The Monty Hall Problem (Observation, Simulations, and Proof Using Bayes Theorem)

We discuss the famous Monty Hall Problem, a probability puzzle inspired by classic game show "Let's Make.a Deal" hosted by Monty Hall. The problem can be succinctly stated as follows:
"Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat. He then says to you, 'Do you want to pick door No. 2?' Is it to your advantage to switch your choice?"
It turns out to be to your advantage to switch, a result that confounds are intuition and expectations. We show how the information and probabilities lead to this answer, provide a Python simulation that demonstrates the result, and a proof using Bayes Theorem.
Original background music by Amanda Chaudhary.
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[@justStardust940]

This is the best explanation on youtube i've found so far

[@CatSynthTV]

Wow, thanks! Please share with your friends who might be curious about this problem 😻

[@rogerknights857]

Another poster commented: The only way to lose when switching is to choose the car, which can only happen 1/3 of the time. The other 2/3 of the time you’ll win.

[@CatSynthTV]

Definitely an interesting observation and way to frame the problem 😺

[@mathanalogies9765]

I jumped here after the Pi Day 2021 video, and it's two-for-two great videos!

[@CatSynthTV]

Thank you! Glad you're enjoying the math videos. I'm planning some more in the near feature.

[@julianguffogg]

Fascinating...but over my head! (We don't have this game show in UK)

[@mintysingularity]

I'd rather have the goat. 💗 And that dress is hella awesome!

[@CatSynthTV]

Thanks 😻 and yes, we had a conversion here at HQ about the value of the goat vs the car as prize...

[@estring123]

so ur muslim?

[@jdanielcramer]

I keep watching these and then forgetting how it works🤔 my only thoughts are: Monty had a lot of goats, and I’m gonna lose out on a new car...🙀

[@gottadomor7438]

I'm not that smart so for the, "Monty Hall Problem" must, "KISS" ie "keep it simple, silly.*) At the start of this, "bet" one side has a 1/3 chance of winning & the other side 2/3. Yes? Irrespective of which door gets opened, that doesn't change the original probabilities. So, if offered the chance to switch, ALWAYS do b/c your likelihood of winning goes from 1 in 3 to 2 in 3. Switching does NOT not guarantee winning the bet but it does improve the probability of it. Re the * above: Some say the second S is for stupid but they're meanies, I'm nice. ;-)

[@CatSynthTV]

This is a good summary. It is very specific to the condition that we outlined to assure that the original probabilities haven't changed.

[@dawyer]

Maybe thinking this way....... Try to understand it from the perspective of the host. The player provides his prediction to the host. The host opens a pair of doors with sheep, leaving only two pairs of doors, one sheep and one car, for the player to choose. If the player predicts the door of the car, what will the host do? of couse Want the player to change their choice. The host asks you to calculate the odds with misleading you already have once choice before. you swapping your choice, 2/3 will win. On the contrary, the player predicted the door with the sheep. What will the host do to make the player still choose the door with the sheep? you can tell. ........ Dawyer's door problem, calculate the chance of the host winning.

[@mrpad0]

It appears to me that mathematics (and probability theories) are being used to describe a version of the observer effect to me. This is why is doesn't seem to be logical to most intelligent people. The observer effect is not logical, but can be seen to be in operation.

[@ryonalionthunder]

No. By no means is that what’s happening.

[@SlinkShady]

Bayes theorem? Really? Look, there's three doors. You choose one. You're then given the choice to swap it for the other two doors, and if the car is behind either of those two doors you keep the car. All the producers did was dress this simple question up and even Paul Erdos was fooled. It's not a complex problem at all. It's a simple choice between having one chance, or two chances.

