Talk:Monty Hall problem/Arguments

Wrong assumptions: 2/3 win probaility only works if change decision is taken A PRIORI of opening door 3

For the 2/3 win probability to work it is necessary to state a 4th hypothesis, which sorry, NEVER happens in this kind of TV shows:

4) The contester must decide if switch or keep the same door BEFORE Monty opens the door 3.

Of course, in this case he has 2/3 probability of winning if he CHANGES because there are 2/3 probability that the car is behing either door 2 or 3. Then, if he decides A PRIORI that he will change to EITHER door 2 or 3 DEPENDING on what door Monty opens (that will be either door 2 if the car is behind 3 or viceversa if the car is behind door 2) THEN by just saying "I WANT TO CHANGE" he/she is choosing the BEST OF TWO OPTIONS OUT OF TOTAL THREE, and he gets a a 2/3 win probability.

BUT....

IF Monty ALREADY OPENED DOOR 3 and THERE IS A GOAT THERE, THEN, with the GIVEN set of information, he KNOWS that the car is EITHER behind door 1 or 2. And his probability will be 50%. The conditional tree is WRONG because it is valid for the moment BEFORE THE DOOR 3 IS OPENED AND THERE IS A GOAT THERE. When the door 3 is opened AND there is a goat there, then the conditional tree changes completely and all assumptions with door 2 being opened must be set to PROBABILITY ZERO. — Preceding unsigned comment added by 186.23.133.28 (talk • contribs)

Excellent, 186.23; I also deduced that solution, as probability 0+1/2, because the first choice is always a loss (never wins car at 1st choice) which shows goat door. People make the false analogy as 2 chances to win, but first choice only shows goat, as reset 0% chance to win. The issue is false assumptions, as in "Kits, cats, sacks and wives, how many were going to St. Ives?" (not how many were met along the way). -Wikid77 (talk) 06:11, 8 May 2017 (UTC)

@Wikid77: Did you read the rest of this section, in particular my reply just below? -- Rick Block (talk) 15:36, 8 May 2017 (UTC)

Yes, I've read the other replies, and all 24 possible game moves must be counted, as in a truth table to examine all 24 moves and ensure complete coverage of all possibilities. The initial choice of doors, to always show a goat, is a "red herring" to distract from the reality of only 2 choices left, a 50-50 option, and any talk of thirds (1⁄3 or 2⁄3) fails to calculate the probability of winning after the first choice as 0, because the probability of being shown a goat after the first choice is 1. Based on the initial move, as showing the goat, then all remaining probabilities are 0.5 as 50-50 chance of winning the car whether switch to the other door or not. -Wikid77 (talk) 21:58, 11 May 2017 (UTC)

is a "red herring" to distract from the reality of only 2 choices left, a 50-50 option. @Wikid77: You're committing the fallacy of begging the question: assuming the result that you are trying to demonstrate. It is true that at the end there are two choices left, but you are asserting that they are equally probable without proof. You're excluding the possibility that the two choices have different probabilities. The probability of a random experiment depends on how it has been set up, and the previous steps in the Monty Hall problem are relevant; the doors haven't been set up as a "car randomly placed behind two doors with equal probability", so you can't assume that they have equal probability. You can't simply discard the set up of a random experiment and call it a red herring. Diego (talk) 10:26, 12 May 2017 (UTC)

You had it right up to the point where you say "And his probability will be 50%". That is a non sequitur to everything said before. The door not chosen at the beginning of the game has double the probability of the chosen door, because it accumulates the probabilities from the two non-chosen doors (since the door opened by Monty can no longer have a car behing it). When the door is opened, it doesn't change the fact that the non-chosen door is the best of two options out of tree - the only difference is that now we know which those two options the user is getting. Diego (talk) 15:59, 14 November 2016 (UTC)

Exactly - everything you say is correct, except for the last couple of sentences. If you throw all the information you know away, and restart knowing only that the car is behind one of door 1 or door 2, you do end with with probabilities of 50%. But that is NOT what happens. You know the original probabilities were 1/3 for each door. You know the host MUST open door 3 if you've picked door 1 and the car is behind door 2. You can assume the host picks evenly between door 2 and door 3 if the car is behind the door you've picked (door 1). You can and should use this information. Let me fix your last paragraph for you:

