|
|
CTK Exchange
Zach Wegner
guest
|
Oct-05-02, 02:20 PM (EST) |
|
|
"Monty Hall problem... again"
| |
There are two solutions that are suggested. Each solution depends upon the representation of data.
For 1/2-From contestants POV, without 'perfect information'-where the tree doesnt rely upon knowing the prize. All outcomes equally likely.
For 2/3-From MH POV, with perfect information. Different outcomes have different chance.
Thats why ALL the simulations show 2/3. No one thought of a simulation without perfect information. I made one, and it shows 1/2. It does this by choosing a picked door, then choosing a door to reveal. The prize is then picked, and it can have 2 places because it cant be in the revealed door. This goes along with if you have a 100 door problem, or say an n door problem. if you take away all but two, the probability of winning by switching is 99/100 or n-1/n. If you take away one, then it is 1/99, or 1/n-1. If you have 3 doors, taking away one and taking away all but two yields the same result, so the amount of doors opened doesnt matter. Which would the final probability be? n-1/n or 1/n-1? Of course, with this problem,
n-1/n = 2/3
1/n-1 = 1/2
If you say perfect information, then the number of doors able to open is either 1 or 2 (or n-1 and n-2, because you know if the prize was picked or not), and it gives 2/3.
If you say no perfect information, then the number of doors able to open is 2 (or n-1 because you dont know if the prize is unpicked or not), and it is 1/2.Since the problem states that it is from the contestants POV, it is 1/2. I would like to see a counterexample. The perfect information seems to be in the mindset of all who say 2/3, and that is the flaw on their part. Zach
|
|
|
Alert | IP |
Printer-friendly page |
Reply |
Reply With Quote | Top |
|
|
You may be curious to have a look at the old CTK Exchange archive.
Please do not post there.
Copyright © 1996-2011 Alexander Bogomolny
|
|
Printer-friendly copy
Monty Hall problem... again
