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Game theory indeed should be a topic in the article MHP, and I will be trying to write a draft including Game Theory. [[User:Gerhardvalentin|Gerhardvalentin]] ([[User talk:Gerhardvalentin|talk]]) 09:13, 14 September 2011 (UTC)
== Monty Hall Disproof?? ==
The Monty Hall problem irritates me profoundly. Mathematics can be used to represent reality, but this is a deliberate con. The mathematics used, does not represent reality.
I see it this way. (And assuming that it is a 'fair test' and Monty does not deliberately mislead the contestant etc etc )
In mathematics, one must always get the same result to a problem. Two people can not get different answers.
Accepting that there were initially three doors. Once one door has been opened all of the previous statistics can be thrown out of the window. In the real world, once one door has been opened, there is now a certainty of what was behind that door. The situation now, is a completely new problem.
There are two doors, one has a car behind it, the other a goat.
Now, what would the odds be if you, or I walked off the street, into the studio and onto the set, and we chose a door. We would stand a 50/50 chance of getting a car or a goat. If mathematics means anything, the original contestant must have the same odds.
I say again, draw the problem at the final denouement. Two doors, one prize, 50/50 odds.
[[Special:Contributions/78.86.43.71|78.86.43.71]] ([[User talk:78.86.43.71|talk]]) 17:07, 20 October 2011 (UTC)
:your answer is the entire point of why this is a problem/fallacy. The intuitive answer is completely wrong. It has been proven multiple times by thousands/millions of times of experimentation, as well as logically proven. You are not just picking between two doors (which would indeed be 50/50). You are picking between 3 doors, with information provided that allows you to get the right door 2/3 of the time. [[User:Gaijin42|Gaijin42]] ([[User talk:Gaijin42|talk]]) 18:18, 20 October 2011 (UTC)
::There never were 3 doors to begin with. The game show host always eliminates one. So there are always 2 doors. To illustrate, supposing you were presented with 10 doors, with a car behind one of them. You select a door. Before opening it, eight unselected doors are opened showing goats. Does switching to the remaining unselected door now increase your chances of winning the car to 90 percent? Of course not, because there were never 10 different selections to begin with. Eight selections showing goats were always going to be discarded. The game is intended to be reduced to a 50/50 chance of winning or losing. On the bright side, the many elaborate arguments here might apply to a Deer Hunter problem where 5 prisoners at a table are given a gun with only 1 bullet in the 6 chambers and are asked to play Russian Roulette and they may select which chamber to fire according to a numbered chart on the wall. The first prisoner is asked to select one of the 6 chambers and he will take his turn last. The other four prisoners select their chambers one-by-one from the remaining unselected chambers. One-by-one the other four prisoners fire their designated chambers and survive their turns. Now the head of the prison camp gives the first prisoner a choice. The prisoner can fire the chamber he initially selected, or he can switch to the sixth unselected chamber. Would you switch? [[User:Safetyweek|Safetyweek]] ([[User talk:Safetyweek|talk]]) 05:18, 27 November 2011 (UTC)
:::I suggest you try an experiment. Take an entire deck of 52 cards. Let the ace of spades represent the car. Shuffle. Pick one and put it aside (as the player). Now shift roles. As the "host" look at the remaining 51 cards and choose 50 to discard that aren't the ace of spades - which will always reduce the number of choices to two. I believe you're saying it's now 50/50 whether the player's card is the ace. Flip a coin. If it's heads lets say the player keeps her original choice and if it's tails lets say the player switches. OK. Now look at the player's card. Record what the player's card was (ace or not), and what the result of the heads/tails choice was (keep or switch), and whether this choice "won". Repeat, say, 20 times. Please report back how many times the player's card ended up being with the ace, how many times the player switched, and how many times switching won. We'll wait. -- [[user:Rick Block|Rick Block]] <small>([[user talk:Rick Block|talk]])</small> 16:43, 27 November 2011 (UTC)