[@rollingstone1784]

@ 1:52. To highlight the set of 2 doors with a white surrounding rectangle is important for understanding. But should distinguish better between the set probability and the door probability. I should be added that the probability of each of the doors is 1/2, not 2/3. Read: if the car is behind the set containing door 2 and door 3 (probability = 2/3, then the chance that is behind either door 2 or door 3 is 50%=1/2. At 2:04, the surrounding rectangle should not disappear and the arrow with the zero should only be drawn inside (!) the set 2:09. At 2:13, no: the probability inside the set 2 is "1", now. read: in case the car is in the set containing both door 2 and door 3; having a chance of 1/2 each, when knowing that the car is not behind door 3 (new probability = 0, it MUST be behind door 2 (probability now 100%=1). BUT because their is only a chance 2/3 that the car is behind both door 2 and door 3, the winning chance is 2/3 * 1 = 2/3. More mathematically: the winning chance of a door is the product of the chance of surrounding set times the chance of the door inside the set. The best way to understand this is to remove door 1 at 1:54 (forget door 1 and calculate with the set containing door 2 and door 3 only. Then you see clearly, that you have a set probability of 2/3 and a car probability of 1/2 (inside the set (!)). The key is that you have split a single set of 3 doors (with set winning probability 1=100% (the car MUST be behind one of the doors) into a set with winning probability 2/3 containing 2 doors with winning chance 1/2 each. If you always think in sets containing doors, the situation becomes obvious. BTW: if you draw a rectangle around door 1, the this be read as: if the door is in the set containing door 1 (probability 1/3) then the car must be behind door 1 (winning chance 100%). Therefore: 1:53 should be changed/improved as follows: draw a rectangle around door 1 and a second one around both, door 2 and door 3. Write 1/3 and 2/3 on top of each set. Write 100% and 50% + 50% on door 1, 2 and 3. Then the situation becomes clear. Final remark: the sum of probabilities inside a set must always be 1, because you must read the set probability as: IF the car is in this set…

[@timmy334]

I still fail to see how the choice between two doors is 2/3 to 1/3. It doesn't matter if he knows what is behind the doors. You are still left with 2 doors to choose from. That is a 50/50 choice no matter how you put it. So his theorem says otherwise. I studied that for my math minor. I thought it was flawed then and still do now.

[@CatSynthTV]

This result is very specific to the assumptions made.

[@RonaldABG]

Two choices don't imply they must have the same chance to be correct. If we put a random person from the street in a 100 meters race against the world champion in that discipline and you have to bet who will win, surely you have two options, but obviously the champion has a clear advantage, so the chances are not 1/2 for each. In the same way, here one door was randomly chosen by someone (the player) that had no idea which the correct door was, while the other was left by someone (the host) that already knew the car's location and was not allowed to reveal it, so had an obvious advantage. The contestant would only manage to start picking the car door 1 out of 3 times, on average, so the host is who would be forced to keep it hidden in the other door that avoids to open the remaining 2 out of 3 times that the contestant starts failing. Therefore, despite there are always two doors left, the player's choice happens to be correct 1 out of 3 times, and the other happens to be correct 2 out of 3 times. In order to be 50% chance, each should happen to be correct 1 out of 2 times.

[@timmy334]

@@RonaldABG I so want to understand this problem, but I just can't see it. To me, regardless of how the contestant gets to the two doors, the end result of two doors is still 50/50. I have no idea how I did a math degree and this never come up. I feel cheated on that, honestly.

[@RonaldABG]

