|
|
Post by omaha on Jun 17, 2009 15:33:51 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 Bingo. The "hold" strategy boils down to "pick one door". The "switch" strategy boils down to "pick two doors". Its as simple as that. |
|
|
|
Post by TDR on Jun 17, 2009 15:34:30 GMT -5
I just gotta say, if I ever go on this show and switch, and it turns out the new car and the hot chick were behind my first pick.... I'm gonna be really pissed I listened to you guys.
|
|
|
|
Post by aquaduct on Jun 17, 2009 15:37:09 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. Let me triple check my math (and these theoretical computer programs) and get back to you. |
|
|
|
Post by millring on Jun 17, 2009 15:38:05 GMT -5
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). . Since we know the game, one in two is the only way one could logically frame it. And that's been the misdirection all along. Every single time, in every single game, in every single instance, no matter how ANY contestant EVER plays the game, and no matter how the doors are numbered, or what is behind what door... the contest will ALWAYS end up with either of two choices -- one car or one donkey. And we know that that will be the case going in. |
|
|
|
Post by Supertramp78 on Jun 17, 2009 15:39:59 GMT -5
nevertheless, the odds of winning after a switch is equal to your odds of initially picking a donkey.
2/3
|
|
|
|
Post by omaha on Jun 17, 2009 15:40:14 GMT -5
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). . Since we know the game, one in two is the only way one could logically frame it. And that's been the misdirection all along. Every single time, in every single game, in every single instance, no matter how ANY contestant EVER plays the game, and no matter how the doors are numbered, or what is behind what door... the contest will ALWAYS end up with either of two choices -- one car or one donkey. And we know that that will be the case going in. But what you are missing is that the decision regarding which door to open is not random. It is informed by the choice the contestant makes first. |
|
|
|
Post by omaha on Jun 17, 2009 15:41:26 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. Let me triple check my math (and these theoretical computer programs) and get back to you. Any time, any place, any stakes. |
|
|
|
Post by millring on Jun 17, 2009 15:42:58 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 The "switch" strategy boils down to "pick two doors". Really? Which two doors, exactly? You have either the one you originally picked, or you have the one that Monte didn't reveal. Which is your other door to pick? |
|
|
|
Post by TDR on Jun 17, 2009 15:45:50 GMT -5
Bingo. The "hold" strategy boils down to "pick one door". The "switch" strategy boils down to "pick two doors". Its as simple as that. Hunh? Nope. It boils down to: Step A: pick one door, at which time you have zero chance of winning anything. Step B: pick one door, at which time you have a 1/2 chance. BTW, in Step A there was a DonkeyA and a DonkeyB. But there was no chance of winning a car. In Step B, there was only a DonkeyA OR a DonkeyB. One Donkey. How is there not, at that point, a 1/2 chance you are wrong for changing your hunch from Step A? |
|
|
|
Post by omaha on Jun 17, 2009 15:46:05 GMT -5
You're getting all wrapped up in the theatrics.
What if Monte just said "pick two doors, and if the prize is behind one of them, you will win".
Under that construction, do you agree that you have a 2/3 chance?
Lets go back to the game as formulated.
You desire to pick doors 2 and 3. So, given your first pick, you select #1, knowing you will have the chance to switch, which you do.
How is that any different from picking doors 2 and 3 in the first place.
The theatrics breaks it down into two steps, but in reality you pick which two doors you want with your first pick.
|
|
|
|
Post by billhammond on Jun 17, 2009 15:47:00 GMT -5
You know, my head is about to explode, so I would just like to say for the record that I ALREADY HAVE A CAR, SO I WILL JUST TAKE THE FREAKING DONKEY, OK!?!??
