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

[doonser]

perfect timing, my probability and stochastic processes professor went over the monty hall problem today and i didn't understand it at all

 

[cobrariver]

What an amazing prize

 

[Yoyojor]

D living up to the title of being an evil god

 

[Borderline]

I like how this duck dude never have many scene in the main series. But, he have contributed a ton of funny thing in side series.

 

[angry_noodles]

i was half expecting the punchline to be "that's right, probability doesn't mean anything in the real woooorld" since there's still a chance of the bad happening, even if "smaller" chance

 

[kuroro86]

Thanks for the chapter

 

[01baku]

Nice chapter

 

[Swagner]

Never thought I'd see the monty hall problem in a gag manga

 

[CabbageGamer]

90% sure door A had Alaba behind it

 

[Kojirremer]

90% sure door A had Alaba behind it

@CabbageGamer Maybe it's sexy Ronant with a rose in his mouth.

 

[max_is]

the faces oh god i'm dead

 

[justanothermanganerd]

@Oeconomist people using statistics poorly or dishonestly doesn't mean the monty hall problem isn't true.

When you're shown 3 equal doors and told that behind one of them is a prize it should be safe to assume they each are equally likely as the other. That and the host knowing which door has the prize are literally the only assumptions required.

 

[Oeconomist]

@justanothermanganerd—

people using statistics poorly or dishonestly

… is not the problem to which I referred. Statistics are collected data, and compressions of that data. The underlying question here is of probability in the absence of data.

When you're shown 3 equal doors and told that behind one of them is a prize it should be safe to assume they each are equally likely as the other.

No. Attempts to use the principle of insufficient reason, or more generally to use what are called “non-informative priors” presume a complete preordering conforming to whatever taxonomy one has adopted. People who merely categorize differently will arrive at different probabilities that are not equivalent. That's why I referred to the problem of reference classes and to Bertrand's Paradox.

The rare proper presentations of the Monty Hall Problem overtly assign an initial probability to each door, rather than mistakenly believing that logic says to assign equal probability.

 

[Oeconomist]

While I'm here, I'll point to another problem in the ordinary presentation of the Monty Hall Problem.

Let's say that there is a rival contestant. Each of the contestants selects independently, and in the case that they should choose the same Door, the outcome would be duplicated. But you chose Door A and the other contestant chooses Door C. Monty then opens Door B to show an outcome that neither contestant desired.

Should you switch to Door C? Should the other contestant switch to Door A?

Well, the answers then depend in part on what rule decided Monty's choice. And that is also true in the one-contestant case.

The usual presentation of the Problem doesn't state Monty's rule for choice.

 

[mahtan]

Pekatots are quite exemplary with the variety of skills they have.

 

[niteman555]

I've always found the easiest way to understand it, without using math, is by recognizing you're less likely to get the answer wrong twice. So you're better off switching doors

 

[Oeconomist]

@niteman555—

recognizing you're less likely to get the answer wrong twice

If that argument worked, then you should change doors when you can even if Monty never opened one; or, in guessing a coin-flip, always switch from heads to tails or from tails to heads when given a chance.

 

[justanothermanganerd]

@Oeconomist obviously the monty hall problem breaks down if there are two contestants at once because that's a different thing entirely. I'm not compelled by an argument that says "maybe jimmy just really likes putting the prize behind door #3 so actually door #3 is 60% likely to start"

 

[Oeconomist]

@justanothermanganerd—

obviously the monty hall problem breaks down if there are two contestants at once because that's a different thing entirely.

No, you are reaching for a cheap solution, and getting things wrong.

As I said all depends on Monty's rule for choosing. Monty could be choosing a Door at random, so that there is a chance that he opens a Door chosen by a contestant, even when there is only one contestant. Or he could have a rule of only chosing a Door not chosen by some contestant, but in the case of one contestant Monty chooses one of the remaining Doors at random, such that in some cases he opens the Door with the most desired outcome. Or he could always mess with you, whether there is one contestant or two, such that if the other contestant had chosen one of the bad outcomes, Monty might have opened her Door, or might not. In that last case, Bayes' theorem applies for you in exactly the same way as in the standard problem (properly presented), even when there is another contestant. So just stop confusing your various presumptions with principles of logic.

I'm not compelled by an argument that says "maybe jimmy just really likes putting the prize behind door #3 so actually door #3 is 60% likely to start"

No one is logically compelled by any argument for any point-value, whether it be .6 or 1/3. It is simply dogma that we are licensed to assign point-values to probabilities and therefore that there must be some appropriate method for doing so. I've pointed you to some of the literature on the peculiar problems of the approach that you adopted; but I can't make the horse drink after leading it to water.

 

[justanothermanganerd]

@Oeconomist no. 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. ALWAYS. that's part of it. you just don't understand the monty hall problem. this is why it OBVIOUSLY doesn't work with 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.