If Monty already opened door 3 (deliberately showing a goat there), then with the given set of information, the player knows the car is either behind door 1 or door 2. What we need to determine are the conditional probabilities given the host has opened door 3. For each door, this is the starting probability (1/3 for each door) times the probability the host opens door 3 if the car is behind that door divided by the total probability the host opens door 3. Door 3 is easy. We know the conditional probability the car is behind door 3 is 0 (because we can see a goat there). But working through the computation is pretty easy, too. If the car is behind door 3, the probability the host opens door 3 is 0, so we get a conditional probability of 1/3 * 0 divided by something - so this is going to end up 0. Door 2 and door 1 are a littler harder, but not much. If the car is behind door 2, the host must open door 3 (cannot open door 1), so the composite probability of the car being behind door 2 AND the host opens door 3 is 1/3 * 1 and the conditional probability is this answer divided by something. If the car is behind door 1, the host can open either door 2 or door 3. Assuming the host is indifferent, then the composite probability of the car being behind door 1 AND the host opens door 3 is 1/3 * 1/2 (and the conditional probability is this answer divided by something). The total probability the host opens door 3 is the sum of the composite probabilities, i.e. 1/3 + 1/6 which is 1/2. Now, we can compute the conditional probabilities. For door 1 we get (1/3 * 1/2) / 1/2 which is 1/3. For door 2 we get (1/3 * 1) / 1/2 which is 2/3.

The key here is that if the car is behind door 2 the host MUST open door 3, while if the car is behind door 1 the host can open either door 2 or door 3. This means the chance the car is behind door 1 AND the host opens door 3 is exactly half the chance the car is behind door 2 AND the host opens door 3. These are the only possibilities, so call them X and 2X. We know X+2X must be 1, so X must be 1/3 and 2X is therefore 2/3. -- Rick Block (talk) 16:55, 14 November 2016 (UTC)

Proposed Solution and Computer Simulation to show evidence

After having an epiphany at 2:00AM and writing computer models and flow charts all night, I have determined that the page https://en.wikipedia.org/wiki/Monty_Hall_problem has several flaws. Mainly that according to the page you should have a 2/3rd chance of winning if you switch, and a 1/3 chance of winning if you don't. I using deductive reasoning and computer models have come to the conclusion that if you don't switch you have a 1/3 chance of winning, and if you do switch you have a 1/2 chance of winning.

https://github.com/McClainJ/Monty_Hall_Simulation http://i.imgur.com/PoJzAoG.jpg <- Yes, I know the flow chart is not the prettiest thing, but the simulation doesn't seem to have any errors so I am forming this as my theory. This simulation does the exact process described by the problem, it does not perform a mathematical simulation but does perform a procedural one. Therefor the results are provably accurate. I ran each simulation 5 times for a total of 50,000 cycles for both switch only and never switch.— Preceding unsigned comment added by Electroninja (talk • contribs) 02:06, 9 February 2017 (UTC)

I haven't reviewed your code, but you're clearly doing something wrong. If you have 1/3 chance of winning if you don't switch and 1/2 chance of winning if you do switch where is the other 1/6? It is perhaps not entirely coincidental that 1/2 + 1/6 = 2/3.

Looking at your flowchart, in the "switch" case you show two doors left where the car might be. Didn't the host open one of these doors? In particular, if you have initially picked a goat (with probability 2/3), isn't the only option available when you switch the car? The flowchart needs to be a little more complicated, perhaps somewhat more like the tree diagram in this section of the article. -- Rick Block (talk) 08:09, 9 February 2017 (UTC)

In regards to the flow chart, I agree it is not the best way to organize it, the top two on the right are not a representation of the top two in the middle, they are meant to represent the 2 unopened remaining doors.

Wait, you didn't review the code, but assumed I did something wrong, do you see the problem with that?

Just check the code, it is not even 130 lines of code, and I guarantee that it plays the game perfectly. (just change the to_switch value when you run the code to true or false for the various simulations.)

And I think the main difference between your analysis and mine, is that you see the choice as a single linear chain, where it is realistically 2 separate decisions, not from which door to pick. but whether to enter into a separate decision. If you don't switch , then you never entered into the second decision, which explains the 1/3rd win rate, but if you do switch then you are entering into a scenario when you have chosen 1 of 2 results, which gives you the 1/2 chance of winning. Both of our lines of logic seem to follow deductive reasoning, and since there are separate results, something is going on. — Preceding unsigned comment added by Electroninja (talk • contribs)

I've had a quick shufti at the code, and admittedly, I'm a beginner with C languages. in enact_Switch When the car is behind Door 2 and the choice door is 2 you have the code Doors[(2 - rand()%2 )].goat_Revealed = true; Which opens the door 1 or 2, it should be Doors[(1 - rand()%2 )].goat_Revealed = true; to open 1 or 0. I think this will account for most of the discrepancy in your switching data.

I've included a VB coding of the game and the results from a hundred million runs that I just ran.