|
|
|
|
Post by millring on Jun 17, 2009 15:47:26 GMT -5
Since we know the game, one in two is the only way one could logically frame it. And that's been the misdirection all along. Every single time, in every single game, in every single instance, no matter how ANY contestant EVER plays the game, and no matter how the doors are numbered, or what is behind what door... the contest will ALWAYS end up with either of two choices -- one car or one donkey. And we know that that will be the case going in. But what you are missing is that the decision regarding which door to open is not random. It is informed by the choice the contestant makes first. Nonsense. Monte will always have a donkey to reveal. That changed nothing. Monte's the one who is not acting on random, but his choice doesn't have any bearing on the other two, except to prove that it was a 50/50 proposition all along. |
|
|
|
Post by millring on Jun 17, 2009 15:48:45 GMT -5
Let me triple check my math (and these theoretical computer programs) and get back to you. Any time, any place, any stakes. Nice "put up or shut up" strategy of debate. But just out of curiousity, what is going to determine who is right? Do you think you can run off enough samples of the game to prove beyond a statistical doubt? |
|
|
|
Post by omaha on Jun 17, 2009 15:49:59 GMT -5
But what you are missing is that the decision regarding which door to open is not random. It is informed by the choice the contestant makes first. Nonsense. Monte will always have a donkey to reveal. That changed nothing. Monte's the one who is not acting on random, but his choice doesn't have any bearing on the other two, except to prove that it was a 50/50 proposition all along. I give up. But my offer still stands. Any time, any place, any stakes. |
|
|
|
Post by millring on Jun 17, 2009 15:50:30 GMT -5
You're getting all wrapped up in the theatrics. . Sorry, but theatrics don't even enter the equation for me. I'm not presuming ANY theatrics. None. I'm just accepting as a given (and I'm right) that Monte will take one goat out of the equation, proving that it was a 50/50 game all along. |
|
|
|
Post by knobtwister on Jun 17, 2009 15:52:44 GMT -5
I surprised this thread has gone this long without mentioning string theory or Hitler.
|
|
|
|
Post by omaha on Jun 17, 2009 15:54:32 GMT -5
Any time, any place, any stakes. Nice "put up or shut up" strategy of debate. But just out of curiousity, what is going to determine who is right? Do you think you can run off enough samples of the game to prove beyond a statistical doubt? I'm not trying to be an ass about this or anything, but I just can't think of any other way to explain this other than what's already been provided. As for proving who's right or not, that's not my intent. The way I've laid out the payouts, anyone who thinks they have a 50/50 chance of winning should jump at this. |
|
|
|
Post by aquaduct on Jun 17, 2009 15:54:48 GMT -5
Nonsense. Monte will always have a donkey to reveal. That changed nothing. Monte's the one who is not acting on random, but his choice doesn't have any bearing on the other two, except to prove that it was a 50/50 proposition all along. I give up. But my offer still stands. Any time, any place, any stakes. And I'm willing to go there after I've pondered a bit more. The math clearly doesn't work from a straight statistical point of veiw (winners in possible outcomes). Now I've got to figure out where the programs are skewing the results, which seems to be the ultimate proof that underlies the whole thing. Even drulee admits he doesn't understand it (page 2 if you've lost track). Hmmm..... |
|
|
|
Post by billhammond on Jun 17, 2009 15:55:32 GMT -5
BESIDES, THE CAR IS PROBABLY A 1999 PONTIAC AZTEK!
|
|
|
|
Post by Russell Letson on Jun 17, 2009 15:56:33 GMT -5
I'm seeing some interesting (or perhaps just dimly-understood-by-the-English-nerd) terminological anomalies here.
First, "odds" generally applies to situations in which there is randomness or indeterminacy on at least one side of the situation: coin tosses, card-draws, and such. In the three-door problem, the indeterminacy resides in the player not knowing which of the three doors conceals the prize, and the chances of finding it are indeed 1/3.
Once a non-car door is opened, that probability has vanished: door #N now has a 100% chance of being a loser and it is no longer part of the problem of "which door has the car?" The question is re-framed as "Do you want to change your choice?", which is actually another way of asking "Which of the remaining doors do you choose?" Not-changing is simply another choice, and the odds are now governed by the new situation of two possibilities, a random chance (there is no new information on which of them might conceal the prize) of 1/2.
The only thing that matters about what Monty/Monte/Monti knows is that he has to know where the car is so he doesn't reveal it in setting up Round 2.
My (admittedly innumerate) take is that everything in this puzzle depends on definitions of what set of actions is being counted. I would welcome an explanation of how my understanding is defective.
|
|
