Posted on 03/18/2009 4:04:29 AM PDT by mattstat
However, once that first door is opened, my chance of picking the correct door (either remain with mine, or choosing a new one) is now 1 to @-1. This continues until we get down to the last two doors - mine and an unknown. Regardless of how many doors Monty has opened - my chance of choosing the correct door (either the one I have, or the one offered) is 1 in 2.
So, I agree that, before Monty starts opening his doors, my chance of having picked the correct on, is 1-@. I also agree that, once the game is over, my chance of having picked it right, from the beginning is 1-@. However, my chance, on the final decision (stay with Door #1, or take Curtain #3) is 1 in 2.
This is not that different than the old coin toss question ...
Or maybe I'm still missing a key piece of the puzzle?
I'm being honest - if I'm missing something I'd like to understand it!
Please pardon my “rude ping” - but you seem to understand and agree with the original premise.
If so, would you mind explaining it a bit ?
You are hung up on the idea that the odds for a choice already made change upon learning more information. What’s done is done. Suppose you pick an apple from a barrel of apples and that apple has a worm in it. Does that worm go away because you are now given the choice of much more reliable apples?
I think I see it - help me just a bit more here.
If I understand you correctly, I need to be playing the game “in reverse” - i.e. betting that I constrained Monty on my first pick as I have better odds (2/3) of doing so. And, if I did constrain him, my first pick is obviously false, thus I must switch.
So, did I find the Big Deal of the Day?
Not really hung up, just trying to see the correct angle.
Suppose you pick an apple from a barrel of apples and that apple has a worm in it. Does that worm go away because you are now given the choice of much more reliable apples?
Not really what we are talking about. Once I know that the aplle has a worm - it is there. We are discussing an unknown condition.
A better example is the coin toss. My odds of flipping "heads" is 1/2. My odds of doing that 4 times in a row is 1/2*1/2*1/2*1/2 or 1/16. However, once I have flipped the first heads (1/2 chance), the chance that the next one is heads is 1/2, and that of continueing the series to the end is 1/2*1/2*1/2 or 1/8. As you said, what is done is done.
After I have done this 1000 times - the chance that the 1001st one will be heads is 1/2. The chance foresseing that this would happen however is extremely small.
Let me try a different approach. Your chance of getting it wrong originally was 2 out of 3. Monty removes 1 of the 3. If you do nothing it’s like you never saw anything Monty showed you. What is the only possible way to use the information Monty just gave you? You have to switch. Then your odds of losing become 2/3 -1/3 = 1/3 (Hence your odds of winning become 2/3)
When you make your original choice, ie., select one of three doors, you chance of selecting the door with the prize is 1 in 3 (33%). That is mathematically irrefutable.
Now, AFTER Monty opens an empty door, there are only two doors remaining that could be the winning door and so your "choice" is now between TWO doors, not THREE, making your chances of selecting the winning door 50%, rather than only 33%. If you stay with your original selection, then you are staying with a selection you made when your chance of selecting the winner was 1 in 3. But if you switch, then you are making a new selection where your chance of winning is 1 in 2. So simply by virtue of switching you are improving your odds.
Again, it has been supported with hard empyrical data that when you switch you will have a 2 in 3 chance of winning. If you have an hour or so to kill, grab a friend and go ahead and run your own test. (Forget about all that "constraint" stuff.)
Why is your chance of winning 2 out of 3 (66%) when you switch? That's easy... think backwards.
When you made your original selection the chance that the winner was one of the OTHER two doors was 2 out of 3, right? Now, "Monty" has done you a big favor by eliminating one of those two doors! AND THAT ONE REMAINING DOOR - - the one you switched to after Monty eliminated one - - STILL has that same 2 in 3 chance of being the winner, only now it's the ONLY door between those other two doors you originally failed to select that could be the winner because "Monty" helped you out (big time!) by eliminating the definite loser between the two!
(I hope you followed this rambling explanation.)
FRegards,
LH
Thanks to each of you for taking the time to explain.
After figuring it out back at 25 - I did some more thinking about it, i.e. “the odds”.
Lets see if this thinking is correct:
Betting on “I picked the winner” involves two choices
1/3 and 1/2 - or odds of 1/6
Betting on “I picked the loser” involves two choices
2/3 and 1/2 or 2/6 = 1/3
Thus, by playing the game that way, may total chance is 1 in 3 of taking home the prize.
Well, yeah, naturally you're chance is 1 in 3 if you pick one of three doors and keep your fingers crossed. But, of course, once you get the additional information ("Monty" revealing a losing door) and are allowed to switch your choice to the one remaining door (which is what you clearly should do), then your "total chance" is 2 in 3 of taking home the prize.
FRegards,
LH
I think you misunderstood (or - more likely I failed to explain well!!).
Let me try it again ...
When I make my initial pick, I have a 1/3 chance of picking the winner and a 2/3 chance of picking the loser.
Since I get a second pick after Monty kindly eliminates one of the choices, my chance of picking the winner *in this choice* is 1/2.
Now, asumming I stay with my first pick, my chance of picking it right on the first choice were 1/3 and then 1/2 - thus my chance *from the beginning* is 1/6.
OTOH, If I switch (because I assume I chose wrong), my
chance is 2/3 and 1/2 or 2/6 (1/3).
IOW, If I start the game with “I will stick with my first pick” - I have a 1 in 6 chance of taking home the prize.
If I play as “I will switch” I have a 1 in 3 (total) chance.
Whether the “math” is correct, or not - I now understand the principle - either way my chances are far greater by switching.
BTW Nateman - this validates your argument from earlier - my problem was the “angle of attack” on the problem.
By assuming from the beginning that I made the wrong choice - statistically MORE probable than assuming I made the correct one - my probability of being correct one the last choice (switch) is far greater.
Again - thanks to all.
Huh? There's 3 choices and you pick 1. If you stick with that choice that's a 1 in 3 chance no matter what, and that's mathematical reality.
If I play as “I will switch” I have a 1 in 3 (total) chance.
Sorry, but you can see right away that that makes no sense either. Why would your (total) chance remain 1 in 3 (ie., exactly the same as it was at the very start) even after you are given additional information and switch choices?
No. If you play the game as described, including getting additional information AFTER you have made your first choice (ie., a definite loser door is revealed to you) and then you switch your choice to the last door, your total chance of taking home the prize is 2 in 3.
My second choice has “odds” as well, and IIRC, those odds must also be calculated into the overall odds.
Idiotic, but good to generate hits.
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