@@timmy334 Ok, let's suppose you made 6 attempts of the Monty Hall game. The expected result (the average) is that each door tends to be correct with the same frequency (1/3), so about 2 times of those 6. For example something like: Car goat goat goat goat Car Car goat goat goat Car goat goat goat Car goat Car goat Let's say your selected door is the leftmost column, which the host does not touch. Now, since he always reveals a goat from the other two, if we only represent the columns of the two doors that remain closed after the revelation, we get: Car goat goat Car Car goat goat Car goat Car goat Car The left column is your original selection (the staying door), and the right column is the switching door. If you notice, despite there are always two doors remaining, it tends to happen with twice frequency that the correct results to be the switching door, not half of the time each, and that's what matters. If the game starts with more doors, like 10, the expected tendency is something like: Car goat goat goat goat goat goat goat goat goat goat Car goat goat goat goat goat goat goat goat goat goat Car goat goat goat goat goat goat goat goat goat goat Car goat goat goat goat goat goat goat goat goat goat Car goat goat goat goat goat goat goat goat goat goat Car goat goat goat goat goat goat goat goat goat goat Car goat goat goat goat goat goat goat goat goat goat Car goat goat goat goat goat goat goat goat goat goat Car goat goat goat goat goat goat goat goat goat goat Car Again, let the leftmost column represent your original selection. The host always removes 8 goats from the nine doors you did not pick, so after the revelation the contents of the two remaining doors in those attempts are: Car goat goat Car goat Car goat Car goat Car goat Car goat Car goat Car goat Car goat Car So, the switching door tends to be correct 9 out of 10 attempts. If you continue extending the number of initial doors, like 1000, it will be harder for your original selection to be which has the prize (1 out of 1000 trials), so easier that the switching door will be in fact the winner (999 out of 1000 trials). Just focusing in the fact that there are two doors left is completely useless if almost always the correct will be the other that the host left, and almost never yours.

[@jayashrishobna]

Hi Timmy. Your comment is almost a year old but I'm gonna try to explain it anyway from a slightly philosophical perspective. Ronald's already given a great answer. I'll try to add to that, because I remember how it felt to finally understand this and hope I can help you get there. First, there is a well-justified reason to be confused here, because there are many different ways the term "probability" is used to describe uncertainty. Depending on how you define it, the probability may or may not depend on your subjective knowledge of the situation! At the start of the game, the player was offered 3 doors. Technically, the car *is already* behind one of the doors. The host has perfect information, so from his perspective, there's a 100% probability of the car being behind one of the doors, and 0% probability for the other two doors! But generally the question is phrased from the player's perspective. Since the player was told that the car was randomly assigned to a door, and he has no other data that could point to one door over the other, it is reasonable to assume that the starting probability is 1/3 each. That is an ASSUMPTION, made based on best knowledge, not a "true probability". Say the monty hall gameshow extensively markets itself as being random, and the host emphasises before each round that the car is randomly assigned to a door. But let's just imagine for a moment that the show has run for 10 years, and the host actually tends put the car behind door 1 only 1/5 of the time, not 1/3, maybe for logistical reasons (eg. easier to position the car near the other doors). So on this fated night, a player comes up on stage, not having watched the show before, and hence doesn't know that historically door 1 is less likely. From this player's point of view, based on the data he *does* have (ie. the host and all the marketing told him it was a random assignment), the chance is still 1/3 to him. Now say that one of the audience members on this particular night is a long-time viewer of the show and has noticed the unequal trend over the years. For that long-time viewer, he has *additional* information on the prior distribution of the chance for each door, and it would be reasonable to assign the chances as 1/5, 2/5, 2/5 instead of 1/3 each. -------- So far, I hope I've demonstrated that the probability from the player's perspective depends on the information he has. It's not that just because there are 3 doors, the chance is 1/3 each. If he has other useful information, then he might use that data to assign the prior probabilities differently. I hope you agree with me on this - if so, we can proceed. -------- Now, the meat of it: Using the same reasoning as above, just because you have 2 doors, doesn't mean the chance is 50-50! It depends on what information you have which could indicate one door is more likely to have the car than the other. And it turns out, you do have extra information about the remaining 2 doors, because of the host's actions! Say you chose door 1 at first. When the host opened one of the other doors and revealed the goat, that was NOT a random choice between the 2 remaining doors. He deliberately chose a door with a goat. Here's the critical part: The host's actions reveal NO new information about your door 1. This is because his actions neither directly nor indirectly depend on or influence what is behind door 1. He could only influence door 2 and 3. He could not show you what was behind door 1. If the car was actually behind door 2, he would have opened door 3. If the car was behind door 3, he would have opened door 2. If the car was behind door 1, he could open either 2 or 3. No matter what, what he does does not tell you anything about or change what is behind door 1. What that means is, doors 2 and 3 are basically linked into a single "bigger" door, a door "system" if you will, and together they carry the probability of 1/3 + 1/3. The host interacts with and influences only this system of 2+3, not door 1. By looking only at this system of door 2+3, we are setting aside door 1 and looking at a "conditional probability" within this system only. Door 2 is opened and revealed a goat. What the host has now told you is, between doors 2 and 3, if there is any chance at all between them, there is now a 0% chance of it being behind door 2, and 100% chance of it being behind door 3. By revealing door 2, he gave you new information about door 3. I emphasise again, nothing he did here told us ANYTHING about door 1. Now let's put all this together. We started with 1/3 each, and player chose door 1. When the host chose to open door 2 (and not door 3), he did not give you any new information about door 1, so the chance remains 1/3 for door 1. But he did tell you that, between door 2+3 which previously had 1/3 each, now none of the probability lies with door 2, and all the probability now lies with door 3. So you now *update* your information about door 3 relative to *door 2 ONLY*, and know that door 3 now has "taken" all of door 2's probability, hence 1/3+1/3 = 2/3. So, door 1 remains at 1/3, and door 3 is updated to 2/3! Hope this helps :)

