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Post by Supertramp78 on Jun 17, 2009 15:57:50 GMT -5
It really is rather simple.
door 1 = car = win
door 2 = notcar 1 = lose
door 3 = notcar 2 = lose
This is your initial starting position. You can swap all the doors around to get more options but you get the same end result.
In that situation the odds of winning is 1 in 3. Even John and Aqua agree to that.
Now let's pick a door and not switch. the scratched out door in the one Monty removes after you make your pick.
pick door 1 = car = win
door 2 = notcar 1 (it could be either noncar. Doesn't matter. You won)
door 3 = notcar 2
door 1 = car
pick door 2 = notcar 1 = lose
door 3 = notcar 2
door 1 = car
door 2 = notcar 1
pick door 3 = notcar 2 = lose
yup, our initial guess was right. Pick and don't switch, you still have that one in three chance.
Now let's switch after a door is removed.
Pick door 1 = car
door 2 = notcar 1
Switch to door 3 = notcar 2 = lose
Switch to door 1 = car = win
Pick door 2 = notcar 1
door 3 = notcar 2
Switch to door 1 = car = win
door 2 = notcar 1
Pick door 3 = notcar 2
Humm. now those odds look different. If I switch, I now win two out of three. the only way to lose is to pick the car initially and the odds of doing that is 1 in 3 while the only way to win is to pick a donkey initially and the odds of doing that is 2 in 3.
As Omaha has been trying to say, switching increases your odds of winning. No argument against that. Well, lots of arguments I guess, just no real argument against it.
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Post by knobtwister on Jun 17, 2009 15:59:11 GMT -5
I bet if we could get Kim Jong iL to start reading this thread our nuclear worries would be over.
Don
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Post by Fingerplucked on Jun 17, 2009 15:59:20 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. There's something wrong with your matrix. If I understood you correctly, you said there's no advantage to switching if you have 10 doors. I made a matrix and tried it with 10 doors. The results are a little hard to read -- I'm not sure how to import the format that I had, but hopefully you can follow (or read) what I posted below. X (cap) = door pic x (little cap)= previous door pick P = Prize O (cap) = Monty door opening o (little cap) = previous Monty door opening n = nothing. This is just a place holder because I don't have Jeff's graphic skills. The first row is your first pick. I picked door#1. The Prize is behind door #8. The second row is Monty opening any non-prize door. A new X shows my new door pick. As long as you don't pick the Prize door on your first turn, and as long as Monty keeps opening doors until all are open except for your current choice and one last chance to switch, he will force you to pick the Prize door. By switching doors, with 10 doors your chances of getting it right (in the end) are 9 in 10. X n n n n n n P n n x n X n O n n n n n x X x O o n n n n n x x x o o O n X n n x x x o o o O x X n x x x o o o o x X O X O x o o o o x x o x o X o o o o x O o O o x o o o o X o o |
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Post by omaha on Jun 17, 2009 16:00:19 GMT -5
BTW, if you want to try this out for free, there is a simulation online. Go there, play 15 games with each strategy. See how you do. |
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Post by billhammond on Jun 17, 2009 16:00:28 GMT -5
Thanks, Russ. You said what I was struggling to put into words before I took delivery of this really smelly and ugly donkey.
On a positive note, for the first time in almost a year, I won't sleep alone tonight. ;D
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Post by Russell Letson on Jun 17, 2009 16:02:17 GMT -5
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. Aha! If it is indeed one process. But what makes the process unitary? Sequential coin tosses aren't part of a single process--they're individual events whose aggregate behavior can be described statistically. In my earlier post I was groping toward this question: what makes this process subject to statistical analysis? Will a large number of iterations finally reveal a 2/3 success rate when the always-switch strategy is used? Otherwise "odds" means something other than what I understand it to mean. |
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Post by millring on Jun 17, 2009 16:04:12 GMT -5
It really is rather simple. door 1 = car = win door 2 = notcar 1 = lose door 3 = notcar 2 = lose This is your initial starting position. You can swap all the doors around to get more options but you get the same end result. In that situation the odds of winning is 1 in 3. Even John and Aqua agree to that. Now let's pick a door and not switch. the scratched out door in the one Monty removes after you make your pick. pick door 1 = car = windoor 2 = notcar 1 (it could be either noncar. Doesn't matter. You won) door 3 = notcar 2 door 1 = car pick door 2 = notcar 1 = losedoor 3 = notcar 2door 1 = car door 2 = notcar 1 pick door 3 = notcar 2 = loseyup, our initial guess was right. Pick and don't switch, you still have that one in three chance. Now let's switch after a door is removed. Pick door 1 = car door 2 = notcar 1 Switch to door 3 = notcar 2 = loseSwitch to door 1 = car = winPick door 2 = notcar 1 door 3 = notcar 2 Switch to door 1 = car = windoor 2 = notcar 1 Pick door 3 = notcar 2 Humm. now those odds look different. If I switch, I now win two out of three. the only way to lose is to pick the car initially and the odds of doing that is 1 in 3 while the only way to win is to pick a donkey initially and the odds of doing that is 2 in 3. As Omaha has been trying to say, switching increases your odds of winning. No argument against that. Well, lots of arguments I guess, just no real argument against it. My mind is starting to melt down here, but aren't two pairs of possibilities you've charted actually just duplications of the same conclusion BECAUSE Monte will always remove a donkey, no matter how the doors are numbered? |
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Post by millring on Jun 17, 2009 16:06:05 GMT -5
Will a large number of iterations finally reveal a 2/3 success rate when the always-switch strategy is used? THAT is the question! |
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Post by TDR on Jun 17, 2009 16:06:56 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. That's obviously not how it works. The theatrics are in the first pick. And the anticipation that you may guess right and lose. Go back to the 100 choices. The first time you have a 1/100 chance to get it right. Doesn't matter if you do or not, because you win nothing and you won't know. Same with 99 left, 98 left, 97, &c. Keep playing till there are three doors. Each time your chance has been one in however many doors. On your next to last guess, the last one that didn't count, its no different than all the others. Still no useful info comes out of the reveal. In the end Monte is going to reduce it for you to a 50/50 shot and that's the one that's gonna count. What went before with DonkeyA and DonkeyB or Donkey brazillion doesn't matter. There was no chance to win with those previous chances and you have no new information that leads you to the right door. And again, whatever a previous hunch may have been is as irrelevant as your hunch on a coin toss. |
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Post by Supertramp78 on Jun 17, 2009 16:08:06 GMT -5
Correct. The initial set up is Monty knows what is behind each door and will always show ou a donkey regardless of where it might be. So the only way to lose is to pick the car first. the only way to win is to pick a donkey first. Since the only other donkey will be removed, whatever you originally picked will be what you don't get. if that makes sense. Since the odds of picking a donkey initially are 2/3, the odds of winning after switching away from a donkey to what will HAVE to be a car is 2/3.
