Kumo Desu ga, Nani ka? Daily Life of the Four Spider Sisters - Ch. 67

[Oeconomist]

@justanothermanganerd—

part of the monty hall problem is that the host ALWAYS reveals a door not chosen by the contestant and without the prize behind it.

What is always given is a sequence of events; what is often not given is why Monty selected that door; but the rule that he uses determines how Bayes' theorem can then be applied. It determines how it applies in the one-person game, and whether Bayes' theorem is useful to one of the contestants in a two-person game.

this is why it OBVIOUSLY doesn't work with two contestants.

You can't transform your mistaken presumptions into truth by shouting that they are obvious. I've already noted how, if in the two-contestant case he uses a rule that also explains his choice in the one-contestant case, then the reasoning does work for one of the two contestants.

crazy how you're apparently reading advanced stats papers and shit but don't even know the monty hall problem which is high school maths.

It's sad that people often think that they learned math (and other things) in high school but really didn't understand it then nor come to understand it later. (Often, the teachers and journalists who supposedly explained it were incompetent.)

The reason that so many mathematicians supposedly got the Monty Hall Problem wrong was that it was mis-stated (specifically in the column of Marilyn vos Savant which appeared in Parade). Initial probabilities and the rule used by Monty were omitted. But people who blithely use the principle of insufficient reason didn't recognize the importance of the first omission, and a large group of people guessed at Monty's rule without realizing that they were guessing.

 

[SeryjVolk]

Pretty sure it goes from 1/3 to 1/2 if you switch, not 2/3 since the door that's opened for free is always 100% wrong.

 

[Oeconomist]

@SeryjVolk—

If the Doors started with equal probabilities, and if D were known to use the rule that most people just assume, then there would be implicit information that wouldn't be the same as in just coming upon two closed Doors and being told that one would open to a good result and the other to a bad result, even if the principle of insufficient reason ever worked. Probabilities would have to account for that information.

And, as I explained in other comments, the principle of insufficient reason doesn't work. There is no measure of probability without information, so you cannot in any case get to 50-50 the way that you think.

 

[Oeconomist]

While I'm here again, let me suggest that D might have used a very different rule from the one that Monty is sometimes said and often assumed to use. Specifically, D never declared that she were presenting the Monty Hall Problem. So, before the final choice, for all that Kumoko knew, D's rule might have been: [ul]If Kumoko chooses a bad Door, then just let her open it; if Kumoko chooses the good Door, open one of the others, to make it seem like the Monty Hall Problem.[/ul] The result would have been to trick Kumoko into abandoning the good Door for a bad Door.

 

[SeryjVolk]

@Oeconomist
If we assume D was set on cheating, any argument on the subject of statistics becomes moot. D could simply have given Kumoko any result she wished regardless of the door Kumoko chose.

 

[Oeconomist]

@SeryjVolk—

Yes, but cheating is violation of the rules, not violation of what your opponent mistakes for the rules, nor otherwise mere violation of the expectations of your opponent. (And I think that D would not cheat for a win when she could win without cheating.)

 

[Sanction]

I had already read about this problem except that in the end it is totally wrong when there are only two choices with a winning choice and a losing choice.
No matter how you look at it, it's 50/50 in the end, no matter which choice you made first. You have a bad choice and a good choice.
The best example is to ask someone from outside who has not made your choices, you ask them to choose for you, they have a 50% chance of getting it right, why not you? Because the theory is wrong.

The theory implies that you make a choice that follows from reasoning when it is impossible, you have chosen an answer, you have not deduced anything before choosing it.

 

[panda34]

@Sanction

The explanation in the chapter is wrong but the Monty Hall problem is valid.

You must choose 1 of 3 doors. What are your options?

1/3 of the time, you will choose Failure 1.
1/3 of the time, you will choose Failure 2.
1/3 of the time, you will choose Success.

So far, this should be pretty intuitive.

Now, the host will open a door.

If you have chosen Failure 1, (1/3 chance) then the host will show you Failure 2. If you switch here, you are switching from Failure 1 to Success. You should switch.

If you have chosen Failure 2, (1/3 chance) then the host will show you Failure 1. If you switch here, you are switching from Failure 2 to Success. You should switch.

If you have chosen Success, (1/3 chance) then the host will show you either Failure 1 or Failure 2. Either way, you should not switch.

Or course, we can't know whether or not our initial choice was Failure 1, Failure 2, or Success. But assuming that our initial choice was equally likely to choose any of the 3, in 2 out of 3 cases switching means that we change from Failure to Success, so you should switch.

 

[Sanction]

@panda34

The problem is that it is invalid when you have only two choices.
If we make a tree of possibilities we have 2/3 for one and 1/3 for the other, proof that there is a proof because in fact if we take the case where you ask someone to make the choice for you, this person has 1/2 of finding the right answer.

The huge problem with Monty Hall is that it assumes that you have made a conscious choice when in fact you have not, you have just randomly chosen an answer. The possibility tree is true as long as there are more than two solutions, as soon as there are only two it is no longer valid.

Here it is totally intuitive not to take into account a possibility tree.

 

[Lynkaster]

@Sanction the thing is, in this problem, the host will show you one wrong door between the 2 you didn't choose. There's a chance he had to select the wrong door rather than choose between 2 wrong doors at random. The chance may not be 2/3 but there's theoretically more chance to win if you change your choice since the host made a decision after your first choice.

 

[Monochrome_Scholar]

While I've heard the logic behind it, I've never really got the Monty Hall problem. To me it will always just be an equal 50/50, because that's how probabilty works. If you have a choice of two unbiased options with no way to inform your choice, it will always be an equal chance of sucess or failure regardless of the circumstances surrounding it. Anything else is simply an illusion to distact or sway your descision, isn't it?

 

[perryvitcon]

@Monochrome_Scholar In the beginning, you're more likely to have picked the wrong door (2/3) so switching choice is better.

 

[Sanction]

@Lynkaster

I know the problem and if we follow a tree of possibilities we have one choice at 2/3 and the other at 1/3 at the end.

You will have to prove this to me:

but there's theoretically more chance to win if you change your choice since the host made a decision after your first choice.

If my choice is random and at the end I am given two choices, one of which is a winner, if in these two choices there is always the choice that I chose first, it is 50/50. This is the basis of probability, the basic problem with Monty Hall is that it assumes that your first choice is not random.

I'll say it again, but there's nothing to stop me asking someone to choose for me, they'll get 50/50 because they'll be excluded from the initial process.

 

[Monochrome_Scholar]

@perryvitcon
But how is switching your choice better? My reasoning is if both remaining doors have the same chance of winning, then it doesn't matter which remaining door you pick, you chances remain the same. In the grand scheme of things, the third door that gets eliminated might as well have not existed. It makes no difference to the end result. The choice was never 1/3, from the very beginning it was only ever going to be 50/50. A choice does not suddenly become better or worse just because the choice beside it goes away, it was only ever affected by itself.

 

[a6503]

It's a lot more intuitive with more than 3 doors. Given 100 doors and 1 prize, let's say you pick one, then the host opens 98 other doors that do not have the prize. Now you are left with the one you picked, and the remaining door. Do you switch?