[@WeWereYoungandCrazy]

I'm pretty comfortable calling bullshit on this.

[@CatSynthTV]

Comforting as that may be, our answer is the correct one. Bayes theorem is one way of approaching it formally, but there are others as well.

[@Magneticitist]

😂😂😂😂😂

[@rhetoric5173]

Baye's disproved this... "Initially, the car is equally likely to be behind any of the three doors: the odds on door 1, door 2, and door 3 are 1 : 1 : 1. This remains the case after the player has chosen door 1, by independence. According to Bayes' rule, the posterior odds on the location of the car, given that the host opens door 3, are equal to the prior odds multiplied by the Bayes factor or likelihood, which is, by definition, the probability of the new piece of information (host opens door 3) under each of the hypotheses considered (location of the car). Now, since the player initially chose door 1, the chance that the host opens door 3 is 50% if the car is behind door 1, 100% if the car is behind door 2, 0% if the car is behind door 3. Thus the Bayes factor consists of the ratios 1/2 : 1 : 0 or equivalently 1 : 2 : 0, while the prior odds were 1 : 1 : 1. Thus, the posterior odds become equal to the Bayes factor 1 : 2 : 0. Given that the host opened door 3, the probability that the car is behind door 3 is zero, and it is twice as likely to be behind door 2 than door 1. Richard Gill[54] analyzes the likelihood for the host to open door 3 as follows. Given that the car is not behind door 1, it is equally likely that it is behind door 2 or 3. Therefore, the chance that the host opens door 3 is 50%. Given that the car is behind door 1, the chance that the host opens door 3 is also 50%, because, when the host has a choice, either choice is equally likely. Therefore, whether or not the car is behind door 1, the chance that the host opens door 3 is 50%. The information "host opens door 3" contributes a Bayes factor or likelihood ratio of 1 : 1, on whether or not the car is behind door 1. Initially, the odds against door 1 hiding the car were 2 : 1. Therefore, the posterior odds against door 1 hiding the car remain the same as the prior odds, 2 : 1. In words, the information which door is opened by the host (door 2 or door 3?) reveals no information at all about whether or not the car is behind door 1, and this is precisely what is alleged to be intuitively obvious by supporters of simple solutions, or using the idioms of mathematical proofs, "obviously true, by symmetry".[43]"

[@dawyer]

The situation is that "decisions without consequences" and "choices with practical benefits" are confused and regarded as having the same meaning. It has been clearly decided. If there are really two choices, shouldn't we get both things? In fact, the rule changes from one of three choices to one of two, misleading you into thinking that you have already chosen once, but in fact you cannot get any of the so-called choices at all. When you finally make the real choice, in order to prove how smart and knowledgeable you are, of course it should be counted as 2/3 when there are only two choices. This is a testament to the Goebbels effect. If scholars and the educational community do not correct the extension of Monty Hall's problem and still believe that 2/3 is correct, then what is the point of educational scholarship? ? It was ruined.

