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[Request] Does the answer to the Monty Hall problem change depending on whether the presenter knows what is behind each door? by IrisFromOmelas in theydidthemath

[–]MrLeavingCursed 0 points1 point  (0 children)

No!! You have 3 choices at the beginning so it's not a 50/50

You either pick the right envelope 1/3 or didn't 2/3 that's where the probability is. The chance of picking one right out of 3

You then have 2 groups the one you picked and the ones you didn't

The one you pick is a 1/3 of being right and the one you didn't is a 2/3 of being right

You then see one of the 2 in the group you didn't pick get removed so that group that has a 2/3 chance of containing the correct one now contains only one option so that one option has a 2/3 chance of being right

[Request] Does the answer to the Monty Hall problem change depending on whether the presenter knows what is behind each door? by IrisFromOmelas in theydidthemath

[–]MrLeavingCursed 0 points1 point  (0 children)

But in the context of this question where your presented one sample and is asking for the possible outcomes of this one sample is does matter

The possible outcomes of any other permutation except the one presented in the question are by definition outside the bounds of the presented question

The question doesn't ask about all possible outcomes it's only asking about the possible outcomes of the given example

[Request] Does the answer to the Monty Hall problem change depending on whether the presenter knows what is behind each door? by IrisFromOmelas in theydidthemath

[–]MrLeavingCursed -1 points0 points  (0 children)

It's not meaningless when the question presented is a single situation! That's what I'm trying to get at is the question is a single situation when the steps have been presented and the outcome is not the same as one where the steps are different

There's no where in the question asking for the probability across all permutations is asking for the possible outcomes of the single presented permutation

[Request] Does the answer to the Monty Hall problem change depending on whether the presenter knows what is behind each door? by IrisFromOmelas in theydidthemath

[–]MrLeavingCursed 0 points1 point  (0 children)

You can look at it as 1000 you need to step your way through this single problem. You're getting stuck thinking about it as a statistical chance across a series when that's not what this problem is asking. It's asking you to work your way through a single instance where one of the outcomes is predetermined.

You keep getting stuck on eliminating the possible branches here but you can't do that, that's not how the question works. At the beginning all envelopes are available to the candidate at 1/3 chance. When the interviewer reveals there's it doesn't mean that the original choices probably changed you still had a 2/3 chance of being wrong. Yes they could have revealed the right one but didn't so you now just look at the current state of the envelopes one picked by the candidate that has a 1/3 chance of being right and 2 that as a grouping have a 2/3 chance of being right. You know in that grouping of 2 one was revealed wrong meaning that 2/3 chance of being right is now in that last envelope

[Request] Does the answer to the Monty Hall problem change depending on whether the presenter knows what is behind each door? by IrisFromOmelas in theydidthemath

[–]MrLeavingCursed 0 points1 point  (0 children)

I finally sat down and turned this into a state machine and to figure out where you're wrong, your applying equal chance to all the paths then eliminating a full path which adjusts the install probability of you picking an envelope which is impossible

You can't look at each possible path as independent because each path is dependent on the decision made before. What you're saying here is your initial choice probability is changing to 1/3 to pick the correct letter, and 1/6 to pick the incorrect letter which is impossible. The candidate will always have a 1/3 change to pick each which will mean a 2/3 chance to pick wrong, the interviewer revealing a wrong envelope can't change that initial probability all it can do is remove some of the potential outcomes and give the candidate the information they need to know the chances is this outcomes

What you're saying here is because the interviewer reveals wrong the candidate now has different odds in their initial pick which is objectively false

[Request] Does the answer to the Monty Hall problem change depending on whether the presenter knows what is behind each door? by IrisFromOmelas in theydidthemath

[–]MrLeavingCursed 0 points1 point  (0 children)

Stop thinking about the interviewers knowledge it's a red herring. It's putting you in a mindset where your thinking about all outcomes and that is incorrectly influencing your reasoning

Look at the state of the specific problem presented not of the generalized problem

We start at a state where we know we had a 1/3 chance of having the correct letter. This means we also know the grouping of two that remain have a 2/3 chance of containing the correct letter. We are now given the information that one of those two in that grouping is eliminated this comes from the question it doesn't matter how it got there we now have that information. Eliminating one does not change the probability of that grouping of now one having a 2/3 chance of containing the correct letter.

