Talk:Monty Hall problem/Arguments

Keep It Simple

If the contestant picks the car and switches, the contestant loses. The contestant has 1/3 chance of picking the car.

If the contestant picks the goat and switches, the contestant wins. The contestant has 2/3 chance of picking the goat.

The contestant doubles the chance of winning by switching.

JAKQ0s (talk) 16:59, 15 November 2018 (UTC)

That's about how I think about it. If you know the host has some algorithm (e.g. always pick the right-most losing door), pick Door 1. If the host picks Door 2, the car is behind Door 3; otherwise the car may be behind Door 2 or Door 1. You have a 2/3 chance of winning at all times. If the selection is random, then you pick a door and the host opens all other doors except one, and never opens the winning door, in which case you have ${\frac {1}{n}}$ chance of having picked the correct door and a ${\frac {n-1}{n}}$ chance of having picked the wrong door, so with more than two doors switching has a large probability of winning. John Moser (talk) 15:38, 17 October 2019 (UTC)

93.106.123.184 (talk) 14:05, 27 May 2020 (UTC) The following question could use a simple answer: Why does sticking with the door you chose originally not count as a new choice in a new situation?

93.106.123.184 (talk) 14:05, 27 May 2020 (UTC)

Law of large numbers proves this (not disproves as originally written)

The below information is accurate so leaving it for prosperity whereas my code was not. However, as a cautionary tale: The code I had originally written was tossing out every attempt where the host picked either the winning door or the contestant's instead of just picking the other door. This is more likely when the contestant picked a nonwinning door. This incorrectly shifted the expected results of the contestant winning to 50%. Now to the original argument.

We have 3 doors.

1 with a car behind it and 2 with goats - Assumed randomly assigned

Contestant randomly selects door - Also assumed random

We have to assume that the host cannot pick the door with the car AND cannot pick the contestant's door but other than that they randomly pick an available goat door (ultimately meaningless as it'll always come down to prize door and a goat door)

So at this stage we have 4 possible states...

1. Contestant picked the right door. Host picks goat door A. Contestant chose correctly

2. Contestant picked the right door. Host picks goat door B. Contestant chose correctly.

3. Contestant picked goat door A. Host must pick goat door B. Contestant chose incorrectly.

4. Contestant picked goat door B. Host must pick goat door A. Contestant chose incorrectly.

Host opens a goat door...

At this point the host HAS opened a door and it's known to be a goat and the problem has completely changed. We're left with the new states of...

1. Contestant picked the right door. If they stay they win but if they swap they lose.

2. Contestant picked the goat door. If they stay they lose but if they swap they win.

This can be programmatically shown by this pseudo-code.

Randomly pick 1 of 3 doors to have the prize.

Randomly pick 1 of 3 doors to be the contestant's pick.

Randomly pick 1 of 3 doors to have the host pick a door to throw out. If it's the prize door or contestant's pick just randomly pick again until it's not.

If the prize door is the contestant's pick we add 1 to the win column. Else we add one to the lose column.

Run it for as long as you want, assuming 1 million+ times to be at least semiaccurate, then then unsurprisingly find out that the tally is approximately 50/50. It cannot be argued against because we did the entire Monty Hall Scenario. We have a random prize door. We had a contestant pick a random door. We had the host pick a random goat door. We then compare if the contestant already picked the right door or not. If they always swap doors or not doesn't impact their chances of winning in the slightest.

If we don't do exactly what the pseudo-code does we don't fit the description of the Monty Hall Problem but rather a situation where we're doing math for before the host opens a goat door but after the host has announced the door has a goat behind it. — Preceding unsigned comment added by 184.166.97.144 (talk) 03:07, 28 June 2020 (UTC)

Nope, it's definitely 2/3 for switching and 1/3 for staying. If you implemented your description of the scenario correctly, a simulation should verify that. See the main article for a more detailed explanation. –Deacon Vorbis (carbon • videos) 03:26, 28 June 2020 (UTC)

Ran the simulation. Worked exactly as I said it did. 50/50 — Preceding unsigned comment added by 184.166.97.144 (talk) 03:38, 28 June 2020 (UTC)

Please indent your replies and sign your posts with 4 tildes; see Help:Talk for more info; thanks.

