Doug McCrae
Legend
It's a tricky one. Highly counter-intuitive. I also came to understand it by considering the hundred doors example.
Explain to me this probability puzzle
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Thanee
First Post
Yeah, it's extremely misleading.
In a way, your way of thinking is actually right, just applied wrong.
The probabilities do not change, when there are only two doors left. You only look at it from the back (only two doors, each with a 50% chance), while you need to look at it from the front (as explained in my post above).
The "problem" is, that even with only 2 doors, it is still a 3 door problem.
Bye
Thanee
In a way, your way of thinking is actually right, just applied wrong.
The probabilities do not change, when there are only two doors left. You only look at it from the back (only two doors, each with a 50% chance), while you need to look at it from the front (as explained in my post above).
The "problem" is, that even with only 2 doors, it is still a 3 door problem.
Bye
Thanee
Enzo Sarlas
First Post
Ok, it took me a LOOOOOOOOONG time to get it, but I got it. It's so counterintuitive at first but it makes complete sense once you see it -- kinda like that old lady / young girl picture.
The problem for me is that I don't trust code, because code can be written on bad logic. What FINALLY convinced me was this...
The doors can only be arranged in 3 ways, shown as cases 1, 2 and 3. There are no other permutations.
Case____A_______B_______C
1_______bad_____*bad*___
2_______bad_____ _____*bad*
3_______ _____*bad*___bad
Let's assume you choose door A to start.
We all agree that originally you have a 1/3 chance of a good prize.
Now... Monty knows what is behind each door, and shows a bad one.
In case 1 he eliminates B. He can't eliminate A or C.
In case 2, he eliminates C.
In case 3, he eliminates B or C.
Case 1: A, A = loser
Case 2: A, A = loser
Case 3: A, A = winner
In only 1 out 3 cases you will win by sticking with A.
Look at it as a 'switcher'
Case 1: A, C = winner
Case 2: A, B = winner
Case 3: A, (B or C) = loser
In 2 out of 3 cases you will win by switching.
Here's hoping that any doubters see this and get it just like I did.
Enzo
The problem for me is that I don't trust code, because code can be written on bad logic. What FINALLY convinced me was this...
The doors can only be arranged in 3 ways, shown as cases 1, 2 and 3. There are no other permutations.
Case____A_______B_______C
1_______bad_____*bad*___
2_______bad_____
_____*bad*3_______
_____*bad*___badLet's assume you choose door A to start.
We all agree that originally you have a 1/3 chance of a good prize.
Now... Monty knows what is behind each door, and shows a bad one.
In case 1 he eliminates B. He can't eliminate A or C.
In case 2, he eliminates C.
In case 3, he eliminates B or C.
Case 1: A, A = loser
Case 2: A, A = loser
Case 3: A, A = winner
In only 1 out 3 cases you will win by sticking with A.
Look at it as a 'switcher'
Case 1: A, C = winner
Case 2: A, B = winner
Case 3: A, (B or C) = loser
In 2 out of 3 cases you will win by switching.
Here's hoping that any doubters see this and get it just like I did.
Enzo
Trainz
Explorer
Ah... the famous Monty Hall Let's make a deal ! "paradox"... it always pops up at one time or another on many message boards. And every time, there is someone who can't accept it and fights it. That is not a sign of stupidity however: if all else it demonstrates an individual who as a functioning mind.
It is just the human tendency to prefer to being right than to be correct, as This message board so eloquently puts it.
Many resources on the "weeeb" can be found about the subject, here's one, and here's a little Java applet that runs the simulation (the results are more convincing if you put a large number of tests, like 1000: the number of successes should be very close to 666 if you switch, 333 if you stay).
It is just the human tendency to prefer to being right than to be correct, as This message board so eloquently puts it.
Many resources on the "weeeb" can be found about the subject, here's one, and here's a little Java applet that runs the simulation (the results are more convincing if you put a large number of tests, like 1000: the number of successes should be very close to 666 if you switch, 333 if you stay).
Last edited:
Thanee
First Post
Enzo Sarlas said:Case____A_______B_______C
1_______bad_____*bad*___
2_______bad_____
3_______
Yeah, and it's important to note, that case 3 is actually only one case, not two (since permutations of the two bad doors are completely irrelevant).
Bye
Thanee
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