::::I see. I got it backwards. Switching in the Monty Hall problem increases your chances to 2 out of 3. Switching in the Deer Hunter problem makes no difference, as your chances of survival have gone from 5 out of 6 to 50/50. The key to my correctly understanding the Monty Hall problem is realizing that the initial selection prevents the host from revealing what's behind that particular door. This limits the all-knowing host's options in eliminating a goat. And so you're right, there's a 2 out of 3 chance the car is not behind the door initially selected, and if we know that one of the remaining doors would certainly reveal a goat, then the other remaining door will reveal the car 2 out of 3 times. On the other hand, in the Deer Hunter problem, the prisoners are not all-knowing. No prisoner knew if their selected chamber had the bullet, and so the odds of the first prisoner selecting an empty chamber go from 5 out of 6, to 1 out of 2. It makes no difference if he switches. Thank you Rick for the lack of derision. [[User:Safetyweek|Safetyweek]] ([[User talk:Safetyweek|talk]]) 05:07, 28 November 2011 (UTC)
:::::If you actually did this experiment it might be useful for anyone else reading this to see your results (they can, of course, do it themselves). Not to belabor the point, but there is a big difference between the odds between two alternatives being 50/50 and the odds of a ''random choice'' between two alternatives being 50/50 (this is the point of the coin flip). The odds of a random choice between two alternatives will always gives you a 50/50 chance of winning, regardless of the actual odds between the two alternatives - so, in the MHP, if you flip a coin to decide whether to switch or not you'll have a 50/50 chance of winning the car. However, if you stick with your original choice you only have a 1/3 chance and if you switch you have a 2/3 chance. The composite given a random choice of whether to switch is 1/3*1/2 + 2/3*1/2, which of course works out to 1/2.
:::::All of this assumes the host chooses evenly when given a choice (in the case the player originally picks the car) - without this assumption there's a whole other can of worms. Suffice it to say that this remarkable little problem has much more to it than meets the eye. -- [[user:Rick Block|Rick Block]] <small>([[user talk:Rick Block|talk]])</small> 05:32, 28 November 2011 (UTC)
::::::Actually, I did the experiment this way: my gorgeous hostess dealt 5 cards face down before me, 2 of them were cars and the other 3 were goats. I selected one and kept my hand over it. Of the remaining cards, my hostess eliminated a goat and teasingly offered to let me change my mind and select from one of her 3 remaining cards, which of course I did. And we did this 60 times, and each time I succumbed to the temptation to select one of her cards. (This was a precondition to our opening a bottle of Chateau Mouton Rothschild which we had acquired for this occasion.) As it turned out, I won the car 32 times, so my overall chances were about 50/50 by always switching. And so with this, my hostess moved my attention to the difference between mystery and mystique.[[User:Safetyweek|Safetyweek]] ([[User talk:Safetyweek|talk]]) 07:49, 2 December 2011 (UTC)
== Monty Hall (super simple) ==
Its just this simple people, chill with your bla bla bla math and 100 dollar words! If you are not going to switch after the host opens the door then you are forever LOCKED into the original odds of winning at the time of your choice, end of story. God All Mighty himself cannot change that number! If its three doors its 1/3, if its two doors its 1/2, if its eight doors its 1/8, do I need to keep going? Everyone keeps harping on the "switch", but it has nothing to do with "not switching" odds!!!! LOL all you "SMART PEOPLE" are amazing!!! LOL!! [[User:Davecross1|Davecross1]] ([[User talk:Davecross1|talk]]) 09:07, 24 November 2011 (UTC)
:Actually, there are two probabilities that may or may not have different values - i.e. the probability of picking one door out of three hiding a car, and the [[conditional probability]] the car is behind the door you've originally picked given that the host has opened the door that he opened (for example the conditional probability the car is behind door 1 which you originally picked given the host has opened door 3). Whether these two probabilities have the same value is the crux of the question, and the answer is it depends on exactly what rules govern the host's behavior. For example, say you pick door 1 and the host forgets where the car is and opens door 3 and luckily reveals a goat. The probability of your original pick being correct was 1/3. But, the conditional probability the car is behind door 1 given the host randomly opens one of door 2 or door 3 and reveals a goat is 1/2 (this is why a player at the end of "Deal or No Deal" with the grand prize still unrevealed has only a 50/50 chance of winning it).