VB simulation

Option Explicit 'SwitchWin(33,334,712) SwitchLose(16,678,126) WinIfSwitch(0.666523104) 'StayWin (16,658,911) StayLose (33,328,252) WinIfStay (0.333263782) Sub MontyHall() 'run counter Dim run As Long 'car and choice and revealed placement Dim car As Byte Dim choice As Byte Dim revealed As Byte 'totals Dim switchwin As Long Dim switchlose As Long Dim staywin As Long Dim staylose As Long For run = 0 To 1000000 'Choose a door for the car car = Int(Rnd * 3) 'Contestant chooses a door choice = Int(Rnd * 3) 'Host opens door Select Case car 'Where car is 0 Case 0 Select Case choice Case 0 revealed = Int(Rnd * 2) + 1 Case 1 revealed = 2 Case 2 revealed = 1 End Select 'Where car is 1 Case 1 Select Case choice Case 0 revealed = 2 Case 1 If Int(Rnd * 2) = 1 Then revealed = 0 Else revealed = 2 End If Case 2 revealed = 0 End Select 'Where car is 2 Case 2 Select Case choice Case 0 revealed = 1 Case 1 revealed = 0 Case 2 revealed = Int(Rnd * 2) End Select End Select 'Randomise switch; 0 stay 1 switch If Int(Rnd * 2) = 0 Then 'If you stay then you win if you're on the car If car = choice Then staywin = staywin + 1 Else staylose = staylose + 1 End If Else 'switching is going to the door which is neither the choice nor the revealed 'all sum to 3 so that door is 3-choice-revealed choice = 3 - choice - revealed If choice = car Then switchwin = switchwin + 1 Else switchlose = switchlose + 1 End If End If Next run Cells(1, 1) = switchwin Cells(1, 2) = switchlose Cells(1, 3) = switchwin / (switchwin + switchlose) Cells(2, 1) = staywin Cells(2, 2) = staylose Cells(2, 3) = staywin / (staywin + staylose) End Sub

— Preceding unsigned comment added by SPACKlick (talk • contribs)

^thank you so much for helping me fix that small typo, I ran the simulation again 5 times after the fix, and the results are the same still 50/50 if you switch, 1/3 if you don't. I have updated github with the more accurate code. I fear I do not know enough about VB to analyze your code, most of it seems to not make sense. I will have to look up operators for VB and see what I am missing. is * your modulus operator? Your code may not have copied correctly. — Preceding unsigned comment added by Electroninja (talk • contribs) 05:22, 10 February 2017 (UTC)

* is the multiplication operator. To roughly explain the code in English:

Define Variables, start a for loop for 100,000,000 runs of the game

Rnd is a random number between 0 and 1. Int is the integer part of it. So Int(rnd * 3) is a random number between 0 and 2

This is used to pick a random car and choice

Select Case picks based on the chosen variable. So it's like your reveal goat function. Where both choice and car are the same it randomly selects one of the other two doors otherwise it picks the only available door and so sets revealed to that door.

It then randomly selects whether to switch or stay. If it switches then it uses the property that the sum of all three doors is 3, the door you can switch to is the one that isn't the chosen door or the revealed door so 3-choice-revealed = remaining door. (say choice was 1, and revealed was 0, 3-1-0 = the door you switch to.

It then checks if the final chosen door is the same as the car. If it is, it adds one to the win, if not it adds one to the loss

The final section of code just prints the results to an excel spreadsheet.

Also, I've spotted a second flaw with your code Where you have

if(Doors[1].goat_Revealed ){ if(Choice == 2){ Choice = 1; }else { Choice = 0;

you should have

if(Doors[1].goat_Revealed ){ if(Choice == 2){ Choice = 0; }else { Choice = 2;

You're switching from door 2 to door 1 if door 1 is open. I've run your code with the correction and it doesn't produce 50/50 any more. It produces a clear 2/3. — Preceding unsigned comment added by SPACKlick (talk • contribs)

Thank you so much for helping me find this dude. Finally, the universe makes sense again. I am gonna make the edits and publish this a Monty hall simulation. Link is posted here in C++ to help those coding simulate this problem. This is what peer review is for, finding code errors that occurred when writing code at 3:00am, and restoring balance to the universe.

https://github.com/McClainJ/Monty_Hall_Simulation/blob/master/main.cpp — Preceding unsigned comment added by Electroninja (talk • contribs) 06:55, 10 February 2017 (UTC)

What if another player picks the other door?

I would know what would happen if another player in the same Monty Hall game had chosen other door (not the one open by the presenter) and given the opportunity to change door also, which would be the probability for him, and the sum of probabilities for both two players?.88.20.162.142 (talk) 03:42, 4 May 2017 (UTC)

Since you don't know which door the host will open until he opens it, the other player only gets to pick a door after the host opens a door. In this case, the other player's door has a 2/3 chance of hiding the car - so this player has a 2/3 chance of winning the car by not switching and a 1/3 chance by switching. The sum of the "not switch" chances are 1/3 + 2/3 = 1 (if neither player switches, one of them will win the car for sure). The sum of the "switch" chances are 2/3 + 1/3 = 1 (if both switch, one of them will win the car for sure). -- Rick Block (talk) 14:59, 4 May 2017 (UTC)

