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 Previous discussions have been archived from the main talk page, which is now reserved for editorial discussions.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]
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)
This has been going on for over 9 months and 100 kilobytes of (badly formatted) text, and while there's leeway on these arguments pages, this has long since stopped being reasonable. Enough is enough. 35.139.154.158 (talk) 01:01, 12 November 2024 (UTC)[reply]
The following discussion is closed. Please do not modify it. Subsequent comments should be made on the appropriate discussion page. No further edits should be made to this discussion.
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
Yes, I did read your reply. All you're doing, which is what you've done before, is giving another example where you purposefully ascribe false odds to the situation. It's not that I don't understand or follow your many examples. If you ascribe false odds to it, of course, it's going to follow whatever odds you ascribe to it.
I didn't feel a need to respond because I specifically asked anyone replying or attempting to rebuff my argument to point out exactly where my argument is incorrect, if it is, and so far no one has done that. All you've been doing is giving yet another example(s) where you purposefully assign false odds to the situation. AI*girllll (talk) 07:06, 13 October 2024 (UTC)[reply]
"All you're ... false odds to the situation."
Are those "false odds" in one or more of the 3 assumptions? Indeed, if those 3 don't all apply, then your argument may be correct, and the answer can easily be 50/50. However, the only other place I see where _any odds at all_ made it in to my previous comment, is in the choice of "6" for "600".
In any case, what odds do _YOU_ ascribe to the 10 things from my above comment?
I see "If it's incorrect somehow, I want to find out exactly where and why it is incorrect.", but I took that as something you wanted to do, rather than as something you wanted attempted-rebuffers to do.
Is there anywhere else you "specifically asked anyone replying or attempting to rebuff" your "argument to point out exactly where" your "argument is incorrect, if it is"?
For the rest of this comment, I assume you are referring to the argument in your "If it's incorrect somehow" comment.
Again, if what I called the crucial assumptions don't all apply, then your argument may be correct, but given what I call the crucial assumptions:
The error is in your "No matter which one ... behind either door." sentence. Although [given that the host opened Door#3], [which door the contestant chose] is the only remaining reason there can be an asymmetry between Door#1 and Door#2, [which door the contestant chose] and [where the car is] are no longer independent given that the host opened Door#3. (So, applying the symmetry to flip the contestant's choice from Door#1 to Door#2 also flips prob(car behind #1) with prob(car behind #2).)
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.
n=number of doors each with a distinct prize, with 'a' (car) the most valuable.
p=number of patterns/arrangements per prize=(n-1)!
h=number of possible doors host can reveal per prize=p(n-1),
which also=number of possible wins for the player.
1 2 3 p h
a 2 2
b 2 1
c 2 1
sum 6 8, win=4/8=1/2, 1/3<1/2.
1 2 3 4 p h
a 6 2
b 6 1
c 6 1
d 6 1
sum 24 30, win=5/30=1/6, 1/4>1/6.
The host alters the number of choices for the player by revealing 1 of the not chosen doors that does not contain a car. This gives the player the option to choose a door different from door 1. His final choice is 1 of n-1doors vs his initial choice. The number of choices does not have to equal the number of arrangements/patterns if the host interacts with the player.
These 2 examples show no advantage with option of host intervention.
The host has a coin and a standard 6-sided die. The host flips the coin. If it lands showing heads, then the host places the die showing side 1, else the host rolls the die. After the host either places or rolls the die, the contestant [wins if the coin is showing heads] and [loses if the coin is showing tails].
What is the probability that the contestant wins the above game?
Do you get the following 7 patterns for the host's 2 objects?
When the guess is c, the host can eliminate a and b two ways,
increasing the number of patterns to 8.
"
What part of all those patterns don't have the same probability to occur you did not understand?
The fact that the host can reveal any of the other two doors when the player's has picked the car mean that those games are divided in two halves, not that they are duplicated.
It's the same reasoning that you would apply if you were sharing pizza. Imagine that there were two pizzas of the same size, call them P1 and P2, but while P1 is entirely eaten by one person, P2 is shared by two persons.
That means that the person that ate P1 entirely ended up eating more than each of those that shared P2, not that P2 suddenly duplicated its size in order that all the three persons could eat the same amount. EGPRC (talk) 04:48, 25 September 2024 (UTC)[reply]
'Monty Hall problem', Wikipedia, Mar 2024
Section 'Conditional probability by direct calculation'
The 'tree' is incomplete. For the general case we know there are 3!=6 possible sequences to arrange 3 distinct objects {a b c} behind 3 doors {1 2 3}. In the list
'c' denotes the car, 'a' and 'b' are prizes of lesser value.
The basic game.
1 2 3
c a b
c b a
a c b
b c a
a b c
b a c
The list includes all possible cases for any 1 door. There are 3 locations but 6 games, 1 per sequence. If the player always chooses door 1 they win a car 2/6=1/3 of the games.
The modified game, and the rules.
The host is allowed to open 1 door per game after the player's initial choice, then offer the player the option to change his choice.
The door opened cannot be the player's initial choice nor contain a car.