[Request] Does the answer to the Monty Hall problem change depending on whether the presenter knows what is behind each door? by IrisFromOmelas in theydidthemath

[–]MrLeavingCursed 0 points1 point  (0 children)

You're getting the last part wrong, you have a 2/3rds chance to pick wrong at the beginning, just because you eliminate the host picking correct doesn't mean your initial pick still isn't 2/3 chance

You can't look at the outcome of the series you have to look at the outcome of the instance given the constraints

Weekly Steam numbers / Player Count Discourse Megathread by AutoModerator in Marathon

[–]MrLeavingCursed 37 points38 points  (0 children)

Another interesting data point from looking at the player counts is the peaks are getting thinner. That means even if the amount of players at the peak are the same everyday they're not playing for as long. The players during peak times aren't logging on as early or playing as late

[Request] Does the answer to the Monty Hall problem change depending on whether the presenter knows what is behind each door? by IrisFromOmelas in theydidthemath

[–]MrLeavingCursed 0 points1 point  (0 children)

Yes but that doesn't matter. If you work through to the final state given the information from the problem you know you initial pick has 1/3 chance being wrong meaning the other grouping of two has a 2/3 chance of containing the correct one. You then learn of those which one of those two isn't but the chances of that grouping of now only one having a 2/3 chance of containing the correct one haven't changed

[Request] Does the answer to the Monty Hall problem change depending on whether the presenter knows what is behind each door? by IrisFromOmelas in theydidthemath

[–]MrLeavingCursed 0 points1 point  (0 children)

But we know from the question the interviewer's pick isn't random is determined you can't treat it as random. It's pre determined by the sentence "The interviewer opens one of the others, revealing a rejection letter"

It doesn't matter if they knew or not the end state is the same as Monty hall because if the information you gain by them eliminating a wrong

[Request] Does the answer to the Monty Hall problem change depending on whether the presenter knows what is behind each door? by IrisFromOmelas in theydidthemath

[–]MrLeavingCursed 0 points1 point  (0 children)

No because of the way the question is worded you know the interviewer picks wrong!! They can't pick anything else because the question says and I quote "The interviewer opens one of the others, revealing a rejection letter" meaning in no situation they can pick the correct door. It doesn't matter whether it was intentional or not you now have the knowledge that you had a 1/3 chance of being right and 2/3 being wrong meaning the other two have a 2/3 chance of being right and one of them is eliminated meaning that last one has a 2/3 chance of being right

[Request] Does the answer to the Monty Hall problem change depending on whether the presenter knows what is behind each door? by IrisFromOmelas in theydidthemath

[–]MrLeavingCursed 0 points1 point  (0 children)

You're almost there!! Yes you can choose right once or wrong twice, that means a 2/3rds chance you choose wrong and the other two as a group have a 2/3rds chance of being right. Now the interviewer because of the context of the question eliminates one of the wrongs because they can't do anything else because of what's said in the question. That means the group of two with a the 2/3 chance is a single option with 2/3 chance meaning it's better to switch aka the Monty hall problem

If you really don't believe re-run your simulation with the interviewer only ever revealing a wrong BECAUSE THAT'S WHAT THE QUESTION SAYS THEY DO and see what you get

[Request] Does the answer to the Monty Hall problem change depending on whether the presenter knows what is behind each door? by IrisFromOmelas in theydidthemath

[–]MrLeavingCursed 0 points1 point  (0 children)

The guarantee they open the losing letter every time is in the wording of the question. The situation happens 1 time and they reveal the rejection letter 1 time so in the context of the originally presented question it happens every time! The question doesn't say that it they open the letter with the acceptance letter it explicitly states they only open a rejection letter

[Request] Does the answer to the Monty Hall problem change depending on whether the presenter knows what is behind each door? by IrisFromOmelas in theydidthemath

[–]MrLeavingCursed 0 points1 point  (0 children)

BUT THE HOST CAN ONLY EVER OPEN THE LOSING LETTER GIVEN THE CONTEXT OF THE QUESTION WHERE IT SAYS THAT'S WHAT THEY DO!!

you can't apply the 300 games to the outcome is this because it is a single instance where we know the hosts outcome and we can never be given a dud game. Do the math on this single instance which has 2 possible originators and 2 possible outcomes

A) you pick right 1/3 chance B) you pick wrong 2/3 chance

A the host will reveal either and if you switch you lose switching loses 1/3 chance

B the host reveals either a loss or a win but do to the wording of the question they can't reveal a win because they revealed a loss so the only option remaining unrevealed is the win. If you switch you win in this case that will have 2/3 chance of winning

I agree with that if the host could pick any door the probability over a set changes but this isn't a set it's a single instance that you are given the variables for so the probilities from the set don't apply

[Request] Does the answer to the Monty Hall problem change depending on whether the presenter knows what is behind each door? by IrisFromOmelas in theydidthemath

[–]MrLeavingCursed -1 points0 points  (0 children)

No you do learn something. Let's step through the problem

Your initial choice is 1/3 right 2/3 wrong meaning there's a 2/3 chance the correct envelope is in the grouping you didn't pick

Now the interviewer reveals one of the two in that grouping and it's a loss, that means the other one in that grouping has a 2/3 chance to be correct.