Explaining switch/stay is 50% not 2/3

Table 1a: Truth Table of Monty Hall Problem
door with car	1st door try	open door	2nd door try	result	Expected cases out of 900
1	1	2	stay	win	25
1	1	2	switch	lose	25
1	1	3	stay	win	25
1	1	3	switch	lose	25
1	2	3	stay	lose	50
1	2	3	switch	win	50
1	3	2	stay	lose	50
1	3	2	switch	win	50
2	1	3	stay	lose	50
2	1	3	switch	win	50
2	2	1	stay	win	25
2	2	1	switch	lose	25
2	2	3	stay	win	25
2	2	3	switch	lose	25
2	3	1	stay	lose	50
2	3	1	switch	win	50
3	1	2	stay	lose	50
3	1	2	switch	win	50
3	2	1	stay	lose	50
3	2	1	switch	win	50
3	3	1	stay	win	25
3	3	1	switch	lose	25
3	3	2	stay	win	25
3	3	2	switch	lose	25
Totals: 12 stay, 12 switch Wins if switch: 6 switch win in 300 of 900 cases. Wins when stay: 6 stay win in 150 of 900. Odds: 300 to 150, or 2:1, 2⁄3 if switch.
Discussion: That is a ridiculous FAKE TABLE, showing only 24 of 36. Missing 12 are: 6 stayings lose, no staying wins, and 6 switching WIN, no swith loses. --Gerhardvalentin (talk) 19:18, 18 May 2017 (UTC) This is a "truth table" of all possible (true) moves in the game. Perhaps create another table, below, of 36 cases (not including 12 which could never occur where host opens door with car). You will realize only 24 possible gameplays. -Wikid77 (talk) 19:14, 28 May 2017 (UTC)

I have been updating the page to explain the 2/3 paradox as a false analogy to a game where contestant gets 2 tries to win, whereas here the "first try" is always a loss (car not awarded) where host shows a goat instead of player's chosen door. One source mentions probability of showing goat is "1" (and probability of winning car is 0% on first choice). The actual probability of both choices is thus 0+1/2 as 50% regardless of switch/stay at door. However, some readers might need more explanation, so I will add a full truth table (of all possible choices) to calculate the precise answer as 6 switches can win car, and 6 stays can win car, as 6 to 6 or 50% regardless of stay/switch choice of door. Similar brain teasers have been solved this way for over 50 years. See truth table at right.

The mindset of the host should also be explained. In fact, game host Monty Hall typically had only 1 zonk door (or "goat"), where the 2nd door was a medium prize (such as TV+sofa set), against a bigger prize behind 3rd door. However more sources would be needed to document the typical middle prize in the original gameshow. -Wikid77 (talk) 05:54, 8 May 2017, Update: However, the probability of choosing the next door is limited by the first choice, and the truth table column of "900 cases" shows the switched door gains double the wins, compared to stay, as the weighted probability of 300 to 150 wins by always switching (as suggested in comments further below). -Wikid77 (talk) 19:14, 28 May 2017 (UTC)

In the table at right, counting possibilities cannot be used to determine probabilities because the events are not independent. Observe that in the eight cases where door #1 has the car, the first try is door #1 four times. There is no justification concluding that the first try is 50% likely to be correct at the outset, before anything else has happened. The fact that there are more choices of door to open if the first guess was right does not imply that the first guess was more likely to be right. ~ Ningauble (talk) 14:00, 8 May 2017 (UTC)

Well, the truth table at right shows all possibilities, based on the dependent events as the game is played, and counting the various wins will confirm there are 6 wins by switch and 6 wins by stay. Hence, by simulation of the entire game, according to the example given, the results are 50-50 as 50% chance of winning car, whether the contestant chooses to switch the door or not. The initial choice of doors, to always show a goat, is a "red herring" to distract from the reality of only 2 choices left, a 50-50 option, and any talk of thirds (1⁄3 or 2⁄3) fails to calculate the probability of winning after the first choice as 0, because the probability of being shown a goat after the first choice is 1 (the 1st choice is never actually a "1st chance" to win the car). -Wikid77 (talk) 21:47, 11 May 2017 (UTC)

You have failed to demonstrate that all entries in the truth table are equally probable. Your argument is incomplete. Diego (talk) 10:31, 12 May 2017 (UTC)

@Wikid77: In particular, fill in the new column I added to your table above. Think about 900 instances of the game. The car is placed randomly, so it will be behind the each door 300 times. The player picks the initial door randomly, so of the 300 times the car is behind door 1 the player picks door 1 100 times. Of these 100 times, the host can open either door 2 or door 3. Picking randomly which door to open (in this case), means the car is behind door 1 AND the player picks door 1 AND the host opens door 2 50 times. This is the number in the top (and second) row. Please try to fill in the rest. Since your table has both switch and stay (for each case), the sum of the numbers in this column should be 1800 (i.e. each case appears twice). -- Rick Block (talk) 15:31, 12 May 2017 (UTC)