[@dienekes4364]

The host presents 3 doors. The player chooses door A The player has a 1:3 chance to win -- We agree up to this point The host opens door B (a goat) -- According to you, the player still has a 1 in 3 chance to win -- According to me, the player now has a 1 in 2 chance to win The host opens the other door (a goat) There is only 1 door remaining: the one the player initially chose -- According to you, the player still has a 1 in 3 chance to win -- According to me, the player has a 1 in 1 chance to win Do you see the failure in your model? Believe it or not, the Earth is a globe, it's not flat.

[@dawyer]

Is the reality of quantum mechanics more logical than what the math? The player never knows which goat is the one revealed.。.Don’t let the Jews fool you. If life is an endless choice, then if we face it with this way of dealing with things, then what aspect of human nature is eternal?

[@nyjsackexchange]

Pure lies- produce the empirical data Til then, pipe down

[@CatSynthTV]

Not quite sure who you are addressing with your comment. But from our end, we did gather empirical data using our computer simulation and it matched the result we got from the equations.

[@nyjsackexchange]

@@CatSynthTV No, empirical data as in the actual contestants' outcomes on 'Let's Make a Deal' -- not some 'computer model analysis' rife with flaws and biases -- everyone knows 'random' number generators aren't entirely random, and most merely mimic preconceived numerical distributions If there was a statistical anomaly with regard to sticking/switching as you claim, surely someone would have pointed it out to the show's producer's or contestants would have caught on long ago

[@Hank254]

@@CatSynthTVYou can ignore him... he goes around trolling all the Monty Hall problem videos. He can't understand it and it drives him nuts.

[@catsynth]

@@Hank254 thanks for the tip, will do.

[@max5250]

@@nyjsackexchange "No, empirical data as in the actual contestants' outcomes on 'Let's Make a Deal' -- not some 'computer model analysis' rife with flaws and biases " You still can't understand this is not "Let's Make a Deal" TV game show, but "Monty Hall Problem", a probability problem disguised in a game show?! "everyone knows 'random' number generators aren't entirely random" Computer algorithms for random numbers a re producing pseudo-random numbers, due to their nature of using a "seed" (initial) number, which is for freshly powered on computer always the same. But, these numbers are still random numbers for all intense and purpose and we can easily prove that. "and most merely mimic preconceived numerical distributions " No they don't. Different computers will generate different sequences of random numbers. "If there was a statistical anomaly with regard to sticking/switching as you claim" The difference in odds of staying/switching is not some kind of "statistical anomaly" but a simple product of the difference in number of doors that are contributing to each of the final two doors (there is only one door contributing to the staying door, while two doors are contributing to the switching door). "surely someone would have pointed it out to the show's producer's or contestants would have caught on long ago" This doesn't happen in 'Let's Make a Deal' for the simple fact that host in that TV game show doesn't behave the same as a host in Monty Hall Problem.

[@dienekes4364]

