This page is for mathematical arguments concerning the Monty Hall problem. Previous discussions have been archived from the main talk page, which is now reserved for editorial discussions.
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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 chance of having picked the correct door and a 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)[reply]
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?[reply]
It IS a new choice, in a new situation. Since the second situation (choosing a door when there are 2 doors, one with a car, one with a goat) is the exact same situation no matter whether the contestant initially chose the car or a goat, __it is not dependent on the original situation or the odds of whether a car or goat was first chosen__ . This problem is not an example of conditional probability, although the wording and storyline ( including the first choice from 3 doors) gives the illusion that it is. If you apply the rules for conditional probability...IE, weighting the odds for second choice (switch or stay) by the odds of the first, unrelated choice, it will result in an inaccurate answer. AI*girllll (talk) 19:09, 4 February 2024 (UTC)[reply]
Consider a different show:
There are only 2 doors: door #1 and door #2. They each start with nothing behind them.
step 1: The contestant chooses one of those 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 as above, what is the probability of the car being behind the door the contestant chose? After step 3, are we in a new situation (there are now no cards)?
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)[reply]
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)[reply]
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)[reply]
There's some discussion of 12 outcomes in the comments here that I believe are relevant, though I haven't read them right through:
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)[reply]
@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)[reply]
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)[reply]
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).
"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)[reply]
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)[reply]
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:
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)[reply]
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)[reply]
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?
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)
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)[reply]
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)[reply]
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)[reply]
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)[reply]
If you are going to accurately represent the percentages for a switch win (instead of just the reverse probability of initially picking a winner), you need to take into account all of the switching scenarios when calculating percentages. When you include all of the switching scenarios for initially picking a winner instead of lumping them together (as exemplified in the 33/66 table, link above), the real-world odds of a switch win reveal themselves to be 50/50 . AI*girllll (talk) 07:31, 3 February 2024 (UTC)[reply]
If you limit the percentage weight of the possible switching outcomes for initially-picking-a-winner to the percentage rate of initially picking a winner (33%), this is simply ascribing/assigning/transferring the percentage weights of initially picking a winner/loser to the problem of the switch win rate (which necessarily must result in the switch win rate equalling the reverse of the percentage for initially picking a winner). This is NOT calculating the actual switch win percentage rate. When you calculate the percentages for the problem of the switch win, instead of simply ascribing the reverse percentages of the simple problem of initially picking a winner/loser, the switch win percentage rate reveals itself to be 50/50. AI*girllll (talk) 08:03, 3 February 2024 (UTC)[reply]
This is dealt with in the section of the article titled 'Conditional probability by direct calculation'. What you've missed is that when dealing with conditional probabilities like this, when you subdivide the scenarios as you have with the 'car is behind door 1, which I picked', the two probabilities are multiplied. So it is 1/3 * 1/2 = 1/6. You have multiplied by two (well, divided by 1/2) where you should be doing the opposite. Tree showing the probability of every possible outcome if the player initially picks Door 1. - MrOllie (talk) 13:43, 3 February 2024 (UTC)[reply]
And, if you are simply ascribing the initial guess probability to the switch win problem, the answer will of course be nothing but the opposite (ie, switch) of that. ANY probability you weight the outcome as will of course be whatever you initially weight it as. That is NOT calculating the probability of the switch win. The probability of a switch win is calculated by counting up all possible iterations of switching, and counting the number of switch wins as opposed to the number of switch losses. The initial guess probability should not be assigned to switch win rate problem, as they are 2 separate problems, and this will skew the odds. if you assign a false weight to these possible outcomes, the answer you get will be, by necessity, be the false weight you assigned it .
At any rate, this discussion point above is exactly where the 33/66 argument and the 50/50 argument differ, and where these arguments diverge. AI*girllll (talk) 15:25, 3 February 2024 (UTC)[reply]
Assigning conditional probability to the switch win rate seems to be incorrect. It would be the same as , taking a coin with a 50/50 chance of coming up heads/tails, flipping that coin 9 times, having it come up tails 9 times in a row, and assigning conditional probability to the percentage of what will happen the 10th time you flip the coin. If you incorrectly assign conditional probability, you will calculate the odds of the coin-flip outcome to be greatly in favor of coming up heads, when in fact the probability of the outcome of the 10th coin flip is still 50/50. AI*girllll (talk) 15:36, 3 February 2024 (UTC)[reply]
Conditional probably must be accounted for - failure to do this is one of the main reasons people have difficulties with this concept. As MartinPoulter says, adding more doors should make the concept clear. The usual formulation is one where you pick from 10 doors and Monty then opens all of the other doors save 1. Or you can try a simulator. MrOllie (talk) 23:00, 3 February 2024 (UTC)[reply]
Adding more doors, as I stated above, only proves my point that the odds remain the same for staying/switching. Unless of course, you incorrectly continue to ascribe a false weight to it, then it will take on whatever weight you ascribe to it. The simulator you link to automatically ascribes conditional probability to the problem. Ascribing conditional probability to the outcome weights the outcome inaccurately . Do you also maintain that a coin coming up heads 9 times on a flip toss changes the odds for the 10th coin toss to be something other than 50/50? Because that, too, is ascribing conditional probability, also inaccurately. AI*girllll (talk) 00:56, 4 February 2024 (UTC)[reply]
Apples and oranges. Coin flips are independent events, this problem is about dependent events. The simulator does not calculate probabilities, conditional or otherwise. It plays the game and notes the results. You could do the same thing with a friend and a few playing cards. MrOllie (talk) 01:06, 4 February 2024 (UTC)[reply]
I just want to say first that I'm not convinced of either position, but my tendency is towards the 50/50 argument. If it's incorrect somehow, I want to find out exactly where and why it is incorrect. The correct or accurate answer you should be able to get to logically no matter which direction it is approached from. I'm starting to think that maybe the 33/66 argument could be correct for a purely hypothetical answer, but just not in the real world, where there is a car and goats behind set doors?
