITCH DOORSDOOR CHOICE, RESULT ......... 1 .............. WIN ......... 2 .............. LOSE ......... 3 .............. LOSE ODDS OF WINNING, 1/3 STRATEGY: SWITCH DOORSDOOR CHOICE, RESULT ......... 1 .............. LOSE ......... 2 .............. WIN ......... 3 .............. WIN ODDS OF WINNING, 2/3 |
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Post by Fingerplucked on Jun 17, 2009 15:13:52 GMT -5
That's the most persuasive explanation of the switch strategy so far. The windows thing didn't do it for you, huh? I could maybe throw in a bottle of Windex if you think that'd help. |
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Post by millring on Jun 17, 2009 15:14:37 GMT -5
let me put it this way. Without Monty opening any door, the odds of you winning are always one in three. We can all agree to that. So lets set up three doors. For the purposes of this explanation we will call these doors CAR, DONKEYA, DONKEYB There are three possible initial setups. You have picked either car OR donkeyA OR donkeyB. With me so far? IF you pick the car, Monty comes along and opens either DonkeyA or Donkeyb and offers you the choice and you take it. this moves you from a car to a donkey. This sucks. So if you start on door CAR, you LOSE. IF you pick donkeyA, Monty comes along and has to show you DonkeyB (the only other non winning door on the floor). You switch and get CAR. You WIN. IF you pick donkeyB, Monty comes along and has to show you DonkeyA (the only other non winning door on the floor). You switch and get CAR. You WIN. So from any starting point where your odds were intially one in three, the odds of winning after a switch become, let's count (lose, win, win), two in three. Those odds are better. Problem is that we know that there is not a DonkeyA and a DonkeyB. We know this because we are told beforehand that Monte will ALWAYS reveal one donkey. Since that is the case DonkeyA = DonkeyB exactly. They are interchangable. They are , for the sake of the way the game is set up, interchangable. They are not two choices because one will ALWAYS be removed. DonkeyA IS DonkeyB because no matter what happens, it will be removed. |
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Post by patrick on Jun 17, 2009 15:15:03 GMT -5
I'm still with Millring on this one. This reminds meof those "word Problems" in algebra, whre they throw in all kinds of extraneous stuff to confuse you about what the real problem is (Two trains, one painted red and one painted plaid, are headed toward each other. If one passes through a city where the football team hasn't won a game in 20 years traveling at... etc.)
Consider this: I'm playing the game with Monte (I'd rather be playing with Carol, but that's another story). So, I pick my first door, then Monte opens another door.
Just then, I get a phone call, and I can buy a Cargo OX at half price! So I run out, telling Omaha on the way, "Go pick a door!" and disappear.
Omaha now has to make a choice between two doors. Why aren't his odds 1/2? And why would switching make any logical sense in that case, since he doesn't know which door I picked vs which one I didn't?
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Post by Supertramp78 on Jun 17, 2009 15:17:45 GMT -5
Let's just say that it is a hard concept to grasp and lots of very smart people don't grasp it. But from a pure math point of view, it is a proven fact. Switching increases your odds of winning.
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Post by omaha on Jun 17, 2009 15:18:19 GMT -5
Ok, you're just screwing with me, right? No, I'm not. Nothing has logically been established except that the choice, the game, was always going to be a 50/50 choice. And when you have a 50/50 choice there's never any statistical advantage to either switching OR keeping the same choice. Lets try it this way. Forget trying to actually process the logic. Lets just look at this as a brute force exercise in the probabilities. There are only three possible configurations to the game, and only two strategies. So there's only six scenarios in the solution space. Here they are (BTW, since I'm constructing the game, and I really don't need a new car, I decided that the prize is to have a hot girl in a bikini come wash your car for you):   It really is that simple. You guys are way over-thinking this. |
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Post by Supertramp78 on Jun 17, 2009 15:21:27 GMT -5
donkeyA doesn't equal donkeyB but it could and still not matter. All donkeyA & B have to be is NOT cars. You start with three choices. If nothing is revealed your odds are 1/3. If a donkey is revealed after you pick, switching results in three options, two of which result in a win. Not switching results in the same odds you started with or you could say you now have a 1/2 chance since the odds have been improved from 1/3 to 1/2 y the removal of one bad choice. Regardless, I'll take 2/3 over 1/2 any day. Switching improves your odds.
