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Post by Supertramp78 on Jun 17, 2009 16:20:59 GMT -5
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Post by billhammond on Jun 17, 2009 16:22:54 GMT -5
Those look totally like zirconium to me, Tramp.  |
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Post by millring on Jun 17, 2009 16:37:55 GMT -5
okay, HERE is my newest revelation:
If you chose one and never switch you will always have a 1 in three chance of winning the car. Statistically speaking, if you do this infinite times, you will come closer to 1/3.
If you always switch you will have a 2/3 chance of winning when the number of games played adds up to an infinite number. Monte's removing of one goat doesn't change it. It's the other two doors and it will always be the other two doors.
I perceive a number of semantic errors made in the defense of the 2/3 proposition, but they don't make it wrong.
I'm sorry, but I'm jazzed! Good workout!
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Post by Russell Letson on Jun 17, 2009 16:38:25 GMT -5
I still don't understand in what way Choices 2 and 3 are different other than in being (unknown to the player) losing choices. They can be labeled "loser A" and "loser B," but all that matters is that they are "not-winners" and thus are identical from the point of view of being not-A (in old-fashioned logic terms).
It does not matter to the player which not-A is revealed--there is still one available to choose. The decision to stand pat or switch is simply a second choice, with no additional information about the state of the available choices (thus their state is indeterminate--not the same as random, but that was the case from the start).
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Post by Russell Letson on Jun 17, 2009 16:41:21 GMT -5
John's way of framing the problem gives me a dim understanding what the problem actually consists of. And I still think the nub of the problem lies in the way it is framed.
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Post by omaha on Jun 17, 2009 16:42:53 GMT -5
"The decision to stand pat or switch is simply a second choice, with no additional information about the state of the available choices "
That's where you are wrong. There is additional information. In revealing the door that he does, Monte is giving you information you didn't have before.
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Post by Supertramp78 on Jun 17, 2009 16:46:58 GMT -5
I see my work here is done.
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Post by Fingerplucked on Jun 17, 2009 17:03:01 GMT -5
Some of you guys are getting hung up on the last step, the last choice to be made from among the three doors, and maintaining that the odds are 50/50. Intuitively and even logically, that seems correct.
Others (like me) are looking at the whole process and keep stressing the strategy of switching doors. Logically, we can see that using this strategy increases your odds to 2/3.
Maybe it makes more sense using 10 doors.
Your initial odds of picking the right door are 1/10. And they remain 1/10 if you don't switch doors.
If you switch doors, your odds become 9/10. As long as you're switching doors at each step, you will win as long as your first choice wasn't the door with the prize.
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Post by Fingerplucked on Jun 17, 2009 17:05:50 GMT -5
Wow. I had that typed but not posted. I got called away to help my boss with the new concept of cut & paste. We went over it a bunch of times. I'll be going over it again, I'm sure.
Now I see that the issue may have been resolved. I have some backtracking to do.
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Post by Russell Letson on Jun 17, 2009 17:11:23 GMT -5
Increasing the number of doors and extending the choice horizon to match changes the nature of the game, or at the very least makes it explicit: it becomes a hunt, which is different from the find-the-lady that is suggested by a three-doors/cards setup. If it's clear that you get to keep guessing, then it pays to spread the guesses around. But what kind of game of chance allows that kind of protocol?
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Post by Fingerplucked on Jun 17, 2009 17:14:48 GMT -5
I don't know. But I've never played a game with a car waiting for me either.
The original game had set rules and one of those was that there are three doors or choices. I only used 10 because it's easier to see a HUGE difference in the outcome of the two strategies. But the principal is the same.
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Post by Supertramp78 on Jun 17, 2009 17:23:27 GMT -5
I work with a guy who plays a lot of bar trivia games. You know the ones where you pick an answer and then the game starts removing wrong choices until only the right choice remains? You can switch your pick as many times as you want but th epoints go down over time. His strategy is that if he has absolutely no idea what the right answer is, and it comes down to what he picked and one other choice - he always switches. "About 60% of the time I win that way". He's probably not that far off.