As suggested by User:Rick Block & User:Diego_Moya, I have added the relative probabilities as truth table column "900 cases" (in Table 1a above) to show the weighted probabilities of each gameplay, as switch 300 to 150 stay, proving the 2:1 advantage of always switching doors, tallied among all possible moves. Thanks for feedback. -Wikid77 (talk) 19:14, 28 May 2017 (UTC)

I haven't read your argument in detail, but I agree with your conclusion. The logic tree used by vos Savant depends on there being no new information added during the process. That isn't the case. The host adds new information by revealing a goat, and the logic tree should be restarted at that point. Your original choice then becomes irrelevant, because a) you are allowed to change it, and, most importantly, b) it was never revealed. You are faced with two doors, one of which hides a car, the other a goat, and you have no information as to which is which. Therefore it doesn't matter whether you change or switch. All statements I have read, and, I suspect, computer simulations too, that say differently, rely on the premise that your original choice still has a 1/3 probability of hiding a car. Once the host has opened a door, based on foreknowlege that there was a goat behind, it doesn't.

I believe the first post in the first discussion thread makes the same point in a different way. We're both saying that you have to be careful to eliminate hidden assumptions.21:56, 14 May 2017 (UTC)Nigelrg (talk)Nigelrg (talk) Nigelrg (talk) 01:04, 15 May 2017 (UTC)

That would be true if the host had chosen any one of the three doors to open at random, and it didn't have the car. In that case, the game could be started again, and the two remaining doors would have equal chance.

But that is not how the game is played. The host takes into account which door was chosen by the player in the first step. Even if that door is not opened, it influences which door is opened (because the host must avoid it), so it carries some information to the second step; and this knowledge cannot be discarded. Diego (talk) 16:12, 15 May 2017 (UTC)

Also, computer simulations do not make assumptions about the probabilities of the outcomes. They play the game as stated in the standard assumptions, and the 1/3 vs 2/3 probabilities of lose/win emerge by the law of large numbers, as this is the natural result of the game. Diego (talk) 18:00, 15 May 2017 (UTC)

Thanks for your reply, which I am still considering. I agree that the host's decision generates new information, which it carries to the second step, and I wasn't proposing discarding it. I would use that, and the other facts I mentioned, as a reason for restarting the calculation, using all the new information. I don't believe you can continue to assume that your initial choice still has the same 1/3 probability of being a car. Nigelrg (talk)Nigelrg (talk) 20:04, 15 May 2017 (UTC)

The main point of your reply that I didn't address was the computer simulations. Because the 2 goats are considered identical, there are only 3 initial scenarios: car behind doors 1, 2 or 3. Jumping to step 2, after the host's intervention, the computer has 2 closed doors, behind one of which is a car, and the other a goat. Any other information relates to situations that no longer exist. Probability 50:50. — Preceding unsigned comment added by Nigelrg (talk • contribs) 23:58, 15 May 2017 (UTC)

Any other information relates to situations that no longer exist. Probability 50:50. This is a non-sequitur, an invalid reasoning step. You have two doors in the simulation, but the car was not put behind them with a 50:50 random process. One door was selected by the player from a set of three, the other was selected as the best of the two remaining doors. You can't calculate their probabilities with a uniform distribution, because you won't open one of them at random; you need to calculate them using the conditional probability that the car is behind the door after you switch, or after you stay. Read Rick Block's reply in the next section, which does this. Diego (talk) 08:07, 16 May 2017 (UTC)

Reductio ad Absurdum of the "vos Savant" Solution

Your first choice had a 1/3 probability of being a car, as did the other two doors. If you continue to say that it has a 1/3 probability, after the host's intervention, then you must say that the remaining unopened door still has a 1/3 probability of being a car, which is clearly absurd.

What has happened is that the host has introduced new information by opening a door, so the original logic tree must be modified or restarted from scratch. Nigelrg (talk) 21:14, 15 May 2017 (UTC)

Indeed, once the host has opened another door to reveal a goat, then the only chance to win the car is a 50-50 choice to stay or switch to other door, as probability 0.5 either way. Conversely, some people claim Monty Hall would show the car if behind the first door chosen; otherwise the car would be behind the other door, as 100% chance of winning car by switching door (not merely 2⁄3). Of course, such a game would be absurd; hence, the only sensible game would include the chance of the car behind first door chosen, as again 50% chance of win, whether stay or switch. There is no other sensible conclusion. -Wikid77 (talk) 03:27, 16 May 2017 (UTC)

@Nigelrg: You are exactly correct. In fact, numerous mathematicians have described vos Savant's reasoning as somewhat less than complete (see https://en.wikipedia.org/wiki/Monty_Hall_problem#Criticism_of_the_simple_solutions). Before the host opens one of the doors, the probability the car is behind each door is 1/3. After the host opens one, the probabilities must now be reevaluated. We're certain the probability the car is behind the door the host opens is 0, but what about the other two?