The door opened by the host is underlined and the 'win' column is the prize if the player changes his choice. In the list g is game ID, door opened by host is underlined, and 'win' column is the prize if the player changes their choice. Player always chooses door 1 1st.
g 1 2 3 win
1 c a b b
2 c b a a
3 a c b c
4 b c a c
5 a b c c
6 b a c c
This supports Marilyn Savant's conclusion that by changing their choice the player will win a car 2/3 of the games.
The error.
When c is behind door 1, the host can open both door 2 and door 3, but only in separate games, per the game rules. The host opening both doors would instantly reveal the location of the car.
This adds 2 additional games.
7 c a b a
8 c b a b
The corrected probability is now 4/8=1/2 of games played.
The probability to win a car is the same with or without the option, 1/2.
The locations are fixed for each game. It is the player choices that vary due to the participation of the host. The host eliminates 1 of 3 locations, thus the player now only considers 1 of 2 locations, i.e. 50% chance. Phyti (talk) 15:16, 7 October 2024 (UTC)[reply]
I could not follow the game with the the die toss.
But will add this to the MH problem.
Setting aside the details of defining the list of games, consider the state of the game in time.
When the host asks the player initially to choose a door, the state was 'the car is behind 1 of 3 doors', and the probability of winning is 1/3.
When the host asks the player to choose a door after revealing a losing door, the state becomes 'the car is behind 1 of 2 doors', and the probability of winning must be 1/2. Phyti (talk) 16:20, 10 October 2024 (UTC)[reply]
This is a good illustration of the difficulty people have with conceptualizing this problem. Because this is a problem of conditional probability, evaluating the 'states' independently is incorrect. The conditions matter, because they provide information that would be lacking if one were to pick the problem up when there are only two doors remaining. MrOllie (talk) 16:25, 10 October 2024 (UTC)[reply]
Monty Hall logic;
modified game:
a=goat1
b=goat2
c= car
p1=player's first choice
p2=player's second choice
h1=host's first choice
h2=host's second choice
if p1=a then h1=b and p2=c
if p1=b then h1=a and p2=c
if p1=c then h1=a and p2=b
if p1=c then h1=b and p2=a
From Wiki article, 'standard assumptions'
Host may open 1 door per game if it is not p1and does not contain a car.
Example expressed as prize won by players choice.
1 2 3 p1 hi h2 p2
a b c a b c
b a c
c a b
c b a
If player stays with p1, 'win car' happens 2/4=1/2 of games.
If player switches to p2, 'win car' happens 2/4=1/2 of games.
There is no advantage.
From the Wiki article, 'Conditional probability by direct calculation'
Here is a corrected flow chart, in agreement with above table.
Nope, you're (again) leaving out the conditional probability, which requires some multiplication, not summation as you are doing here. Try out a simulator (many are on the internet) or play the game with a friend and a deck of playing cards and see what happens. MrOllie (talk) 17:35, 12 October 2024 (UTC)[reply]
First, it would be nice if word processing could be standardized. Posting here is like cooking an egg with a nuclear reactor! The values should be aligned, but they are not.
When the host asks for the player's first choice, the car is in 1 of 3 locations.
When the host asks for the player's second choice, the car is in 1 of 2 locations.
The host has eliminated 1 door,thus the players choices are 1 of 2. The probability changes due to the interaction of the host. Equivalent to asking the player heads or tails.
No need for complex mathbabble of probability theory.The game has its own dynamics. Knowing all possible games allows calculating the win frequency.
MrOllie, conditional probability does not apply in this case -- that is the illusion. The wording and storyline of the problem makes the listener assume that it does apply, when it in fact, does not. AI*girllll (talk) 07:15, 13 October 2024 (UTC)[reply]
Again, play the game and record what happens. Without applying conditional probability the mathematics will not match observed reality. MrOllie (talk) 11:54, 13 October 2024 (UTC)[reply]
@Phyti :
Flip a coin. If it shows heads, then place a 6-sided die so it shows 1, else roll that die.
What is the probability that, after doing the above, the coin is showing heads?
For states, consider the following:
There are 2 doors, and a standard 6-sided die. The host rolls that die. If it shows 1, then the host puts the car behind door 1, else the host [puts the car behind door 2, and then picks up the die and places it showing 1].
What is the probability that, after the host does the above, the car is behind door 1?
I believe in constant refinement.
'Monty Hall problem', Wikipedia,
There are 3!=6 sequences for arrangement of 3 prizes behind 3 doors.
The prizes are {goat1, goat2, car}.
For any 1 sequence there are 4 possible games (series of choices) by the player 'p' and host 'h' including an option for the player to change their choice.
For {g1 g2 car}:
If p picks g1, h picks g2, opt is g1 or car.
If p picks g2, h picks g1, opt is g2 or car.
If p picks car, h picks g1, opt is car or g2.
If p picks car, h picks g2, opt is car or g2.
If player stays with first choice, car is prize 2 of 4 games.
If player switches their choice, car is prize 2 of 4 games.
There is no advantage to switch.
The reason is the game rule restricting the host from revealing the car location.
(That would end the game)!
The host can only reveal the location of a goat, eliminating 1 door.
The player's choice is now 1 of 2 locations, a gain of 1/3 to 1/2 probability.
To JumpDiscont, it is not the math or the solving of equations that your argument is incorrect on -- your argument is modeling the word problem incorrectly. Your argument is mis-using (or perhaps mis-defining) conditional probability.