FAILURE. This is very disappointing for this channel. The ONLY thing that the idea that the game is played out this way is that the odds for round 1 are meaningless because the round will NEVER be resolved and it will be forced into round 2, where the odds remain 50/50. Nothing in round 1 changes this. I can prove, beyond any doubt (to anyone who is intellectually honest) how this video is complete and utter garbage. What is presented here is called, in mathematical and scientific nomenclature, a "model". There are 2 ways to prove a model: direct testing and observation and applying it to similar yet slightly different scenarios and having it be consistent. Since testing isn't possible in this venue, I will prove this model is wrong by the second method. Core Scenario: A contestant is presented with 3 doors. They get to choose 1. For simplicity, we'll say they pick door 1. Each round, the contestant get to switch to a different door or stay with the door they have. There is only 1 door that has a car, but 1 door always has a car. So, at this point, each door has a 1:3 chance of winning. If the round is resolved, the contestant has 1:3 chance of winning because they only got to choose 1 out of the 3 doors. If anyone disagrees with this, please comment below. The host opens one of the doors the contestant didn't choose (door 3). If the car is behind the door the host opened, the contestant loses. If the host opens a door with a goat, then we move to round 2. The contestant is allowed to switch to door 2 or stay with door 1. I think we are all in agreement up to this point. According to the model presented here, the odds of the eliminated door gets shifted over to door 2. This is based solely on the fact that the contestant picked a door in round 1 (an action that played absolutely no role in establishing the odds in the first place). According to the standard model, the odds of the eliminated door get evenly distributed between the existing doors. This is based on the number of doors remaining. In order for a given model to be correct, the outcome must equal 100%. There is a 100% chance that the contestant will win or lose. All odds needs must equal 100%. So, in round 2, according to the presented model, there is a 1:3 chance you win if you stay with door 1 and a 2:3 chance you win if you move to door 2. This equals a 3:3 chance, which is equivalent to 100%. We're good so far. But what about the standard model? In round 2, with an eliminated door, since there are 2 doors, the odds of either door having the car is 1:2 (also described as 50%). Since each door has a 1:2 chance, that adds up to 2:2 or 100%. So, both models are seemingly viable at this point. So, let's apply these models to 2 slightly different scenarios: Alternate Scenario 1: In round 2, the round isn't resolved. The host opens door #2, which has a goat. What are the odds of the door the contestant has picked having the car? Let's go back to our models: In the presented model, remember that the ONLY criteria of the odds for being correct are the doors that the contestant DIDN'T pick. So, according to this model, if the contestant stayed with door #1, there is a 1:3 chance of winning. But the 2 other doors, having been opened and showing goats, is clearly 0:3. What does that add up to? 1:3 + 0:3 = 1:3. So, if the car isn't behind door 2 or 3 and is only behind door 1 1/3 of the time, where is the car the other 2/3 of the time? The total for winning and losing here is 33%. So this model fails. What about the standard model? Once door 2 is eliminated, the odds of each remaining door get recalculated and distributed evenly. So, if there is only 1 door, that means that the chance of that door being correct is 100%. So the standard model is superior and proven correct so far. But let's move to alternate scenario 2 and see what happens: Let's play the exact same game except with 2 contestants: In the 1st round, contestant A chooses door 1. Contestant B chooses door 2. The host opens door 3 with a goat). Where do those odds get shifted? In the presented model, the odds can't be shifted because there is no door that hasn't been picked, so the odds are: Door 1 -- 1:3 Door 2 -- 1:3 Door 3 -- 0:3 (shown to be a goat) This adds up to 2:3 or 66.6%. So this model fails because it doesn't add up to 100%. 1/3 of the time contestant 1 wins, 1/3 of the time contestant 2 wins, and 1/3 of the time they both lose. If they both lose, where is the car? OR The contestants both switch to the other door. According to the model, "the other door" gets the additional odds. Okay, so the odds end up like this: Door 1 -- 2:3 (because contestant 2 didn't pick it, so it gets the odds from door 3) Door 2 -- 2:3 (because contestant 1 didn't pick it, so it gets the odds from door 3) Door 3 -- 0:3 (exposed and eliminated as a possibility) So, the odds (2:3 + 2:3 + 0:3) equals 4:3, or 133%. Once again, a failure. 1/3 of the time contestant 1 wins, 1/3 of the time contestant 2 wins, and 1/3 of the time they both win. We've established that there is only 1 car behind 1 door, so it is impossible for both contestants to win. So, what does the standard model say? The odds get evenly distributed to the remaining doors. How do those odds look? Door 1 -- 1:2 (one out of 2 remaining doors) Door 2 -- 1:2 (one out of 2 remaining doors) Door 3 -- (not in the calculation because it was eliminated by being exposed as a losing door) 1:2 + 1:2 = 2:2 or 100%. Once again, the standard model is victorious. So, in both of these scenarios, the presented model fails miserably and the standard model stands. Can anyone come up with an alternate scenario where the presented model works, but the standard model fails? I doubt it, but please present it if you can think of one.