:You might argue that in the MHP the host MUST open a door and must reveal a goat - but even this is not enough to say that the conditional probability has the same value as the original chance of picking the car from the three doors. For example, say the host always walks across the stage from door 1 to door 3 and opens the first door he runs into that is not the door you picked and not the door hiding the car. In this case, if you pick door 1 the conditional probability the car is behind door 1 given the host opens door 3 is 0 (not 1/3).
:For these two probabilities to have the same value, the host has to open a door, must reveal a goat (is not opening a door randomly), and if the player has originally picked the door hiding the car must randomly choose which of the other two doors to open. -- [[user:Rick Block|Rick Block]] <small>([[user talk:Rick Block|talk]])</small> 17:44, 24 November 2011 (UTC)
== Think about the doors as the players. They all have equal chances ==
The fallacy rests with the assumption that after the host opens one door, we are still playing the same game. In reality, that is a new game, a game with two doors, not three. While in the beginning there were three doors with one in three chances to pick the right door, after one of the doors in eliminated, the game has only two doors, with one out of two chances to get it right. If changing your pick would give you two out of three chances to get it right, then your new pick could have been any of the other two doors, including the eliminated one, which is not possible as you know that one is the wrong one.
Another way to look at this is considering three contestants holding three boxes, with only one having inside a prize. Every contestant has equal chances to be holding the price. The host eliminates one of the contestants with the empty box. The price is now within the other two contestants. In this instance, they still have equal chances to be holding the prize and that cannot be changed by any external observer’s personal gamble.
In probability, regardless if you have 2 or 1000 choices, every choice is equal. By eliminating one of the three doors, while you are making it more probable to get the right door, it is still equally probable to get it right, regardless on which of the two doors you will pick.
[[Special:Contributions/121.44.184.159|121.44.184.159]] ([[User talk:121.44.184.159|talk]]) 12:36, 13 December 2011 (UTC)
:The MHP is different than your scenario with three players. In your scenario, the two remaining players are indeed equivalent but any one of the players ends up a remaining player only 2/3 of the time. For example, look at the player who holds box 1 over 300 trials of this game. In (about) 100 trials this player's box will have the prize and in these 100 trials this player must be one of the remaining two. In (about) 200 trials this player's box will be one of the two that does not have the prize and - everything else being equal - this player will be eliminated (about) 100 times. The player is in the game (at the end) only 200 times and in 100 of these this player's box has the prize. The analysis holds for any player, so any player in the game at the end has a 50% of winning (their box has the prize 100 out of 200 times).
:In the MHP, the one and only one player is ''always'' in the game at the end. To make this analogous to your 3-player game, one player would be designated "off limits" to the host and one of the other two would be eliminated. Think about 300 trials of the MHP (where the player has picked door 1). Please answer the following questions:
:1) In how many of these 300 trials will the car be behind door 1?
:2) In how many of these 300 trials will the host open door 3?
:3) In how many of these 300 trials will the host open door 2?
:4) Thinking about only the trials where the host opens door 3, in how many of these is the car behind each door (1, 2, and 3)?
:5) Thinking about only the trials where the host opens door 2, in how many of these is the car behind each door (1, 2, and 3)?
:Hints: The sum of the answers to #2 and #3 should be 300. The sum of the times the car is behind each door in #4 and #5 should be 100, and the car is behind door 3 zero times if the host opens door 3. "Everything else being equal" means that the door 1 answer for #4 and #5 is the same (if the car is behind door 1 the host must open one of the other two doors, picking which one at random).