@Wikid77: Wikid77 - please pay attention to this.

If the player picks door 1, and the car is behind door 2, then the host MUST open door 3, so the composite probability the car is behind door 2 AND the host opens door 3 is 1/3 * 1, i.e. 1/3.

If the player picks door 1, and the car is behind door 1, then the host can open either door 2 or door 3. Let's say the host doesn't care which door he opens in this case (e.g. he flips a coin to decide). This makes the composite probability the car is behind door 1 AND the host opens door 3 equal to 1/3 * 1/2, i.e. 1/6.

If the host opens door 3, these are the only two possibilities. To express these as conditional probabilities, we divide each by their sum. 1/3 + 1/6 = 1/2, so the conditional probability (given that the host has opened door 3) that the car is behind door 1 is 1/6 / 1/2 = 1/3. And the conditional probability (given that the host has opened door 3) that the car is behind door 2 is 1/3 / 1/2 = 2/3.

As it turns out, the probability the car is behind door 1 doesn't change. But we really can't just assume that. We should compute the conditional probabilities. As one source puts it "The host can always open a door revealing a goat and the probability that the car is behind the initially chosen door does not change, but it is not because of the former that the latter is true." -- Rick Block (talk) 04:09, 16 May 2017 (UTC)

@ Rick Block. I believe that your analysis only holds true if the player has decided whether to stay or switch before the host opens the door. The first post of the first discussion thread makes this point, and the main article addresses it in some depth, as stated in the unidentified post above: https://en.wikipedia.org/wiki/Monty_Hall_problem#Criticism_of_the_simple_solutions. No such stipulation is made in the description of the game, so we should assume that the player makes his decision after the host opens the door. I prefer to use Occam's Razor, and cut out unnecessary detail. Because the player has no information about the location of the car, other than the fact that it's behind one of two closed doors, he/she has a simple 50:50 choice.Nigelrg (talk) 02:43, 18 May 2017 (UTC)

@Nigelrg: First, the analysis I presented above is not "mine", but it is what the most reliable sources in the field of probability use. It most definitely does NOT hold true only if the player decides whether to stay or switch before the host opens the door. It evaluates the probabilities explicitly AFTER the host opens a door. Yes - you don't know for sure which of the two remaining doors the car is behind, but you have "partial" information. This partial information (the fact that the host MUST open door 3 if the car is behind door 2, but has a 50/50 choice which door to open if the car is behind door 1) is what lets you conclude that the chance of winning the car by switching (AFTER the host opens a door and there are only two doors remaining) is 2/3.

Lets take this a bit further. Say we have a deck of 52 cards, and the ace of spades "wins". If I (the "host") shuffle the cards, give you one (face down), and now (looking at the rest) discard all but one making sure I'm not discarding the ace of spades is it now 50/50 that you have the ace of spades? Or is it 1/52 that you have it and 51/52 that I have it? There are only two choices - you have it or I have it. You don't know for sure which one it is. In case it's not obvious the analogy to the MHP is this - instead of putting a car behind one of 3 doors we're putting the ace of spades among 52 cards. Instead of the player picking a door, we're dealing a random card to you. Instead of the host opening one "losing" door resulting in only two doors being left - the host host is discarding 50 losing cards resulting in only two cards being left. Please try this (really, no kidding) - say 20 times and let us know how it turns out (how many times do you end up with the ace of spades vs. how many times does the host end up with it). The notion that "you don't know for sure" and "there are only two choices" necessarily leads to the chances being 50/50 is simply incorrect. This is the entire point of the Monty Hall problem. It goes against a deeply held belief. The bottom line is that humans suck at conditional probability (or, perhaps, schools suck at teaching elementary probability concepts). -- Rick Block (talk) 06:49, 18 May 2017 (UTC)

I'll think over your interesting post later. I had some further thoughts myself, but your post may have negated them. Re: your last sentence, it's not just schools in general. Your views are opposed by post-graduates in math from some of the world's better universities. They're not infallible, of course, but it's a bit like climate change. When a large body of experts say one thing, there's a high probability that they're right.Nigelrg (talk) 18:05, 18 May 2017 (UTC)

I'm dealing with your post piecemeal :-) The deck of cards example is only relevant to one option in the Monty Hall problem - the case in which the player has pre-decided to stick (once you've picked a card, you can't change it). Therefore I don't think it's relevant to the problem as a whole≈Nigelrg (talk) 20:05, 18 May 2017 (UTC)

That's not right. In the formulation of the card game aboive, it does not preclude that the player could still switch cards with the dealer, turning around the probabilities of having the ace of spades. This would make the game equivalent to a Monty Hall game with 52 doors, where the host opens 50 doors which don't have the car, and the player can decide to switch or not after that. Diego (talk) 23:06, 18 May 2017 (UTC)

Re: "partial" information. I think you've neglected the situation when the car is behind door 3, so the host must open door 2. Therefore the host has 2 possible actions when the player's door hides the car, and only 1 possibility when it doesn't. Either way, the player gets new information. I came back to this post months later and corrected it. I made a typo/brain fade when I originally wrote it.I hope my correction didn't change any responses.≈Nigelrg (talk) 21:00, 18 May 2017 (UTC)Nigelrg (talk) 06:07, 11 November 2017 (UTC)

My views are NOT oppposed by post graduates in math (well, not any in probability or statistics) anywhere in the world. This is a classic problem - it appears in many elementary probability textbooks (with the exact answers I'm giving you). So, yes, it's a little like climate change in that the science is settled. But this is math, so not only is the science settled it's actually proven.