Then you made a mistake in implementing it; there are references in the article that describe doing exactly this and that it supports the analysis showing a 2/3 win chance for switching. –Deacon Vorbis (carbon • videos) 03:49, 28 June 2020 (UTC)

I gave exactly the code I used. Random numbers for everything. If you see a problem with the pseudo-code point out exactly where I made an incorrect assumption.184.166.97.144 (talk) 06:52, 28 June 2020 (UTC)

No, you gave a very very rough description of an algorithm, certainly not rising to the level of pseudocode, let alone actual code. What you described seemed more or less in line with the usual version of the problem, so if you got a different result, you must have erred in its implementation. Without seeing that, it's impossible to say how. –Deacon Vorbis (carbon • videos) 12:45, 28 June 2020 (UTC)

3 random ints [0,2]. They stand in for prize door, chosen door, and the door the host shows to have a goat. Ensure that the host hasn't picked either the prize door or the contestant's door but rather a free goat door, and then just a comparison if the prize door is the chosen door. It's that easy.184.166.97.144 (talk) 19:19, 28 June 2020 (UTC)

But if you're simply checking to see if the player's choice matches the car as you say, then you're simply checking to see if those two random numbers were equal (the value of the host pick is irrelevant), which is 1/3, just as it should be. Again, if you're getting 1/2, then you're implementing something that doesn't match what you've described, because what you've described plainly would result in a 1/3 win chance. –Deacon Vorbis (carbon • videos) 23:18, 28 June 2020 (UTC)

Found the mistake. It didn't help that the the error shifted it to ~50% at large numbers which exactly fit my misconception but it did. I apologize for the inconvenience.184.166.97.144 (talk) 01:00, 30 June 2020 (UTC)

No worries; it happens. –Deacon Vorbis (carbon • videos) 02:27, 30 June 2020 (UTC)

using the same logic with 2 players

I can't figure out what’s wrong with this line of reasoning: You choose door #1, host reveals door #3. A viewer watching the show at home had picked door #2. If your odds to win by switching are 2/3, then the odds for the viewer to “win” are also 2/3 by switching. So each of you will win 2/3rds of the time by trading doors. How is that possible? 71.162.113.226 (talk) 16:31, 19 September 2020 (UTC)

The host is guaranteed to open a door that the contestant in the show didn't choose. This doesn't apply to the viewer at home. It's possible that the viewer chooses door 2 and then the host opens door 2. So it's trivial that a lot of the time the viewer is better off changing their choice. The information the host is giving is the same: out of the two doors that the contestant didn't choose one doesn't have the prize behind it. The viewer can use this information to see that the unchosen, unopened door is more likely to have the prize. MartinPoulter (talk) 14:43, 20 September 2020 (UTC)

What’s wrong with that line of reasoning is: the viewer has 2/3 chance of “winning” the car by NOT switching (from door #2 to door #1). The information they and the contestant both have about those doors is the same. The viewer is observing the contestant’s probability set, not generating one of their own. Freddie Orrell (talk) 19:32, 28 September 2020 (UTC)

You said: "It's possible that the viewer chooses door 2 and then the host opens door 2." With all due respect, I'm not talking about that case. I'm only talking about the case where it is in the interest of both the contestant and the viewer to switch. The contestant would be better off with the viewer's door and vice versa, as each will win 2/3 of the time by switching. 71.162.113.226 (talk)

What's your basis for "2/3 of the time" if you're only talking about a specific case? The probabilities are completely different if you get to specify which door wins. Put another way, if you want to say there is a certain probability of winning when you explicitly exclude the case where the viewer chooses the door that will be opened by the host, you need to provide a calculation of that probability. MartinPoulter (talk) 13:12, 23 September 2020 (UTC)

For both the contestant and the viewer, staying wins 1/3 of the time, switching wins 2/3 of the time. This is the logic of the “correct” solution. Obviously, if the viewer is eliminated by Monty’s door opening, then the viewer’s chance of winning is 0. But when both contestant and viewer are still in it after the door is opened, each has a 2/3 chance of winning by switching, since their original 1/3 chance of winning by staying hasn’t changed. 71.162.113.226 (talk) 18:44, 24 September 2020 (UTC)

'their original 1/3 chance of winning by staying hasn’t changed' you're making a false assumption there. When the host acts, he adds information about doors unchosen by the TV contestant. The contestant should then switch to the door he has more information about. In your example the at home player has learned information about a door he has already chosen, so he needs to stay with what he's picked. There is no contradiction. - MrOllie (talk) 18:50, 24 September 2020 (UTC)

Multipe Game Approach

There are 24 possible courses of the game, 12 where the contestant switches and 12 wherethe contestant stays with his or her original choice.