:At the final stage of the MHP there are definitely two doors. And if you pick randomly between them you'll have a 50% chance of winning. However, this doesn't mean that the chances the prize is behind the two remaining doors is the same. I could (for example) roll a 10-sided die and put the car behind door 1 if the die comes up 7 and behind door 2 otherwise, and then give you the choice between these two doors. There are only two doors, but the car has a 10% chance of being behind door 1 and a 90% chance of being behind door 2. The MHP is very similar. Two remaining doors, but unequal chances. -- [[user:Rick Block|Rick Block]] <small>([[user talk:Rick Block|talk]])</small> 17:22, 13 December 2011 (UTC)
::My problem with your theses is that you still calculate the variables from the first choice into the second, which for me it appears to be false.
::The idea of looking at the problem as those three doors are the players is a way to point out that every time, there is equally chances between the available doors/players to have the prize. The variables are exactly the same; it is the same problem with the only difference being the way of looking at the problem.
::My theory is that after the one change on the variables of the game (one door opens) we end up with a new game and therefore the variables from the first game (2/3) cannot be applied in the later (1/2): the maths will be wrong. To make it obvious, add another irrelevant variable to the game, i.e. the players sometimes arrive to the studio with a taxi and sometimes they arrive with their own car, or half the times they wear black and the other half they wear their lucky shoes. If you try to calculate in to the probabilities those variables, you end up with superstition, not real maths, because it is irrelevant their mode of transportation or what they wear with which door has the prize.
::The confusion is that in this problem the variable that changes is not something outside the studio, like the mode of transport, but something that appears to be in the game, the door. But the variable that changes is not the door itself; it is the available sum of the doors.
::Another way of looking at my theory is to consider a second player being introduced into the game after the door opens. Initially, player one has 1/3 chances. If we stop there, he has 1/3 chances to win. After the door opens and before the player makes the second choice, we bring another player and ask to pick one of the two remaining doors. If they both pick the same door, then according to the wiki theory, the first player will have 2/3 and the second players will have ½ chances on the same door. But their individual choices and past experience is irrelevant to the fact that there are two doors and one car, so ½ chances.
::[[Special:Contributions/121.44.184.159|121.44.184.159]] ([[User talk:121.44.184.159|talk]]) 20:31, 13 December 2011 (UTC)
:Did you work through the questions I asked? What are your answers? -- [[user:Rick Block|Rick Block]] <small>([[user talk:Rick Block|talk]])</small> 20:46, 13 December 2011 (UTC)
:: "My problem with your theses is that you still calculate the variables from the first choice into the second, which for me it appears to be false." Therefore, I believe that your formula is wrong from the start, regardless the outcome. I question your variables, not the logic after you apply them.
:: Here is another diagram. In this one, we first calculate every available option, including the host revealing the door with the car (12 different scenarios), in an effort to separate the two games, the "first" game with 3 doors and the "second" game with 2. Then we exclude the options where the host reveals the car and we see that there are equal chances on every door.
::[http://4.bp.blogspot.com/-ATj11G1ofzo/TufOhkS0_II/AAAAAAAAAIA/DDImWCQxAwk/s320/MHP.png Diagram]
::[[Special:Contributions/121.44.184.159|121.44.184.159]] ([[User talk:121.44.184.159|talk]]) 22:19, 13 December 2011 (UTC)
:The game does not start anew after the host opens a door - it continues from the initial start. Let's take this in steps. The rules are:
:1) the host hides the car behind one of three doors (picking which door to hide the car at random)
:2) the player picks a door (which remains unopened)
:3) knowing what's behind the doors the host opens one (not the door the player picked), always revealing a goat - and if the player's initially picked door is hiding the car the host randomly picks which other door to open
:To make sure we're on the same page here, do you agree that one iteration of the game involves all three of these steps and that the question is what is the probability the car is behind the initially picked door as opposed to the other (unpicked, unopened) door after following these steps - for example, if you pick door 1 and the host opens door 3 we're interested in the probability the car is behind door 1 and the probability it is behind door 2 (having followed the steps above)? -- [[user:Rick Block|Rick Block]] <small>([[user talk:Rick Block|talk]])</small> 00:38, 14 December 2011 (UTC)
:: Yes, I understand that there are three stages and I propose that the probabilities change when the variables change, so when the door opens, the game/variables/probabilities are changing. Just because the problem is one sentence, it does not mean that all the proposition in that sentence are related to each other. When my hand throws the dice on the table and it brings a six and then I do it for the second time, that is my same hand throws the same dice on the same table, the probabilities are exactly the same to bring a six as it was the first time. No more or less.