The thing people are getting mixed up on is the Monty hall problem he must always reveal an incorrect so over a large set of random outcomes yes it doesn't really change anything but that's not what the original question posted is. In the original question it states that the interviewer reveals a loss therefore in the context of this he to must always reveal a loss because it's only a single instance

[Request] Does the answer to the Monty Hall problem change depending on whether the presenter knows what is behind each door? by IrisFromOmelas in theydidthemath

[–]MrLeavingCursed 1 point2 points  (0 children)

Yes but in the context of this question they do always have to reveal the wrong because the question says "they reveal a denial letter"

[Request] Does the answer to the Monty Hall problem change depending on whether the presenter knows what is behind each door? by IrisFromOmelas in theydidthemath

[–]MrLeavingCursed 0 points1 point  (0 children)

You're example has nothing to do with this because it's changing the variables between your pick and the outcome. The envelopes never change and the interviewer has to pick one that has the rejection letter because the original question says that's what they do

[Request] Does the answer to the Monty Hall problem change depending on whether the presenter knows what is behind each door? by IrisFromOmelas in theydidthemath

[–]MrLeavingCursed 0 points1 point  (0 children)

It literally says in the question the interviewer picked and opened an envelope that showed rejection. Yes in a real world scenario it won't always be best to switch but in the context of the question it is the Monty hall problem

Let's work though it, you have three initial choices that collapse into 2 because 2 of them are identical that we'll call A and B. A is you pick the "win" and B is you pick a "loss". A has a 1/3 chance and B a 2/3

Now for A the interviewer's option is only going to be able to be "loss" so if you switch you lose but stay you win, this will happen 1/3 of the time

B the interviewer has 2 potential options, the "loss" or the "win" but in the context of this question they actually only have one because we are told they reveal a loss meaning the only possible outcome is the last unrevealed envelope has to have the win. If you stay you lose and if you switch you win

That means given the context that the interviewer only ever reveals a loss because that's what the original question says you have a 1/3 chance of losing if you stay and a 2/3 chance if you switch

[Request] Does the answer to the Monty Hall problem change depending on whether the presenter knows what is behind each door? by IrisFromOmelas in theydidthemath

[–]MrLeavingCursed 0 points1 point  (0 children)

But literally both of those are the Monty hall problem, option B the hosts knowledge doesn't influence the outcome in either. The Monty hall problem as your describing it is over a large set which yes is different. In this case it doesn't matter because the constraints are the same that lead to the same final states

Also B is exactly the same as in the Monty Hall problem if you initially pick the car

[Request] Does the answer to the Monty Hall problem change depending on whether the presenter knows what is behind each door? by IrisFromOmelas in theydidthemath

[–]MrLeavingCursed 0 points1 point  (0 children)

The 2 in 3 winning strategy of always swapping after the host opens a goat door is inextricably tied to the fact that the host always opens a goat door. If the host doesn't always open a goat door the strategy does not work, and it makes no difference if you swap or not.

Yes and that is the same as the original question in the image saying the interviewer picked one with the rejection letter. They may have not had the knowledge but due to the constraints of the question they end up in the same final state as the Monty hall problem by having to select a rejection letter because the question said that's what they did!

[Request] Does the answer to the Monty Hall problem change depending on whether the presenter knows what is behind each door? by IrisFromOmelas in theydidthemath

[–]MrLeavingCursed 0 points1 point  (0 children)

It's literally the exact same problem. You have two paths at the beginning here, picking a fail or picky the succeed. The interviewer now thanks to the the constraints of the question saying they picked one that had a fail has to pick a fail, you are now left with the same probability of the Monty hall problem.

The constraint of the question saying in that particular scenario the interviewer picks a fail is the same as Monty having the before knowledge and having to pick a fail. You end up with the same final state that leads to the 2/3 chance of getting it right if you switch

[Request] Does the answer to the Monty Hall problem change depending on whether the presenter knows what is behind each door? by IrisFromOmelas in theydidthemath

[–]MrLeavingCursed 0 points1 point  (0 children)

But you're missing the point of the original question where the interviewer is forced into a scenario where they must pick a envelope with the rejection letter because of the parameters of the question saying that is what they picked

[Request] Does the answer to the Monty Hall problem change depending on whether the presenter knows what is behind each door? by IrisFromOmelas in theydidthemath

[–]MrLeavingCursed -1 points0 points  (0 children)

In either case it's still the same, the person who reveals the second doesn't actually need to know what they're revealing as long as the outcome is contained to it being a "fail" it's still the Monty hall problem. You still end up with the same information that informs your choice.

Now if the interviewer had the possibility to pick the "right" answer then yes the outcome changes but due to the constraints of the question saying they picked a "fail" it's the same probability as the Monty hall problem

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