From the table pictured, you will see that over a number of games, the chance of winning or losing is the same, 50%, whether the contestant stays or switches.

Initially, the contestant has 1/3 chance of picking the car, the chance of any door hiding the car is 1/3. When the host removes a goat door, the chance of any door hiding the car is 1/2. Commomsense reigns. Rjtucker (talk) 11:03, 11 September 2022 (UTC)

So you're saying the situation "Prize in 1, contestant picks 2, host opens 3" is equiprobable with "Prize in 1, contestant picks 1, host opens 3". But it should be clear that those to outcomes aren't equally probable. You're saying there's a 50/50 chance of the contestant making the right initial choice when they are choosing randomly from three options. "Prize in 1, contestant picks 2, host opens 3" should be equiprobable with "Prize in 1, contestant picks 1, host opens either 2 or 3". It's up to you to explain why you expect the contestant to get a well-above-chance success rate in the no-switch scenarios. MartinPoulter (talk) 12:30, 11 September 2022 (UTC)

I'm saying there are twelve equally probable ways the game can go starting from just before the contestant makes his first choice until just before he makes his second choice (stay or switch) provided there is no prejudice to any particular door(s) by either contestant or host, and 24 equally probable ways the game can go if the contestant makes his stay or switch choice at random (which is not likely currently if s/he searches the topic on the Internet).

"Prize in 1, contestant picks 2, host opens 3" is an equally probable scenario, sequence of events in any one game show, as "Prize in 1, contestant picks 1, host opens 2" and "Prize in 1, contestant picks 1, host opens 3". In a very large number of game shows, one would expect each of those three selection sequences equally evident.

I think there's a fallacy in the logic of the 1/3 | /2/3 idea somewhere. To have a 2/3 probability of winning, you need to be allowed to open up to two doors. You lose that possibility when one door is taken away. Rjtucker (talk) 16:03, 11 September 2022 (UTC)

There's some discussion of 12 outcomes in the comments here that I believe are relevant, though I haven't read them right through:

https://statisticsbyjim.com/fun/monty-hall-problem/

Thinking further, perhaps the game can be considered as one where there are two chances to win a prize hidden in one of three boxes, behind one of three doors, but the first choice is a forced failure. The choice is therefore, ultimately, between two boxes, doors.

Unless I've made some horrendously incorrect misinterpretation of the original problem – I've never seen the show – I believe I see grounds for calling for deletion of the article (possibly to be replaced by one along the lines of the Monty Hall Hoax).Rjtucker (talk) 08:03, 12 September 2022 (UTC)

@Rjtucker: Your claim that these situations are equiprobable is made without argument. You are circumventing an argument based on conditional probability by refusing even to calculate the conditional probability; that's not a failing of the original argument. You do not deny that your description of the problem means that the contestant has a 1/2 chance of making a correct initial guess, when they are making a three-way choice. This obviously doesn't fit with common sense nor with probabilistic reasoning. You do not need to see the show to understand the problem: lots of people writing about this problem have never seen the show, but have seen the descriptions of the problem. It's pretty clear that you've found a problem with your understanding, not with the mathematical puzzle. As for the implications for the article, the article is a summary of what's been publishing in reliable sources by experts. You've no grounds for calling for a deletion of the article (those sources still exist) and even less for claiming a "hoax" (which would imply somebody having an intention to mislead). What is thought to be established knowledge can sometimes be overturned, but not by someone who insists on an obviously absurd conclusion without argument. MartinPoulter (talk) 12:04, 12 September 2022 (UTC)

I didn't read the full article before posting, noting it maintained the increased chance of winning by switching. Going back to it, I find that even Nobel physicists stick adamantly with what the article calls the "wrong" answer.

You offer no argument as to why each line in the table I wrote should not be as likely a description of the events of a game as another line (given no host or competitor biases).

I strongly felt I did understand the puzzle, but given the amount of work and publications I seem to be disagreeing with, I felt some reassurance on the matter might be in order.

I said along the lines of "hoax". Again given the amount of knowledgeable input to the problem, I'm a little uncertain what is going on. Perhaps "revisited" ... but I've no intention of writing it!

It would perhaps be interesting to know how many shows there were and how many cars they gave away.

I have started to look at the mathematical input there is for the problem, but am currently uncertain of the necessity to make the problem more complicated than it is.Rjtucker (talk) 14:49, 12 September 2022 (UTC)

Oh, I see, a show where the host has a choice and chooses a particular box is going to be half as frequent as a show where he has no choice (the contestant has not chosen the door with the prize in his first choice).

