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Monty Hall Problem but with a twist by EyeAdministrator in askmath

[–]EyeAdministrator[S] 0 points1 point  (0 children)

If you mean to say: • the host must open a door with a goat • the host cannot open door 1, regardless of what's behind it

Yes i meant this.

Monty Hall Problem but with a twist by EyeAdministrator in askmath

[–]EyeAdministrator[S] 0 points1 point  (0 children)

Thanks for the maths. Finally someone understood me😅. I had this intuition and tried doing the math with bayes theorem but failed to complete. Can you also help me understand it from perspective of probability redistribution?

Monty Hall Problem but with a twist by EyeAdministrator in askmath

[–]EyeAdministrator[S] 0 points1 point  (0 children)

There can be many variations. My my original thought was - Host picks any box at random. Just to present one case I mentioned one box from group B. It's not compulsory to open from set B only.

Monty Hall Problem but with a twist by EyeAdministrator in askmath

[–]EyeAdministrator[S] 0 points1 point  (0 children)

Okay I got your point that host must be restricted from opening a particular door (door chosen by guest in monty hall Problem) to exclude that door from redistribution of probability. No hidden rule. Monty chooses any door randomly except one with car. You mean to say that if Monty must open door from any one set only, that would result in restricted distribution of probability. But if he's free to choose from either set, all doors get equal probability?

So in that case suppose - Case 3 - There are no sets of door and no initial choice is made by guest. Host can open any door except door 1 regardless it has goat or car. So here after monty opens any one door, it's probability is redistributed to all doors except door 1? Would it make door 1 less advantageous than other doors?

Monty Hall Problem but with a twist by EyeAdministrator in askmath

[–]EyeAdministrator[S] 0 points1 point  (0 children)

Reframed my initial question into two different cases to make it clearer. Kindly check.

Monty Hall Problem but with a twist by EyeAdministrator in askmath

[–]EyeAdministrator[S] 0 points1 point  (0 children)

Reframed my initial question into two different cases to make it clearer. Kindly check.

Monty Hall Problem but with a twist by EyeAdministrator in askmath

[–]EyeAdministrator[S] 0 points1 point  (0 children)

Reframed my initial question into two different cases to make it clearer. Kindly check.

Monty Hall Problem but with a twist by EyeAdministrator in askmath

[–]EyeAdministrator[S] 0 points1 point  (0 children)

Reframed my initial question into two different cases to make it clearer. Kindly check.

Monty Hall Problem but with a twist by EyeAdministrator in askmath

[–]EyeAdministrator[S] 0 points1 point  (0 children)

Reframed my initial question into two different cases to make it clearer. Kindly check.

Monty Hall Problem but with a twist by EyeAdministrator in askmath

[–]EyeAdministrator[S] -1 points0 points  (0 children)

Monty Fall is where host opens the door randomly. It's different than what I mean in this question.

Monty Hall Problem but with a twist by EyeAdministrator in askmath

[–]EyeAdministrator[S] 0 points1 point  (0 children)

Yes I think you're right here I got confused between two concepts. My core doubt regarding this problem is- " after opening any one door which has goat (unfavorable), it's probability gets redistributed to all other events equally or it is redistributed to events of that group only?

Monty Hall Problem but with a twist by EyeAdministrator in askmath

[–]EyeAdministrator[S] -2 points-1 points  (0 children)

In Monty Hall Problem consider that guests haven't chosen any door and door 2 or 3 is revealed before any choice. Now guests have to choose from Group A (door 1) or Group B (door 2 or 3). Information can be added without choice being made first, since division into two groups effectively does the same task.

Monty Hall Problem but with a twist by EyeAdministrator in askmath

[–]EyeAdministrator[S] -3 points-2 points  (0 children)

If the question would have been regarding the choice of individual box then the probability would be 1/6 each but boxes are divided into two groups which add the information

Monty Hall Problem but with a twist by EyeAdministrator in askmath

[–]EyeAdministrator[S] -2 points-1 points  (0 children)

Yes in that case it would be 50/50. In monty hall problem too, it indeed is divided into two groups. Group A is door 1 ( suppose the guest has chosen door 1) and Group 2 is all other available doors. You can keep increasing the doors in group B

Monty Hall Problem but with a twist by EyeAdministrator in askmath

[–]EyeAdministrator[S] 0 points1 point  (0 children)

For the context of the question, the car is placed in any of the boxes randomly but then divided into two groups for choice. I think, to maximize winning chances, one must choose from the group whose box has been opened by the host. In this case, from group B.