Let's look at a real-world example., with set prizes. A car is behind Door#1. A goat is behind Door#2. A goat is behind Door #3. Say the contestant initially picks Door #1. Monty opens Door#3, revealing a goat. There are 2 doors left, one with a car and one with a goat. No matter which one the contestant picked initially, #1 or #2 (as contestant has just as equal a chance of initially picking Door#2), there is an equal (ie, 1 out of 2, or 50/50) chance that the car is behind either door. Switching (or staying) does not confer an advantage either way. The contestant is faced with the same scenario of picking between 2 doors (one with a car and one with a goat), **no matter whether there was a car or a goat behind the initial choice** .
So, the 33/66 percentage weighting that was present in the initial choice no longer applies to the current second decision of whether to switch or stay, and it is incorrect and inaccurate to assign it to this second decision of whether to stay or switch. IF the initial choice (whether a car or a goat was initially chosen) mattered, and affected or caused the second situation to be different DEPENDING on the first choice, THEN and ONLY THEN would assigning the reverse odds of initially picking a winning door to switching, be accurate and correct. But, since the situation for the second decision of whether to stay or switch isn't affected by the win/loss rate of the initial choice (it is the same situation either way), it is not dependent on the initial odds of choosing a winner/loser. Therefore, the initial 33/66 odds of picking a winner should NOT be assigned to the second decision of whether to switch or stay.
Assigning the reverse of the odds of initially picking the one winner out of 3 doors (33/66) to the odds of a switch win is simply stating that switching your initial choice results in the reverse odds of winning. This is not true. It is misequating 'switching your choice' to mean 'the opposite odds ', when they are not identical. That is the illusion behind the 33/66 argument.
I think I just articulated what other ppl before me , who also disagree with the 33/66 argument, have not been able to previously articulate. AI*girllll (talk) 19:33, 4 February 2024 (UTC)[reply]
Since you mention "for a purely hypothetical answer, but just not in the real world", I am bringing up the crucial assumptions for the 33/67 answer:
The host _never_ opens the door the contestant chose. The host _knows where the car is_, and _never_ opens that door. If the contestant chose the door with the car, then the host chooses 50/50 which other door the host opens.
if one or more of those don't apply, then the answer can easily be 50/50. However, given the above 3 assumptions:
IF the initial choice was the car, and a goat was behind #2, then there is only a 50% chance of the second situation being #1 vs #2 (there is a 50/50 for which of #2,#3 the host opens). IF the initial choice was a goat, and the car was behind #2, then there is a 100% chance of the second situation being #1 vs #2 (since here, the host has a 100% chance of opening #3). That is how the initial choice matters, and affects the second situation DEPENDING on the first choice.
If what we learned was _just_ "#3 has a goat", then 50/50 would be correct (this is "Monty Fall"). However, we know more than that: We know #3 is the door Monty chose, by following specific rules.
It's an illusion. You are condensing some of the possibilities down into one possibility when you shouldn't be, which is weighting the probability wrong. When you give each equal possibility equal weight (as you should / as it is in reality) , the probabilities for picking a winner (out of 2 doors , one with a prize one not) return to 50/50. When you weigh the possible outcomes incorrectly or dismiss some of them as outcomes, that changes the probability percentage. The word problem glosses over this omission, so it gives the illusion the probabilities are skewed in favor of switching. AI*girllll (talk) 12:56, 9 July 2024 (UTC)[reply]
What one possibility do you think I should have as more than one?
Under the 3 assumptions I stated,
The host _never_ opens the door the contestant chose.
The host _knows where the car is_, and _never_ opens that door.
If the contestant chose the door with the car, then
the host chooses 50/50 which other door the host opens.
out of 600 shows in which the contestant initially picks Door #1,
in approximately how many will it be the case that
Firstly, if you don't believe it, you can simulate the game to corroborate switching wins in fact 2/3 of the time and not 1/2. Secondly, to illustrate your mistake, imagine another scenario in which you have a job on Fridays, Saturdays and Sundays. Every Friday you go to a place that we will call A, every Saturday you will go to another place that we will call B, and on Sundays sometimes you will go to place A and sometimes to place B (only a place per day). To make it easier, suppose the Sundays are intercalated: the first you go to A, the second you go to B, the third you go to A, and so on.