I think the full chart:

https://cdn.analyticsvidhya.com/wp-content/uploads/2020/04/Screenshot-from-2020-04-20-10-34-07.png

from this exposition of the puzzle:

https://www.analyticsvidhya.com/blog/2020/08/probability-conditional-probability-monty-hall-problem

is rather clearer than the current Wikipedia article presentation.

Rjtucker (talk) 07:59, 13 September 2022 (UTC)

"where the host has a choice and chooses a particular box is going to be half as frequent as a show where he has no choice" Yes, you got it. I hope you realise the burden of proof was on you to show why the lines in your table are equiprobable. The "Statistics by Jim" link you've provided seems a good explanation. MartinPoulter (talk) 16:06, 13 September 2022 (UTC)

I wonder what proportion of people would see the problem and its solution most easily and clearly from the tree diagram that it should not be in full somewhere across the centre of the main article. I suspect software exists that will create these diagrams. Rjtucker (talk) 08:46, 14 September 2022 (UTC)

I know you understood it with the diagram tree, but I have also created an area diagram that provides another way to look at it. The whole rectangle is assumed to have area 1, which represents the total probability. It is divided in three sectors of equal size (1/3) that represent the three possible locations of the prize. Now, since the host has two possible doors to open when the player's selection has the car, thoe two possibilities are subdivisions of the 1/3 in which that door has it, so each constitutes (1/2)*(1/3) = 1/6 of the total. The diagram is for when player picks door 1, but the other two are analogous:

EGPRC (talk) 17:28, 1 November 2022 (UTC)

The diagram I eventually created for myself. It is, I believe, complete; I don't know if it's any clearer to others than those of its genre already available. I can supply the .odg file if anyone wants to tidy, use it. Rjtucker (talk) 18:39, 1 November 2022 (UTC)

Misconception

The Monty Hall Paradox ist a misconception. In terms of chance there is no relation between the 3-door and 2-door systems. I believe that the problem lies in the fact that the wrongly calculated numbers will match empirically observed statistics because the misconception carries over.

The problem is that what actually occurs are 2 completely different scenarios for each of the 2 participants, i.e. the host and the player. That makes 4 different systems of probability. The illusion is that they would somehow have a relationship in terms of probability because they are related in space and time.

Let’s first look at the host-side. He knows that the door with the prize has 0 chance to be revealed by him. For him the three doors have the chances 0, 0.5, and 0.5 of being revealed by him at the beginning.

The player chooses a door. Yes, the chances were each 1/3 to be picked but now we are faced by a new situation. The game has rules that are known. These rules do influence the game as much as the participants’ choices. The host now looks at 2 doors of which he will remove 1. He either has two goats or one prize and one goat. In the latter case there is no probability. The prize has a 0 chance of being revealed and the goat has 1 chance, which means there is no chance. Only if he sees 2 goats the 2 doors each have a 0.5 chance of being picked. But we do know that according to the rules, this chance does not matter.

After the revelation of the door with absolutely no chance at all, that is, with certainty we are facing a new system. There is a goat and there is a prize. For the host who knows the doors the chances are 0 and 1, which of course means that there is no chance involved. The player faces a 50/50 chance. That’s basically it.

From the player‘s side we are faced by a system of 1/3 chance for each door and later with this 50/50 chance system.

Of course you can try to make calculations between those systems and you will get numbers as result but this case is like adding apples to pears. In terms of chance these 4 systems are not related. People may believe that there must be a relationship because we are looking at the very same doors and we are playing one game session. But there are manipulations by the player and the host as much as there are rules that break the string of chance.

The only way you could successfully refute my statement is by pointing out that the the host only faces 1 system of chance. And that only in case the player chooses the prize. If the player chooses a goat, the host has no choice of which door he reveals. That means that there is a 1/3 chance that the host and player only face 2 different chance-systems and a 2/3 chance that there are 3. As we know 0 or 1 means that there is no chance. RK20030 (talk) 03:30, 17 November 2023 (UTC)

Consider a different show, with only two doors, A and B:

step 1: ​ ​ ​ The player chooses one of those 2 doors.

step 2: ​ ​ ​ The host takes 1 ace and 2 twos from a deck of cards, shuffles those 3 cards, and then looks at the top from those 3 without revealing it. ​ If that card is the ace, then the host puts the car behind door the player chose, else the host puts the car behind the door the player didn't choose.

step 3: ​ ​ ​ The host burns the 3 cards.