This becomes easier to see when you pare down the word problem and state it more simply, as follows:
"There are 3 doors. Behind one of them is a prize, behind the other 2 are duds.
Contestant says the number for the door he/she is hoping has the prize, but doesn't open it.
One of the doors with a dud is removed from play, leaving 2 doors: one with prize and one with a dud. We don't know whether the door the contestant picked has the prize or not, because neither door has been opened. What are the chances that the door that the contestant *didn't* pick has the prize behind it? "
With this simplification, it becomes more obvious:
Since we have no additional information about what is behind the door the contestant picked, the only change is that the contestant said the number, we can not and should not ascribe the odds from the contrstant's choice from the previous situation to the second situation.
Conditional probability is defined as when the odds of a first known choice (event) change the odds of the outcome of a second choice (event).
The odds are changed in this second selection (ie, the odds from the contestant's choice in the first situation -- what you are calling conditional probability -- do not apply) because the host may *only* remove a door with a dud, and never one with the prize. If the host were allowed to remove the door with the prize as well, then and only then might the 1/3 odds from the first situation (or conditional probability) accurately apply to the second situation.
Because we *don't know* what is behind the contestant's chosen door, and *do know* what was behind the door the host removed (a dud), the only conditional probability that can be applied to the second situation is based on the known host's choice.
Since we know that there is one less 'dud' (and one less door) in the available choices, the odds for the second situation are no longer based on 1/3...they are based on 1/2...as there is now one less door and one less dud.
Conditional probability in this problem is based on the known host's choice, which removes a known dud door from being calculated in the correct odds for the he second situation.
Your argument is using the term 'conditional probability ' incorrectly, or at the least modeling the 'conditional probability ' in this problem incorrectly.
That is the illusion in the Monty Hall Problem problem -- it alludes to readers that the unknown contestant's choice should affect the odds instead of the known host's choice.
You are using the term to say that the odds of the first situation should be carried over 'because conditional probability means the second situation is based on the first situation's odds' . What conditional probability is defined as when the odds for a second situation change based on a known first choice. In the Monty Hall problem, the contestant's pick is NOT the 'known first choice ' that the second situation's odds' are modified by -- the 'known first choice ' is that the host removed a dud door from the pool of choices.
So while there *is* conditional probability in this problem, your argument is incorrectly attempting to use the unknown contestant's choice as a basis for it instead of the hosts known choice. AI*girllll (talk) 04:26, 19 October 2024 (UTC)[reply]
Conditional probability is when the odds are modified by asked on the *outcome* of the first situation. The contestant's choice is not an outcome, as it is never revealed what the door hides.
A better example to explain conditional probability correctly is with playing cards. To calculate the odds of drawing 2 spades in a row from a full deck, with 13 of the 52 cards being spades, the correct equation is *NOT* simply 13/52 * 13/52 = .0625 as one might think at first glance. If one spade is drawn first as required by this problem, that card and spade is no longer in the deck to be drawn, so it changes the odds for a spade to be drawn a second time. Once the first spade is drawn, there are now only 12 spades in a deck of 51 cards. So once the first spade is drawn (as required by the problem) , the odds for the second draw are modified to be 12/51 instead of 13/52. So, the correct equation for two spades being drawn from a full deck is 13/52 * 12/51 = .05882 , and not .0625 . This is conditional probability.
How is it that no one else has said this before? I did a somewhat involved Google search and nothing of the sort of what I just said is even mentioned.
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If anyone wants to invite me on their podcast or talk show, DM me for my contact info. :)
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And, sorry to the owner of this page for disabusing you of certain illusions, it seems though you were/are rather attached to them. But, fear not -- it's still an immensely interesting word problem and tricky puzzle and is worthy of its own page just for the discussion and history surrounding it. I'm happy and proud that you thought it was interesting enough to inform ppl about it, and to spark this discussion! AI*girllll (talk) 04:55, 19 October 2024 (UTC)[reply]
Your pared-down version omits how the door-to-be-removed was chosen: If the contestant's door has a dud, was it a 50-50 which happened to select the door the contestant didn't, or was it already guaranteed that the contestant's door would not be the one removed?
The former goes against what I called the crucial assumptions - this is another chance for you to say so if you don't think your 1/2 answer applies even when those assumptions apply - so for the rest of this comment, I assume it was already guaranteed that the contestant's door would not be the one removed. With that in mind:
"Since we have ... previous situation to the second situation."
This is "Since [condition that makes X correct], we can not and should not [do X].".
If [the conditional probability parts of your block of comments] are due to the "given"s in my "The error is in your ..." comment, then pretend I wrote "after" instead of "given that" those two times: Although I indeed got the word "given" from conditional probabilities, that also works when one insists on using probabilities-in-situations instead.
As far as I can find, I have not used conditional probability anywhere else on this page. If you find another place I did use it on this page, then I can go further into the issue, but for now, the above "If your ... probabilities-in-situations instead." is my response to all the conditional-probability parts of your block of comments.
Would the following give you a 1/2 chance of winning the lottery?