@Nigelrg:I thought you wanted to talk about a case where the host has already opened a door and there are only two possibilities for where the car is - for example, the player picked door 1 and then the host opened door 3. In this case the car is manifestly not behind door 3. I'm not 'neglecting" the situation where the car is behind door 3 - we're explicitly talking about only a subset of cases where the host has opened door 3, which means the car is not there.

Here's yet another way to think about it. Imagine 300 shows where the player has initially picked door 1. We'd expect the car to be behind each door about 100 times (right?). So, now the host opens door 2 or door 3. If we want to think about only the shows where the host has opened door 3 we're not talking about 300 shows anymore - but only some subset of the entire 300. Can you answer the following questions (thinking about 300 shows where the player has picked door 1 and the car is behind each door 100 times)? -- Rick Block (talk) 14:58, 19 May 2017 (UTC)

In how many of the shows where the car is behind door 1 does the host open door 2? ______

In how many of the shows where the car is behind door 1 does the host open door 3? ______

In how many of the shows where the car is behind door 2 does the host open door 2? ______

In how many of the shows where the car is behind door 2 does the host open door 3? ______

In how many of the shows where the car is behind door 3 does the host open door 2? ______

In how many of the shows where the car is behind door 3 does the host open door 3? ______

In how many shows overall does the host open door 2? ______

In how many shows overall does the host open door 3? ______

In how many shows where the host opens door 3 is the car behind door 1? ______

In how many shows where the host opens door 3 is the car behind door 2? ______

If you pick door 1 and the host then opens door 3, are you more or less likely to win the car if you switch?

This is a good example, Rick. Answer: It is completely impossible that you ever can be less likely to win the car if you switch.
This is valid not only in that (given?) case if you pick door 1 and the host then opens door 3 (in order to show a goat), but this is valid in any case, regardless which door you may pick, and regardless which other door the host may be opening (in order to show a goat).

Once more: You never are (nor can be) "less likely" by switching. This is valid in any situation given.

Your chance to have picked the car by luck is 1/3, so on average the risk to lose the car by switching is still 1/3, but
your chance to have picked one of the two goats (=wrong guess scenario) is 2/3, so on average the chance to win the car by switching doors is 2/3. And the host's opening of a door (to show you the SECOND goat) does not alter the scenario you're actually fixed in.
The chance to have picked the car by luck (lucky guess scenario) is only 1/3, and the chance that you are fixed in the wrong guess scenario, having picked a goat (thereafter the SECOND goat has already been shown to you !) consequently is even 2/3.
So in 2/3 of all cases you will win the car by swithing doors.

As to the host's behavior, he will never give you any hint on the scenario you're actually (unchangingly) fixed in (Henze, Mladinow et al). --Gerhardvalentin (talk) 12:20, 20 May 2017 (UTC)

Nigelrg seems to think that the analysis of the situation AFTER the host opens a door is that there are two doors and we don't know where the car is, so therefore (??) the chances are 50/50. The point is that the structure of the imaginary show (the real show wasn't actually run like this) gives us some information. -- Rick Block (talk) 15:50, 20 May 2017 (UTC)

Some readers believe in miracles, may be. Or they forget about the significant progress of the advancement made in course of the imaginary "show". But reality never will. --Gerhardvalentin (talk) 17:02, 20 May 2017 (UTC)

Bottom line is that the idealised Monty's behaviour is precisely equivalent to his telling you, "If you have picked the wrong door, then the prize is HERE". And you had 2 chances in 3 of being wrong. What more needs saying? Fredd169 (talk) 13:12, 23 December 2017 (UTC)

This post is very useful to settle the topic. — Preceding unsigned comment added by 190.199.242.101 (talk) 23:15, 16 August 2017 (UTC)

In probability calculation no single case has any meaning; 'control groups' need to be used

There is a way to explain that several answers -and most logic behind it- can be right, using the 'different control groups' idea. Which is based on the simple fact that in any single case, probability has no meaning. So in probability 'thinking', we always have to use bigger amounts that we also use in statistical evidence, to make it clear that if we follow certain rules (consistent behavior) and repeat it enough times to rule out 'coincidence', the calculated probability is actually exactly that of repeating rules. For that, we first have to decide what control group of consistent behavior the players are in.