In this way, there are 4 types of days of work that you can have:
1) A Friday in which you go to A.
2) A Saturday in which you go to B.
3) A Sunday in which you go to A.
4) A Sunday in which you go to B.
But obviously this is not going to make the weeks have two Sundays in order that the 4 types of days occur equally. They will continue having the same amount of each day (one). Therefore, in the long run you will go to place A more times on Fridays than on Sundays, and also you will go to place B more times on Saturdays than on Sundays.
I mean, by the time that two weeks have passed so you have gone to A one time on a Sunday and to B one time on a Sunday, you have already gone to A two times on a Friday, and to B two times on a Saturday.
In general, if you have to share something, like a pizza, with another person, you end up getting less amount of it than if you took it completely.
In the Monty Hall game, it occurs similar. In the first selection you are equally likely to select each of the three possible contents: GoatA, GoatB and Car, so their choices are like the three days of the week, as they tend to occur with the same frequency 1/3. But the games that you pick the Car are like the Sundays, because they are shared between two possible revelations, meaning that each revelation occurs in a portion of them, not in all, getting less amount. EGPRC (talk) 00:22, 15 July 2024 (UTC)[reply]
Reference 'Monty Hall problem' in Wiki.
There are 3 distinct objects, {car, goat1,goat2}, which could all differ as
{a, b, c}, with c the car, a and b of lesser value.
The 3 distinct objects can be arranged in 3! = 6 patterns, with 1 object behind each door {1, 2, 3}.
1 2 3
a b c
a c b
b a c
b c a
c a b
c b a
If the initial choice is door 1 for the simple game with 1 guess, the probability to win c is 1/3.
If the initial choice is door 1 for the complex game with an option to change their guess, the host cannot reveal the door for c nor the door selected by the player per the rules. This restricts the patterns to
1 2 3
a x c c
a c x c
b x c c
b c x c
c a b >c x b b
c a b >c a x a
c b a >c x a a
c b a >c b x b
where x is the door eliminated by the host.
When the guess is c, the host can eliminate a and b two ways,
increasing the number of patterns to 8.
For the complex game and no change the probability to win c is 1/2 (column 1).
For the complex game with a change the probability to win c is 1/2, as shown in the rightmost column, consistent with a choice of 2 doors.
There is no advantage with the option of change.
The difference of 1/3 vs 1/2 results from the player having more knowledge of the patterns resulting from the host actions.
If the lesser prizes are identical as in the 'goat' example, they exist simultaneously, thus require a unique ID for the purpose of gaming. Phyti (talk) 15:10, 10 September 2024 (UTC)[reply]
Need to add this.
Marilyn most likely considered the 2 goats as equal which reduces c a b and c b a to c a a, reducing the configurations to 6. On that basis her claim is true, but it’s a special case. The above explanation is the general case with 3 distinct prizes. Phyti (talk) 17:18, 13 September 2024 (UTC)[reply]
Let "the "coinplex game" be the result of adding "if the door selected by the player is door c, then the host chooses with equal probability which other door the host reveals" to "the complex game with an option to change their guess". Do you also get 1/2 for the coinplex game? JumpDiscont (talk) 15:15, 16 September 2024 (UTC)[reply]
refinement
game 1.
There are 3 distinct objects, {a, b, c}, with c the car, a and b of lesser value.
The 3 distinct objects can be arranged in 6 different patterns, with 1 object behind each door {1, 2, 3}.
1 2 3
a b c
a c b
b a c
b c a
c a b
c b a
The player chooses door 1. His chance of winning c is 2/6=1/3.
game 2.
There are 2 distinct objects a and c. The 2 distinct objects can be
arranged in 3 different patterns
1 2 3
a a c
a c a
c a a
The player chooses door 1. His chance of winning c is 1/3.
game 3. A variation of game 1.
There are 3 distinct objects, {a, b, c}, with c the car, a and b of lesser value.
Objects a and b exist simultaneously in the game requiring different identities even though they may be identical in properties.
The 3 distinct objects can be arranged in 6 different patterns, with 1 object behind each door {1, 2, 3}.
1 2 3 2 3
a b c c
a c b c
b a c c
b c a c
c a b b
c a b a
c b a a
c b a b
The player chooses door 1. Then the host (knowing the location of each prize) opens door (2 or 3) if it does not contain c, then offers the player the option of choosing the closed door of (2 or 3) instead of door 1.
If door 1contains a or b, the host can open 1 door per pattern,
If door 1contains c, the host can open 2 doors per pattern, increasing the number of choices for the player from 6 to 8.
His chance of winning c is 4/8=1/2 for his initial choice (door 1).
His chance of winning c is 4/8=1/2 using his option (door 2 or 3).
The host opening a door that does not contain c decreases the choices for the player to 2 doors, i.e. 1/2.
There is no advantage if the player changes his choice.