You buy a ticket and show the ticket's sequence to Bob. You don't watch the drawing, and Bob does watch the drawing. Bob gets 2 pieces of paper, and writes your sequence on one of those pieces. If your sequences won, then Bob [chooses at random from all other sequences the drawing could've produced] and writes Bob's choice on the other piece of paper, else Bob writes the winning sequence on that other piece of paper. Bob shuffles those 2 pieces of paper, and then brings and shows them to you.
I looked and it was Mr. Ollie who kept saying conditional probability , not you, so you are correct on that, and I apologize for mistaking Mr. Ollie for you. But, that doesn't invalidate my argument. And, you keep giving different examples which dont prove your point and only muddle/offshoot the discussion.
I have time rn so I'll go into your lottery ticket example, and because I think it actually proves MY point. Say there is a lottery with 100 tickets. One, and only one of them, is a winner. I have a ticket and so does Bob. Neither Bob nor I have checked to see if our ticket is a winner, but the other 98 ticket holders have, and none of the other 98 tickets are a winner. What are the chances Bob is holding the winning ticket? It it 50% (as I am arguing, for the 50/50 odds)? Or does Bob's ticket have a 98% chance of being the winner simply because it is not mine[the one I originally picked] (as you are arguing, for the 33/66 odds [from the original problem, we could do this example with 3 tickets as well]) ? And, if you are still arguing that Bob's ticket has much much greater odds of being a winner than my ticket, how do you explain this when we look at this problem from Bob's perspective? Wouldn't that mean that, for Bob, *my* ticket has a 98% chance of being the winner? How do you explain the 2 remaining tickets in that lottery BOTH having a 98% chance of being the winner in that lottery?
.
Both of the two remaining tickets *cannot* both have a 98% chance of winning, because that violates one of the initial givens of probability which says that all probabilities add up to 100%. Probabilities only work with 'unknowns'; it's a shortcut used for practical purposes. But, another property inherent to probabilities is that they are fluid and shared among all unknowns. 'Knowns' , however, are discrete and not fluid. If we know what is behind a door, and that door/contents enters a new situation, it remains discrete -- what is behind it is still what is behind it. Just because someone 'picked' a door does not turn it into a known: what is behind it is still unknown. So it cannot keep its probability from the first situation as a known would. It is still an unknown, so it must share probabilities fluidly with the other unknowns when it enters the second situation. AI*girllll (talk) 15:40, 20 October 2024 (UTC)[reply]
That's not the problem here. Bob isn't offering you one other ticket, he is offering you every other possible ticket in aggregate. It's the same as the 'more doors' version MartinPoulter mentioned further up the page. If there are 10 total doors, Monty wouldn't be opening one door, he'd be opening eight. MrOllie (talk) 15:45, 20 October 2024 (UTC)[reply]
But, it's *not* in aggregate. That is incorrect: a false assumption. When you collapse the odds down due to removal of one unwinning door(or multiple unwinning tickets), the door that was originally picked *is still an unknown*, and must share in the unknown odds, instead of being kept separate/constant as a 'known' would.
That's what the stated problem is. One of the issues here (and indeed throughout all your attempts to argue on this page) is that you are answering problems that are not actually the ones stated. MrOllie (talk) 17:11, 20 October 2024 (UTC)[reply]
Your argument is adding that to the problem: your argument is assigning those increased odds incorrectly only to the door/ticket the contestant doesn't have. This is an incorrect assumption/assignment, and it is why that argument gives an incorrect answer and the illusion of increased odds. AI*girllll (talk) 17:15, 20 October 2024 (UTC)[reply]
That's the illusion: that the remaining door/ticket (the one you didn't pick) = the odds of all of the remaining doors except yours. You could just as easily assign those odds to the door you did initially pick. AI*girllll (talk) 17:12, 20 October 2024 (UTC)[reply]
That you misread the problem statement is not an 'illusion'.
To restate the 'More doors version'. You chose 1 door out of 10. Monty then closes eight of the non-winning remaining doors. He offers you a chance to switch to the remaining door.
And to rephrase that problem (math is exactly the same): You chose 1 door out of 10. Monty then offers you the chance to get the prize behind any of the remaining 9 doors. MrOllie (talk) 17:15, 20 October 2024 (UTC)[reply]
That is *your* addition or change of the original problem, lol. Monty is NOT offering the contestant all of the other doors/tickets in aggregate. The problem stated that Monty only offers the contestant to switch to ONE other door, not ALL/BOTH of them . That is where your argument is modelling the problem incorrectly. AI*girllll (talk) 17:19, 20 October 2024 (UTC)[reply]
When you chose that initial door, your odds of being correct were 10%. They do not subsequently improve regardless of the host's subsequent actions. The 'Remain' option will forever be 10%. Thus the 'switch' option is 100% - 10% = 90%. MrOllie (talk) 17:28, 20 October 2024 (UTC)[reply]
The error for that argument (I only call it 'yours' because it is the position you are arguing for) isn't in the math after the problem is modeled. The error of that argument is that the problem is modeled incorrectly. AI*girllll (talk) 17:29, 20 October 2024 (UTC)[reply]
Mr. Ollie, I hope you are not taking this personally -- it is not intended to be. I am trying to follow the argument for switching that you are defending, to see where the actual point of contention is, to try to better establish which argument is correct.
.