-

Between step 2 and step 3, what is the chance the car is behind the door the player chose? ​ After the burning of the cards with absolutely no chance at all, what is the chance the car is behind the door the player chose?

JumpDiscont (talk) 04:47, 8 December 2023 (UTC)

For both of the above questions, I mean, what chance does the player face of the car being behind the door the player chose? ​ (i.e. this is from the player's side, not the host's side)

JumpDiscont (talk) 04:53, 8 December 2023 (UTC)

Note: Moved from main talk page. Aristippus Ser (talk) 07:56, 8 December 2023 (UTC)

Monty Hall 33/66 argument is based on an illusion

The Monty Hall 33/66 argument is based on a statistical illusion. Odds/percentages are accurately represented by looking at each possible scenario, counting up each and it's outcome, and dividing by the total number of outcomes. However, the 33/66 argument doesn't count each outcome separately; specifically, it lumps 2 separate scenarios for initially choosing the winning door together as one outcome. This weights it as only one occurrence, when it should be weighted as 2. Weighting it as 2 occurrences (as it actually is), results in the odds being 50/50 instead of 33/66.

My thought process is: What I noticed that the table for the 33/66 position is that there is a small difference between how the table organizes it and how I had started to break it down myself (and, if the 50/50 position is incorrect, this will be exactly where and why it is incorrect) .

If you break it down to each possible scenario in real life, there are actually 2 real-world scenarios scenarios after a winning door is picked: one option where the first losing door is removed, and another scenario where the second losing door is removed.

The table that explains the 33/66 position organizes it so that , if the winning door is initially picked, these two real-world scenarios are lumped together into one row, instead of two. This attributes very different (and possibly inaccurate) weights to them when calculating odds. If each of these possible real-world scenarios are listed separately in their own row, the odds become 50/50. One could also list the 2 different scenarios for after 'choosing a losing door' together in one row, which would also weight them differently.

The table for the 33/66 position lists the different scenarios for after ' choosing a losing door' separately, while lumping the different scenarios for after 'choosing a winning door' together as one scenario, giving it less statistical weight than it should have. If you list each of the two scenarios for after 'choosing a winning door' separately, as is being done for the 'choosing a losing door' scenarios, the odds are 2 out of 4, or 50/50, for staying/switching instead of the 1 out of 3 or 2 out of 3, or 33/66, for staying/switching.

This becomes even clearer if you alter the game to include more doors.

I would like to post the tables because it's much much easier to see the difference that way, but I'm not sure how to post pictures here in this section. AI*girllll (talk) 10:53, 2 February 2024 (UTC)

What is the basis for allocating equal probability to the four scenarios you describe? It's more clear when you have more doors, so please set out for us a worked example with 100 doors. I personally think that considering more doors shows why the 50/50 answer is incorrect. MartinPoulter (talk) 19:53, 2 February 2024 (UTC)

Is there some way to add a picture, screenshot, or table here? If you list it out in a table, it becomes much easier to see, or at least discuss! AI*girllll (talk) 23:05, 2 February 2024 (UTC)

The basis for allocating equal probability is that each is a possible scenario that through random chance, has an equal probability of occurring.

Lumping the different scenarios of after a winning door is picked, together, would be the same as lumping the different scenarios of after a losing door is picked, together. But the 33/66 argument initially weights the 2 different options of how picking a winning door could occur, together into one, while listing out the different scenarios for after an initial losing door separately. This is an incorrect premise of the argument, that weights the possible outcomes incorrectly. Once that premise is accepted, any logically correct deduction based on that premise is just an extrapolation of the falsely weighted premise. Any experiment designed on this inaccuracy will also skew the results.

If you break it down into a table of all possible real-world scenarios, you can follow it step-by-step and see.

Really wish I could post a pic of the 33/66 position table, and then a pic of the real-world scenario table , because it shows the difference very clearly. AI*girllll (talk) 23:23, 2 February 2024 (UTC)

I would love to show you a worked example, and the tables for the 3-door game. How do I post pics and/or tables here? AI*girllll (talk) 23:27, 2 February 2024 (UTC)

Here is a link to the tables. Tables make it easier to discuss because you can point out exactly what/where you agree/disagree with.

https://www.instagram.com/p/C23YLgpOrl3/?img_index=1 AI*girllll (talk) 00:35, 3 February 2024 (UTC)

If you disagree with the 50/50 argument table, please specify exactly what you disagree with and why. AI*girllll (talk) 00:40, 3 February 2024 (UTC)