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Post by Russell Letson on Jun 17, 2009 17:32:28 GMT -5
I'm trying to figure whether the three-door game is just a case of a successive-iteration hunt with N-1 chances. Certainly increasing the number of doors without allowing a corresponding number of chances would make it a different situation. And there aren't any points in the three-door game--just a win-lose proposition after two rounds--there's no revelation of success in Round 1 if you're right, and you don't get half a car if you guess right in Round 2. Again, I'm stuck with thinking in words, but the situations strike me as different in non-trivial ways.
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Post by Russell Letson on Jun 17, 2009 17:48:08 GMT -5
I'm gonna keep picking at this until it bleeds.
Round 1A: I don't know where the car is, so I choose a random door--say, Door 1. There's a 1/3 chance that I'll be right.
Round 1B: Instead of opening Door 1 and declaring me a winner or a loser, Monte (who does know where the car is), reveals a no-car door--call it Door 3. His revelation is not an option but a requirement.
Pause for thought: I now know a) that Door 3 is a loser and b) that Monte, who sees all, didn't show me Door 1. But I still don't know whether the car is behind 1 or 2, which means that there is no reason to change--or to stand pat. There is still no data about the status of 1 and 2 other than that one of them has the car.
Round 2: Monte asks whether I want to change my choice. From my point of view, it could be either 1 or 2 (a 1/2 chance of either being right). I don't see how changing my mind in this single game changes my chances of success.
Would the game be different if I were allowed two choices in Round 1 and given the whole car if either of them proved right? Is this the same as eliminating an empty door and letting me switch? If that's the case, then I understand, kind of.
Maybe my hangup is epistemological (if not metaphysical)--I'm standing in front of doors with no data other than the eventual revelation of a wrong choice that I didn't make.
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Post by Fingerplucked on Jun 17, 2009 18:08:08 GMT -5
Maybe my hangup is epistemological ... I think it probably is, if that word means that if you didn't have your head cluttered with words like that, you'd have room for the simple math required to solve this problem. |
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Post by Russell Letson on Jun 17, 2009 18:32:43 GMT -5
It's not a matter of data storage; it's a matter of processor capability. Remember when you needed an 8087 math coprocessor to run a good CAD system?
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Post by Fingerplucked on Jun 17, 2009 18:45:39 GMT -5
Not me. I picked door number two, then switched to door number three and wound up running women's lingerie. (We did have a few issues with push-up bras, but I convinced them that the manual system was better.)
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Post by billhammond on Jun 17, 2009 20:04:06 GMT -5
Russ -- I don't think men can have episiotomies.
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Post by TDR on Jun 17, 2009 20:28:42 GMT -5
This one is like the missing dollar puzzle in that you have to do more than state the right answer. You have to explain how you got there in a way that makes sense to the next guy.
How did your odds flip from 1/3 to 2/3 by switching your pick? Like Russell, I had a hard time figuring how the second chance at picking a door was anything but unrelated to the first try, or how you had any new information. Only Monte has inside info, right?
But no. Monte is forced by the rules to show you a goat. Not only that but he's forced to let you remove one door from the selection set before he does. That's crucial.
If you only said, "Monte show me a goat" before you did anything else, and he did, your odds would move from 1/3 to 1/2. But with the exclusion of your first pick they go to 2/3.
Here's why. Your first pick wasn't a bet and wins or loses you nothing. But it does force his hand and it does affect the odds if you calculate like this: If you picked goat A, then the car and goat B are left. He has to reveal goat B, he has no other choice. He's showing you the car. If you picked goat B, then the car and goat A are left. He has to reveal goat A, he has no other choice. So he's showing you the car again. One of those things will happen two times out of three.
At this point you could stay with your first guess. In which case you have the same 1/3 odds of getting it right you had going in. But if you swap to the door Monte didn't reveal, out of the selection set you handed him, two out of three times its the car.
Yeah, if you had it right in the first place and you change your pick, you're tweaked. But the odds say flip it.
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Post by Fingerplucked on Jun 17, 2009 20:36:12 GMT -5
I think it's only fair to add that after 9 pages, Goat B is really sick of being second choice and is wondering what makes Goat A so damned special. I tried talking to him about it, but all I could get him to say is "kerrik, kerrik, kerrik."
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