Because of the way of questioning: 'if you are the contestant, what should you do?', the contestant has to try to determine what control group he is in, to understand the (behavior) rules of the group, and then decide what to do himself in this single case. Which is of course nonsense, because 1.) his deviation in a single situation has nothing to do with probability calculation, which applies to (endless) repeating, and 2.) his decision about his own behavior automatically creates the same control group, in which all cases follow exactly the same rules. Let me explain by examples, starting with the unexpected.

Assumptions in all cases are the obvious, which is not the problem. The actual problem is in the two questions being asked:
1.) "Do you want to pick door No. 2?"
2.) Is it to your advantage to switch your choice?

Example winning probability = 1/2

Most people think that switching does not change the probability of winning. Let's suppose the contestant is one of those. Let's also suppose that the contestant chooses randomly to switch or not. Now it becomes clear that his control group will answer question 1 randomly with yes or no. Just like the host randomly chooses between two doors that have goats. Rule = random. Now we also know that half of them switches and uses the 2/3 probability, while the other half uses the 1/3, equaling to 1/2. Because we cannot apply probability calculation to a single case, we have to apply these common rules of the whole group (both halves) to this one contestant, because in repetition he is sometimes switching and sometimes not. The correct answer to question 2 is that there is no advantage of switching, because it happens randomly, which is the rule. (And which is very counterintuitive.) The contestants will give the right answer, but not with the right reason.

Please notice that these two questions can be seen as really different questions, or actually the same. When seen as different questions, the first one is about the personal preference of the contestant, while the second one necessarily is a logical one, which can be answered right or wrong. It seems obvious (because of the missing quotes in the second question) that these questions are meant to be one: 'please estimate your advantage and then make your choice to switch or not'. The reason leading to the first answer is a given fact that must apply to the entire control group, while the correct answer to the second question is the result of that reason and thus behavior, which has opposite dependency between both questions as is meant to be.

1/3 < Example winning probability < 1/2

In this situation the contestant still thinks that switching does not matter, but he tends to stick to his initial choice, let's say he does this in 70% of the cases. Chances of winning are (7/10 * 1/3) + (3/10 * 2/3) = 0.4333.. He will answer question 2 with 'no', which is incorrect, because he still switches in 30% of the cases. Given his behavior, the correct answer is 'yes, my winning probability increases from 0.3333 to 0.4333 switching as often as I do', but he is not giving that answer of course. This situation is probably most realistic and also makes it very well clear that a single case or choice does not mean anything in probability; we need to know the consistent processes behind it. In one case he will switch doors, in another he will not, which is not even distributed equally.

Note: the counterintuitive thing about this whole 'one case has no probability' approach, is that it seems that it doesn't matter whatever you choose, having only one chance, which seems to happen a lot in everyday life. Vos Savant helps us to understand: suppose that in 999,999 of 1,000,000 cases you are stupid enough to randomly switch, and only in one case you have a short moment of seeing the light, convincing you to switch. This will not significantly change the odds. However, if this would be a consistent reality (it happens to you once in a million), the only way to calculate it right, is to use the 0.500001 probability of switching. But what if you, when you have this bright moment, simply are aware of that brightness? How could it not be significant? There is only one way to be (probably :)) sure about that, which is your ability to recognize such bright moments, proven by experience. This completely changes the control group of situations, because now you have to add a second group of cases that overall have a high rate of brightness. Which shows us again that only the extent of consistent behavior matters, but also that the odds are really relative and depending on the control group(s) used!

Example winning probability = 2/3

First of all I'd like to emphasize that the key is not in the knowledge, but in the behavior (by knowledge). (Then again it's the knowledge of behavior that enables us to create a realistic control group.) If we check the standard assumptions from the chapter in the article with the same name, we see that the three doors are distinguished as 'originally chosen', 'opened' and 'remaining closed'. Not as numbers. We can safely make the assumption that host nor contestant is consistently paying attention to these numbers. Because they have no reason for that. We only know that the host, when offering the switch, is explicitly using the name of the number of that remaining closed door. It may well be that he is simply reading the number displayed on it, at that moment. It does not mean that any player is aware of the number of the originally chosen door at any time. Moreover, since the possible knowledge of the number of the chosen door does not change the behavior of the host, it does not make sense to use this information in any scenario. Also the information from the puzzle: "You pick a door, say No. 1" may be interpreted as: "this may also be No. 2; we only use the numbers as relative distinction to each other, not as absolute information to recognize one certain door." Because of this, conditional solutions using the specific numbers as absolute information are not only unnecessarily complex; they are even wrong when this interpretation is right. Only if (additional) knowledge changes behavior, it should be used in calculation. We have no clue that this is the case, so the simple solution is most appropriate.

--Heptalogos (talk) 23:59, 7 January 2018 (UTC)