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Post by omaha on Jun 17, 2009 16:08:32 GMT -5
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. Aha! If it is indeed one process. But what makes the process unitary? Sequential coin tosses aren't part of a single process--they're individual events whose aggregate behavior can be described statistically. The only "random" event in the game is the player's initial choice. Technically, I would call it "uninformed" rather than "random", but that's sixes. Once that choice has been made, its like playing tic-tac-toe: The game unfolds deterministically. This is not a case of trying to string together multiple coin tosses, or anything like that. And yes, if you run the simulations, you will get 1/3 winners with the "hold" strategy and 2/3 winners with the switch strategy. |
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Post by Russell Letson on Jun 17, 2009 16:08:43 GMT -5
I took delivery of this really smelly and ugly donkey. You got a donkey? All I got was a damn goat, and he's not only smelly and ugly but too small to serve as transport. |
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Post by omaha on Jun 17, 2009 16:08:56 GMT -5
Will a large number of iterations finally reveal a 2/3 success rate when the always-switch strategy is used? THAT is the question! And the answer is yes!!! |
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Post by omaha on Jun 17, 2009 16:10:01 GMT -5
whatever you originally picked will be what you don't get Best summation yet. |
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Post by Fingerplucked on Jun 17, 2009 16:12:53 GMT -5
Thanks, Russ. You said what I was struggling to put into words before I took delivery of this really smelly and ugly donkey. On a positive note, for the first time in almost a year, I won't sleep alone tonight. ;D Oh wait, I was getting all excited and happy for you and thought that everyone was so wrapped up in being wrong (except for a few of us).... But you were talking about the donkey, weren't you? |
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Post by billhammond on Jun 17, 2009 16:14:05 GMT -5
Meanwhile, I have spoken with the game sponsors and verified that this IS the car, so what in thee hell are we doing wasting all this time and energy and bandwidth??!?!  |
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Post by millring on Jun 17, 2009 16:15:09 GMT -5
In that situation the odds of winning is 1 in 3. Even John and Aqua agree to that. Guess what I would conclude is: You start with a 1/3 choice to pick the car right off You don't know it at the time, but your odds will change. If you want to consider that 50/50 is better that 1/3, then you will have a better chance in the next round -- but not because you switch. You will not EVER have a 2/3 odds conferred upon your next choice (as the experts say). You will, if you choose, get to make another choice with 50/50 odds this time. Thus, in response to Russell's question, I would guess that you will not end up with 2/3 cars realized in a switch-every-time scenario played all the way out to statistical meaningfulness. |
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Post by billhammond on Jun 17, 2009 16:17:25 GMT -5
Boils down to semantics vs. statistics.
Or, put another way, it's between We Of the Beautiful Minds and All You Tight-Assed Number Nazis.
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Post by Supertramp78 on Jun 17, 2009 16:18:16 GMT -5
nope. There are three initial options, not two. One option for each door. Pick it and keep it or pick it and swap it. Pick it and keep it will result in a win 1 out of 3 tries. Pick it and swap it will result in a win 2 out of 3 tries. The ODDS of your first choice being correct is 1/3. The ODDS of your SECOND choice being correct improve to 2/3.
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Post by Russell Letson on Jun 17, 2009 16:19:35 GMT -5
It occurs to me that this is one of those situations that is not easily explained in words--partly because the language that is optimized for dealing with is mathematical. My understanding of mathematical concepts is about 98% visual: if it can be graphed, I'll get it. If it's a matter of computation or symbolic logic, I won't. This is why I crapped out at the analytical algebra level and never could get a handle on trig or calculus. (And maybe why structured programming languages like Pascal and C never stuck, while assember and machine code made [very slow] sense.)
Two people could run this problem, sans goats, cars, and sexy letter-turners (oops, wrong show) with two people, three Scrabble tiles, the tile tray, and a pad and pencil to keep score. I'd be interested in seeing how many iterations it would take to establish the pattern. Then I'd like to have a long think about why it works.
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