You stated that the argument for switching converting a great chance of winning, is because in the problem, Monty offers the contestant both of the other unknowns (which includes their 66 percentage of winning). I am saying that this is NOT the case, as anyone who re-reads the actual problem will see. In the problem, Monty opens one of the doors (turning it into a known instead of an unknown) , and offers the contestant the chance to switch to the [one] other unknown door.
Mr. Ollie, what I am trying to say is that the way the equation is set up is wrong, the equation is not modeling the actual problem presented here. The actual solving of the equation (ie, the math) is correct, but the equation is set up incorrectly for the problem presented. AI*girllll (talk) 17:36, 2 November 2024 (UTC)[reply]
Bear in mind that some of my examples were not at you: I believe that
[those which were not at you] either [did prove my point] or [would've if whoever they were at had answered].
regarding my examples which were at you:
I was trying find the relevant parts of your understanding of probability, since I figured you would have a realistic-enough understanding of probability that I could prove my point with a pair of very-similar examples. We probably would've gotten where we are now in fewer rounds of interaction if you had answered, rather than leaving me to try inferring what your answers likely would've been.
For your example, 50%. Whether-or-not I am "still arguing ... my ticket" depends on whether you're still referring to your example, or are back to referring to mine.
If you were still referring to your example, then:
No, because [which ticket is Bob's] was also independent of which was the winning ticket, and there was only a 2/100 chance of getting into the second situation (other 98 checked and none of those won). Before that, there was a 98/100 chance of the winning ticket being among the other 98: This corresponds to the host being "allowed to remove the door with the prize as well".
Accordingly, for the rest of this comment, I
assume that the rest of your comment was instead referring to either
my example exactly, or the simplification to the 100-ticket lottery
and
phrase all but the last paragraph of this comment, I explain for
the 100-ticket lottery, since real-world lotteries are analogous
.
with that in mind:
Yes, I am "still arguing ... my ticket", because Bob's choice of ticket depends on which ticket won, whereas yours did not:
Your ticket was chosen before [which ticket wins] was chosen, and the drawing was done with probability 1/100 each no matter which ticket is yours, so the winning ticket was independent of yours. Thus, if the winning ticket was independent of Bob's choice, then one would have
You are misunderstanding my point, and my example. I tried to post earlier, but kept getting the error msg that my ip address had been blocked and so I was not allowed to comment. It appears that has been resolved. I do not have the time to adequately read through your response rn, but will when I have time. AI*girllll (talk) 16:39, 21 October 2024 (UTC)[reply]
My understanding of probabilities is that , somewhat like *i* , they are used as a non-discrete placeholder in math -- a mathematical shortcut -- for the likelihood of an unknown ..and as such, they are uncertain , relative, fluid, and non- discrete... as they are a placeholder for an unknown (or quantum) state which, by definition, varies situationally. Probabilities are an imprecise mathematical shortcut to condense a quantum state into a variable so that it can be used in common mathematics.
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Probabilities differ from known, discrete numbers, as they are uncertain and defined as changing situationally according to the circumstances involved. A probability should not be treated as a certain known number.
Do you not have this understanding of probabilities? Or, do you have a different one?
.
.
The argument for an increased chance of switching keeps/carries-over the probability of the initial choice to the second situation as if it were a known (a discrete number that doesn't change situationally), when it is instead merely a probability, a fluid unknown that does change situationally. The chance of the door containing or not contacting the prize is merely a probability, which in a new situation does not remain static (as the switching-is-better argument assumes), but changes according to each new situation. Therefore, it is incorrect to consider the number from the probability of the contestant's choice in the first situation as static discrete and unchanged in the second. The 'probability of the contestant's choice containing a prize ' is by necessity defined by each specific situation....is by definition situational.. and cannot be staticly carried over (at least not correctly) to a new situation as an absolute, discrete, known value is.
By definition, probabilities are dependent and variable according to the specific situation.
The situation I described is the same as if there were 100 doors. But since you say you don't understand how it is similar (and I assume there are others who don't as well), I will go through my example with only 3 tickets so it is easier for you or others to see how it is the exact same situation as the Monty problem.
.
You buy a raffle ticket. So does your friend Bob. Good news: there are only 3 tickets in this raffle, so your ticket has a 1/3 chance of winning! [Ie, EXACTLY the same as the Monty Hall problem]. The raffle owner knows which ticket is the winner and before revealing which ticket is the winner, reveals that the third ticket is not the winner [ie, exactly the same as the Monty Hall problem], and asks you if you want to switch tickets with your friend Bob [ie, exactly the same as switching to the other door in the Monty Hall problem][ie, you originally had a 1/3 chance of winning, the other 2 tickets collectively had a 2/3 chance of winning, one of the other 2 tickets was revealed to not be the winner, do you increase your odds by switching tickets?]. According to the 'switching is better ' argument , your odds of winning would increase from 1/3 chance, to 2/3 chance of winning, if you switch tickets [according to that argument, the collective odds of the unknowns are transferred only the unknown you did not pick, . Also according to the 'switching is better ' argument, Bob as well would increase his odds from 1/3 chance of winning to 2/3 chance of winning by switching tickets with you, so he should be more than willing to switch.