[@DennisNogueira1]

A month has passed, but in case you have not figured it out, in your examples you are not representing the fact that in this game the revealed door did not happen to have a goat just by chance. The host knows the positions and deliberately goes to reveal that goat from the non-selected options, that is, he avoids to reveal the player's door and the winner door. Think about this: if you repeat the first part of the game with the three doors lots of times, you ignore your single selected door and instead you open the other two to see their contents, how many times do you think you will find the car in any of them, 1/2 or 2/3? As they are two doors, it is reasonable that you would find the car in that group 2/3 of the time, twice as much that in your single original one. As the host knows the locations, it is like if he was already seeing the contents of that group of two doors. He sees that the group of your non-selected doors happens to have the car 2/3 of the time, and as he deliberately avoids to reveal it, always going for the goat, then he leaves the car in his closed door 2/3 of the time.

[@dienekes4364]

@@DennisNogueira1 _"The host knows the positions and deliberately goes to reveal that goat from the non-selected options, that is, he avoids to reveal the player's door and the winner door."_ -- First, yes, I know this. It's part of the scenario. I included it in my explanation simply to be thorough. The ONLY thing this does is force the game into round 2. It has absolutely no other impact on the results.

[@dienekes4364]

@@DennisNogueira1 _"Think about this:"_ -- This is just regurgitating the premise. It doesn't add anything to the conversation except to show that you are completely ignorant of how probability works. _"instead you open the other two"_ -- Yes, exactly my point. You don't get to choose 2 doors. You only get to choose *_ONE_* door. The other door, as I have explained multiple times in multiple different ways, has already been eliminated. It's not part of the equation anymore. How hard is this to understand? _"how many times do you think you will find the car in any of them, 1/2 or 2/3?"_ -- _IF_ you got to open 2 doors instead of 1, then you'd have 2/3 of a chance of being right. But didn't you start this comment with the idiotic accusation that I was "ignoring the rules"? And then you go on to completely change the game. But you don't see the childish idiocy of this? Really?

[@dienekes4364]

@@DennisNogueira1 _"As the host knows the locations, it is like if he was already seeing the contents of that group of two doors."_ -- Once again (for the slow kid in the class), the ONLY thing this does is prevent the player from losing in the first round and pushes the game to the second round. It doesn't change the odds for the second round in the least bit. That is, unless you can explain how the odds change in the second round from the host knowing that the door they open is a loser vs the host just happening to open a door that is a loser. Can you explain what the difference is? How *_EXACTLY_* does that chance the odds? Or did you just pull this assertion out of your ass? _"He sees that the group of your non-selected doors happens to have the car 2/3 of the time, and as he deliberately avoids to reveal it, always going for the goat, then he leaves the car in his closed door 2/3 of the time."_ -- Yes, I understand the premise. As I have explained several times. Just regurgitating the flawed logic over and over again isn't going to magically make it correct. In the FIRST ROUND the player would lose 1/3 of the time if the host opens a random door. The player wins 1/3 of the time _IF AND _*_ONLY_*_ IF_ the game is resolved in the first round. But, according to the rules in this scenario, the game is NEVER resolved in the first round, so you can't use that to calculate the odds. The odds of the player winning in the first round is 0% because the round is never resolved.

[@dienekes4364]

@@DennisNogueira1 But here's the thing: I know that _YOU_ know that I'm right. Do you want to know how I know that? Because you didn't address anything, that is ANYTHING that I actually said. That's because you know if you even acknowledged it, it would prove you wrong. But you are so mind-numbingly stupid, you actually thought you could waltz in here, regurgitate the original premise, and think you could actually win the debate just with your self-important, glowing presence. You get how pathetic that is, right?