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Mar
28
awarded  Editor
Mar
28
revised Why do we limit the definition of a function?
added 8 characters in body
Mar
28
answered Why do we limit the definition of a function?
Dec
18
comment Monty hall problem extended.
I'm really sorry I'm confused you even more with my (now-removed) comment. I was wrong. It would be still "flat" if Monty had performed the elimination at random. Let's say he could eliminated the winning one. Then you'd be still left with 1/3 that what you picked is good, but now the other one left has 1/3 to be good or 1/3 to be bad. Now switching gives truly nothing: 1/3 vs 1/3. The odds of winning are SAME, so there's no point in switching. But notice they are NOT 50-50. They are 33-33 and the 33 left is that NONE of them is good. So even the 50-50 idea is a huge misconception.
Dec
18
comment Monty hall problem extended.
The most important part is, that at the SECOND pick, you know something so the distro is not flat. You KNOW that earlier you picked A with 2/3 chances that it is bad. So, you cannot evaluate the A-or-B and A-or-C as 50/50. You have previous knowledge of events. It's not purely random anymore.
Dec
18
comment Monty hall problem extended.
So, you have 1/3 chance that A is good, and 2/3 chances that A is bad. Nothing changed since the original pick! But now you have only 2 options: A/B or A/C, depending on what Monty has eliminated, and in each of those cases you KNOW that 2/3 is that A is wrong..
Dec
18
comment Monty hall problem extended.
A is good: 1/3*1/2 + 1/3*1/2 == 1/3. A is wrong = 1/3 + 1/3 == 2/3.
Dec
18
comment Monty hall problem extended.
Now, summarize the outcomes and see, what's the P that A is good.
Dec
18
comment Monty hall problem extended.
At (1/3)*(1/2) your A is good and he dropped B. At (1/3)*(1/2) your A is good and he dropped C. At (1/3) your A is wrong AND B is good and he dropped C. At (1/3) your A is wrong AND C is good and he dropped B.
Dec
18
comment Monty hall problem extended.
He NEVER opens the winning window is very important. He knows something we don't. He never opens the door you picked, and he never removes the winning one. 3 doors, one prize. At first you pick door A. BC are left. Now, if A was winning (1/3), Monty will drop one of B or C at random so both have (1/2) of being dropped. If B was winning (1/3), Monty will always DROP C(1/1). If C was winning (1/3), Monty will always DROP B(1/1). Now, write down the final probs for each case.
Dec
18
comment Monty hall problem extended.
At the final point of the game, the probability that the prize is at option A and probability that the prize is at option B is NOT 50%/50%. It would be 50-50 if we had NO KNOWLEDGE of the past events. At the start of the game, the window-prize setup was assumed to be perfectly-flat, because we knew nothing. Hence, for 3 options, 1/3 1/3 and 1/3. The issue is that Monty not eliminate options at random, and this causes the distribution to cease to be flat.
Dec
18
comment Monty hall problem extended.
I've just reevaluated myself all the facts about the problem, and now I'd like to cancel what I just said. Everytime I get to this problem I fall in the same pitfall: when considering the probabilities of the second pick, I always forget to apply conditional corrections that come with new knowledge obtained from opening X windows. Then, the core of the problem looks more complicated, and again related to 'common sense' and 'natural language expressiveness'. Sopuli: double, or triple-check what probability you are talking/thinking about. There's where your error lies!
May
8
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