But, as should be apparent, both you and Bob cannot realistically both have a 2/3 chance of winning just by switching tickets. In reality, only one ticket can be a winner. This reveals the error in the 'switching is better ' argument -- that you can't treat the probability of your door /ticket as discrete and certain while treating the probability of the remaining unknown choice as fluid and changing. You have to treat both probabilities the same: as the uncertain, situationally changing variables they are. AI*girllll (talk) 11:03, 22 October 2024 (UTC)[reply]
Apparently the blocking error message can happen randomly. If someone on the same ip address (ie wifi, or cellular service) was a jerk, they will just block that address, so anyone else using it can't post. AI*girllll (talk) 11:07, 22 October 2024 (UTC)[reply]
I realize I was I was imprecise:
I was simply using the situation/circumstances just
[before "probability" and the event] or [after "probability" and the event],
rather than putting the situation / those circumstances between "probability" and the event.
That said, it seems to me that altough you've not done so with "probability" specifically,
you've similarly left the situation outside for "chances" - "What are the chances Bob is
holding the winning ticket? It it 50% (as I am arguing, for the 50/50 odds)?" - although this
may have just been you accidentally following how I was typing my probability claims/questions.
It's like I said back this spring, you are weight the choices incorrectly, you are weighting 2 choices as 1, instead of 2 as they should be weighted. So you're getting 2/3 (= 67/100) odds to switch when you should be weighting them individually and getting 2/4 (=50/50) odds. AI*girllll (talk) 16:31, 25 October 2024 (UTC)[reply]
Another way to put it simply:
If you remove the possibility of the third door (the one Monty opens) (or the ticket revealed to not be the winner) containing the prize, there are now only 2 options for the prize to be in.
The probability/percentage of possibility that the third door previously contained when it was unknown gets transfers equally between all remaining unknowns. The door that was 'picked' remains an unknown (it does not retain any special status by being picked), so its probability doesn't remain static from the previous situation. AI*girllll (talk) 16:58, 25 October 2024 (UTC)[reply]
What are your answers to questions (a) , (b) , (c) from my "3-ticket raffle" comment?
I have a guess, but my writeup in that case will be fairly long, so I'm not
typing it up without confirmation that your answers are what I'm inferring.
There has to be a reason for Marilyn Savant's solution.
(This material in this article was originally published in PARADE magazine in 1990 and 1991.)
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 #1, and the host, who knows what’s behind the doors, opens another door, say #3, which has a goat. He says to you, "Do you want to pick door #2?" Is it to your advantage to switch your choice of doors?
Craig F. Whitaker
Columbia, Maryland
Her reply.
Yes; you should switch. The first door has a 1/3 chance of winning, but the second door has a 2/3 chance. Here’s a good way to visualize what happened. Suppose there are a million doors, and you pick door #1. Then the host, who knows what’s behind the doors and will always avoid the one with the prize, opens them all except door #777,777. You’d switch to that door pretty fast, wouldn’t you?
Arrangement as presented by Marilyn Savant. Door opened by host prefixed with '-'.
1 2 3 opt
g –g c c
g c –g c
c –g –g g
list.1
Player wins car 2 of 3 games.
When car is behind door 1, host can open doors 2 AND 3.
She considers the 2 options for ‘g’ as 1 game,
which violates the ‘standard assumption’ rules. That is her error.
The host cannot open 2 doors in 1 game, since that would reveal the location of the car, the game would end, and the player would not get his option.
1 2 3 opt
g –g c c
g c –g c
c –g g g
c g –g g
list.2
When car is behind door 1, host opens door 2 OR 3.
There are 4 possible games.
Player wins car 2 of 4 games.
If there were 100 doors, the probability of the player to choose the door with the car is
1/100. After the host opens 998 doors revealing goats, the remaining closed door cannot have its original probability of 1/100, since there are now only 2 doors in play. For the player’s second choice, they win a car or a goat, a random choice since they still don’t know the location of the car. phytiPhyti (talk) 16:41, 26 October 2024 (UTC)[reply]
AI*girllll should see my previous comment.
The rest of this is @Phyti .
for my "rolls a 3-sided die" game:
When die shows 1, coin can land showing heads AND tails.
Do you think those 2 should be considered as one game?
if no: What if the "The host comes ... coin is showing." part was removed?
Phytu -- yes, that is what I was saying,--- that argument for switching (Marilyn vos Savant's argument), is wrong because it is modeling the problem incorrectly. All the examples and computer programs that support her argument are also copying her incorrect model, so of course they come to the same incorrect answer. AI*girllll (talk) 23:44, 26 October 2024 (UTC)[reply]
Phytu -- apparently I'm the first person in 50 YEARS to not just say she's incorrect but follow it through to say WHERE and WHY her argument is incorrect. :) AI*girllll (talk) 23:47, 26 October 2024 (UTC)[reply]
@AI*girllll : In case "Phytu" was something other than a typo for you trying to indicate
that you were just replying to Phyti - rather than also or instead to me - I am pointing out
that you have not yet answered questions (a) , (b) , (c) from my "3-ticket raffle" comment.
JumpDiscont -- the validity of my argument (and my pointing out the error in Ms. vos Savant's) is not dependent on answering questions on situations you created which have nothing to do with the original problem. I picked one as an example because it was simple to transfer and it matched the original situation exactly. If you want me to spend 100 hours in the next week walking you through what should be a simple exercise for you to do by yourself, I'd love to for a fee. Otherwise, I am doing this in my spare time, which currently has many demands on it. Or, as I said towards the beginning of this thread, if you can find an error in my logic or argument as to why Ms. vos Savant's argument is incorrect, please don't hesitate to point it out. AI*girllll (talk) 16:11, 30 October 2024 (UTC)[reply]
"the validity ... with the original problem."