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Post by omaha on Jun 17, 2009 15:21:41 GMT -5
I'm still with Millring on this one. This reminds meof those "word Problems" in algebra, whre they throw in all kinds of extraneous stuff to confuse you about what the real problem is (Two trains, one painted red and one painted plaid, are headed toward each other. If one passes through a city where the football team hasn't won a game in 20 years traveling at... etc.) Consider this: I'm playing the game with Monte (I'd rather be playing with Carol, but that's another story). So, I pick my first door, then Monte opens another door. Just then, I get a phone call, and I can buy a Cargo OX at half price! So I run out, telling Omaha on the way, "Go pick a door!" and disappear. Omaha now has to make a choice between two doors. Why aren't his odds 1/2? And why would switching make any logical sense in that case, since he doesn't know which door I picked vs which one I didn't? But the thing your scenario misses is that Monte chose which door to open based on which one you selected. Assuming I knew which door you selected, I would always switch. If I don't (under your modification) know which door you chose first, you've fundamentally altered the game, and it is indeed 50/50 to me. But that's a different game than the one we are discussing. |
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Post by Fingerplucked on Jun 17, 2009 15:22:19 GMT -5
STRATEGY: SWITCH DOORSDOOR CHOICE, RESULT ......... 1 .............. LOSE ......... 2 .............. WIN ......... 3 .............. WIN ODDS OF WINNING, 2/3 If you don't switch doors, the only way you can win is to pick the right door. If you switch doors, the only way you can win is to pick one of the wrong doors. But since there are more wrong doors than right doors, it's easier to win. |
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Post by omaha on Jun 17, 2009 15:22:43 GMT -5
BTW, I see that while I was putting together my visual aids, Jim posted essentially the same info in text form. I think, with all due modesty, that my version was worth the wait! :-)
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Post by aquaduct on Jun 17, 2009 15:23:02 GMT -5
Oops. Just put pen to paper and I think I've proved Millring right.
The probabilities are the straight number of winning scenarios from the pool of available outcomes.
So write down a matrix of available outcomes, i.e.- choose door 1, cars behind door 1; choose door 2, car's behind door 1; etc. You'll wind up with 9 possible outcomes.
Now pick any door to open and cross out the outcomes that either have you choosing that door or have the car behind that door.
You're left with (if I did it right) 4 possible outcomes, 2 of them win with your original guess and 2 of them win if you switch.
Dang. Hate to admit it, but this appears to be an urban legend that I've swallowed as hard as anyone.
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Post by millring on Jun 17, 2009 15:24:20 GMT -5
The chart is a misdirection. It doesn't matter how the doors are numbered. One door will always be removed, so you might as well make a two door chart. It doesn't add to the number of possible outcomes just because the car is to the left or to the right of the contestant's original choice.
There's still only a fifty-fifty choice. A donkey or a car. One donkey is ALWAYS removed from the choices.
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Post by knobtwister on Jun 17, 2009 15:28:14 GMT -5
I think what is throwing some folks off is breaking it down into the 2 steps;
1.Pick a door
2. Switch or don't switch
If you pick it apart the odds are different depending on where you are in the process. But it's all one process so you can't pick it apart.
Don
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Post by aquaduct on Jun 17, 2009 15:31:13 GMT -5
Did I mention that Millring is right?
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Post by Fingerplucked on Jun 17, 2009 15:31:54 GMT -5
BTW, I see that while I was putting together my visual aids, Jim posted essentially the same info in text form. I think, with all due modesty, that my version was worth the wait! :-) I like yours better too. I was going to draw naked women, but they looked just like my clothed women: %$$ sdf-0df 3033 4340954p98 See what I mean? The only problem with your visual is that you kept switching the contents behind the doors. |
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Post by omaha on Jun 17, 2009 15:31:56 GMT -5
Tell you what.
Anyone who thinks the odds are different than what Jim and I are saying, I will gladly play this game for money.
Here's the deal. It costs $100 per game to play. If you win, I'll pay out $250. If you believe this is a 50/50 game, then you should love those odds.
The only stipulation is that you have to stick to the "hold on to your initial choice" strategy. No switching.
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Post by Supertramp78 on Jun 17, 2009 15:33:23 GMT -5
You could look at it this way. The odds of you winning AFTER switching depend on what you pick initially. If you pick a car, you will lose since switching will result in a donkey. if you pick a donkey initially, you will always win if you switch because only a car will remain after the other donkey is removed. So if you stick with your original pick, you still have your one in three (or one in two if you want to think that way). But if you switch, you win based on the inverse of the original odds. Pick a donkey = win. Pick a car = lose. Since there are two donkeys, the odds of winning now go to two chances out of three.
Listen, there are lots of three doors problem simulators onmath sites all over the internet. there are java aps and mathematica programs and freshman college students banging on this all over the place and it just is what it is. Switching increases your odds.
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