Indeed, and the same applies to problems that
are almost identical to the original problem:
Rather, that is a way of testing validity of
someone's arguments - to get valid or invalid -
by seeing if their reasoning can give coherent answers.
But since you say the situations I created have nothing
to do with the original problem, I will go through (c)
from my "3-ticket raffle" comment so you can see how
it is a similar situation to the Monty Hall problem.
You buy a raffle ticket. Good news: There are only 3 tickets in this raffle, so your ticket has a 1/3 chance of winning! [Ie, EXACTLY the same as the Monty Hall problem]. Robert knows which ticket is the winner and instead of revealing which ticket is the winner, puts the 2 you didn't choose in a pile so that the bottom ticket is not the winner [even if Robert doesn't reveal that the pile's bottom ticket is not the winner, it nonetheless can't be the winner, and although I do claim the problem has the same answer under what I call the crucial assumptions, even under those I'm not claiming this is exactly the same as the Monty Hall problem]. According to the 'switch-or-not doesn't matter' argument, what Robert did increased your odds of winning from 1/3 chance, to 1/2 chance of winning. [According to that argument, the odds are now equal for the ticket you bought and the pile's top ticket, just like for the door you picked and the closed door from the other 2 after Monty opens one of those other 2.]
Figuring out what I think are the correct answers
to my questions certainly is a simple exercise
for me, but I'm asking for your answers:
Whether-or-not your's are the correct answers
turns on the validity - namely, valid or invalid -
of your argument, which is what this discussion is about.
Note that before your most recent pair of comments,
my guess was that your answers would be 1/3 , 1/2 , 1/2
in that order, due to you thinking (b) and (c)
were essentially just rephrasings of the MHP.
In particular, it now seems to me that the writeup
I was planning for that case would've been a waste.
(though you've still not answered them, so ...)
"if you can ... point it out."
Is this my [logic or argument] as to ...
or my logic or [argument as to ...] ?
(i.e., are you referring to your logic separately, or to
those two things both applying "to why ... is incorrect"?)
If it's my [logic or argument] as to ... , then:
which of her arguments?
the [logic or argument] from which of your comments?
If it's my logic or [argument as to ...] , then:
(This material in this article was originally published in PARADE magazine in 1990 and 1991.)
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 #1, and the host, who knows what’s behind the doors, opens another door, say #3, which has a goat. He says to you, "Do you want to pick door #2?" Is it to your advantage to switch your choice of doors?
Craig F. Whitaker
Columbia, Maryland
Her reply.
Yes; you should switch. The first door has a 1/3 chance of winning, but the second door has a 2/3 chance. Here’s a good way to visualize what happened. Suppose there are a million doors, and you pick door #1. Then the host, who knows what’s behind the doors and will always avoid the one with the prize, opens them all except door #777,777. You’d switch to that door pretty fast, wouldn’t you?
Using truth tables
The national census issues a statement that the average family has 2.3 children.
No one would expect to visit a home and find .3 of a child. It's an abstraction in the world of mathematics and not a literal fact.
Thus we use a different method of finding the truth.
Instead of 'What is the probability of an event?'
we ask 'Is the event possible?'
If 'the prize is {a car}' is true, its value is 1.
If 'the prize is {not a car}', its value is 0.
The player's initial choice is always door 1. The host can only open a door if its value is 0 and it is not the player's choice. 'op' is the optional closed door.
The error
Arrangement of prizes as presented by Marilyn Savant using truth values 1 for a car and 0 for a goat.
1 2 3 op
0 x 1 1
0 1 x 1
1 x x 0
list.1
When a car is behind the initial choice (door 1), the host can open doors 2 and 3, which requires 2 games. Savant considers this case as 1 game.
This excludes 1 win a goat result and distorts the win a car ratio.
The host cannot open 2 doors in 1 game, since that would reveal the location of the car, end the game, and deny the player their option.
Using probabilities she concludes the ratio of (win a car)/(games played) is 2/3, for an average of 2/3 of a car or 1/3 of a goat per game. This contradicts the reality of possibilities as 1 car or 1 goat.
1 2 3 op
0 x 1 1
0 1 x 1
1 x 0 0
1 0 x 0
list.2
When a car is behind the initial choice (door 1), the host can open doors 2 and 3, which requires 2 games. Thus there are 4 possible games.
Player wins a car in 2 games and wins a goat in 2 games, with a win car ratio of 2/4.
Conclusion
There is no advantage to switch choices.
The value of any sequence of the set {0, 0, 1} is 1 since it contains a car.
The value of any element in the sequence is 0 or 1, i.e. there are no fractional values.
As in the case of the census, no door can have 1/3 or 2/3 of a car.
This is the extended version as offered by Savant with 100 replacing 1M.
If there were 100 doors, the representation would be a string of 99 0's containing a single '1'.
000000000010000000000000...
The staff could use a 100 cycle method to place the car left to right for each game, ensuring an equal opportunity for each door.
The (truth) value of the set of 99 goats and 1 car is the same as for 1 goat and 1 car, {0, 1}, since n 0's = 0 .
After the host opens 98 doors revealing goats, there are 2 possible ordered sequences remaining, 0 1 or 1 0, each containing a car and a goat with a value of 1. The player wins a car in 1 game or wins a goat in the other game, with a win car ratio of 1/2. Adding more doors with a value of 0 does not change the value for the set, it just requires a longer (and boring) game to progress to a 2-door game.
The player’s choice is always random since they never know the location of the car when choosing.
The issue is the properties of a set of elements vs the properties of its elements.
Example:
Actuarial statistics as used for insurance purposes predict 10% of a population will die within 10 yrs. It cannot predict which individuals. Each person can only be alive or dead within that period. No one will be 10% dead. Phyti (talk) 18:19, 28 October 2024 (UTC)[reply]
for my "Flip a nickel" game:
When the nickel shows tails, the dime can land showing heads and tails.
Does this require 2 games? Do you consider this case as 1 game?
If [no,yes in that order] and [your explanation is that MHP
["is a dynamic process with changing factors as it progresses"
or "evolves over it's duration" or something similar]
whereas my "Flip a nickel" game isn't/doesn't], then:
Do you regard my "rolls a 3-sided die" problem as
Phyti -- yes, no one can be 10% dead, they have to be with alive or dead. Probabilities are not real concrete numbers -- and you cannot carry over the imaginary, situationally -dependent number from the first problem, into the second, different problem, without introducing gross errors. AI*girllll (talk) 16:16, 30 October 2024 (UTC)[reply]
Monty Hall game
difference between possible and probable
1__2__3__set
0__0__1___1_____________possibilities for each door and the set
0__1__0___1
1__0__0___1
1___2___3____set_______player chooses door 1
1/3_1/3_1/3___ 1________equal probabilities for each door
1/3_1/3__0____ 1________host opens door 3 revealing a goat
1/2_1/2_______ 1________set=2 doors with equal probabilities for a car or a goat
The host can only open a door containing a goat and not the door chosen by the player.
When he does, that door is no longer available leaving 2 closed doors for the player.
The player can only choose a closed door.
The reality (not probability) is the set still contains a car which has a truth value of 1, which is also the value of the set. The 2 doors must have probabilities of 1/2 each. It's the same case for the player of H or T.
Since rigged game shows were investigated in the 1950's, they are less likely to use deceptive tactics today. Viewership, sponsorship, and ratings are more important.
Today my answer to Whitacker's question of strategy would be 'there is no strategy'. It is a game of chance since the player has no knowledge of the car location relative to the 3 doors, thus they can only GUESS. Phyti (talk) 18:01, 2 November 2024 (UTC)[reply]
The latest analysis.
Monty Hall game
An analysis of the general game.
Set of prizes = {a, b, c}, with c a car, a & b of lesser value.
'g' is game number.
x = prize behind player choice of {door 1}
y = prize eliminated by host opening {door 2, door 3}
z= prize behind remaining door
g x y z
1 a b c
2 b a c
3 c a b
4 c b a
Ratio of car prize/games played = 1/2, does not change using the option.
In the case of considering a and b identical prizes, the list of games becomes
g x y z
1 a a c
2 a a c
3 c a a
4 c a a
Ratio of car prize/games played = 1/2, does not change using the option.
Marilyn Savant considered games 3 and 4 as a single game, but not games 1 and 2, introducing a bias which alters the win car ratio to 2/3 using the option. Phyti (talk) 18:21, 8 November 2024 (UTC)[reply]
Your 4 games are not equally likely:
See my "Flip a nickel" example, and the part of my
After analyzing the problem using different methods, I try to present the solution in the simplest possible form.
If the statement 'this location contains a car' is true, its value=1.
If the statement is false, its value=0.
For any game there are two 0's and one 1.
There are 3 events per game.
e1. Player who does not know location of car, makes 1st choice as door 1.
e2. Host who knows location of car, opens door 2 or 3 with value 0 and not the player's choice.
e3. Player makes 2nd choice between door 1and 'op'.
With g the game number and 'op' the other closed door, the list is
g 1 2 3 op
1 0 0 1 1
2 0 1 0 1
3 1 0 0 0
4 1 0 0 0
Game 3 requires a 2nd host opening as game 4.
The ratio of (win a car)/(games played) is 1/2, with or without the option.
If the solution depended on all 3 possible locations of the prizes, games 1 thru 3 create a bias which alters the win car ratio from 1/2 to 2/3 using the option.
The 3 events per game resulting in 4 possible choices for the host, decide the question of strategy asked by Whitaker.
If the door numbers are changed to {231, 321, etc.} the choices remain the same. Thus the player's 1st choice is irrelevant.
1M doors would be represented as a sequence of (1M-1) 0's containing a 1 at a random location. If the host opened all doors containing zeros except one, the player's 2nd choice would be between (0 1) or (1 0). A 50% chance of winning a car. Thus number of doors is irrelevant. Phyti (talk) 21:06, 11 November 2024 (UTC)[reply]
The discussion above is closed. Please do not modify it. Subsequent comments should be made on the appropriate discussion page. No further edits should be made to this discussion.