:: I think the reason why the MHP is still under debate is not because some get it and others don’t. It is because the problem itself is not well defined, so one can make different assumptions from others, coming to a different conclusion.
::[[Special:Contributions/121.44.184.159|121.44.184.159]] ([[User talk:121.44.184.159|talk]]) 10:04, 14 December 2011 (UTC)
:I agree the probabilities are changed by opening a door - for example the probability the car is behind the opened door is now (clearly) 0! Lets try another approach. I assume you agree the probabilities the car is behind each door after step 2, for a player who has initially picked door 1, are 1/3, 1/3, and 1/3. There are only two potential outcomes from this position - the host opens door 2 or the host opens door 3. So there are a total of six probabilities that might be interesting: the probability the car is behind door 1 AND the host opens door 2, the probability the car is behind door 1 AND the host opens door 3, the probability the car is behind door 2 AND the host opens door 2, etc. Lets call these P12, P13, P22, P23, P32, and P33. We know that the sum of these must be 1 (they cover all the combinations of where the car might be and what door the host opens), and we know (following step 2) P12+P13=1/3, P22+P23=13, and P32+P33=1/3. With me so far? -- [[user:Rick Block|Rick Block]] <small>([[user talk:Rick Block|talk]])</small> 15:56, 14 December 2011 (UTC)
:: Sure, 6 options, coupled together by where the car is = 1/3 each... and if there were 100 doors (or a large enough number for ones memory capacity), the 1/100 chance will be competing with the 99/100 collated chances if you choose to switch. I understand, no need to explain it. I have read the article and many others. The point I am making is that those calculations are assuming variables that are not true or have not a calculated connection. We calculate doors and cars along with human cognition on new information. For that one choice, the answer is predetermined, regardless of what you will do.
:: Maybe this paradox has been created because stats and probabilities have a meaning on events that have many occurrences, like going to the casino and playing the game many times. When there is only one occurrence, like when you can only be once in your life a contestant on the Monty Hall game and you can make that choice only ones, all (equal) options available have exactly the same weight. This would explain why people using common sense believe that it doesn’t matter if you change, while mathematicians, who look at repetitive applications of the paradox, think that it is better to switch.
:: [[Special:Contributions/121.44.184.159|121.44.184.159]] ([[User talk:121.44.184.159|talk]]) 20:58, 14 December 2011 (UTC)
:Ah, so you do understand where I'm going with this but you apparently don't believe it. You're essentially saying you don't believe in [[conditional probability]], since clearly where I'm going with this is
:1) P12+P13+P22+P23+P32+P33 = 1
:2) (regrouping by the door the host opens) (P12+P22+P32) + (P13+P23+P33) = 1
:3) (assuming it's equally likely the host opens door 2 or door 3) P13+P23+P33 = 1/2
:4) (the host can't open the door the car is behind) P22 = P33 = 0
:5) So, P13+P23 = 1/2
:6) (the probability the car is behind door 2 before the host opens door 3 is 1/3) P22+P23 = 1/3
:7) (from #4, P22 = 0) P23 = 1/3
:8) So, P13 + 1/3 = 1/2
:i.e. P13 = 1/6, meaning the probability the car is behind door 1 (given the host opens door 3) is exactly half the probability the car is behind door 2 (also given the host opens door 3). Expressed as conditional probabilities, this means after the host opens door 3 the probability the car is behind door 1 is (1/6) / (1/2) = 1/3 and the probability the car is behind door 2 is (1/3) / (1/2) = 2/3.
:If you're not going to believe this (which has the form of a formal proof using extremely elementary probability theory and simple math) I'm not sure where we can go with this discussion. Do you understand the playing card analogy - i.e. use the ace of spades to represent the car and two other cards (say, the red twos), shuffle, draw one, and looking at the remaining two discard one that is not the ace? If so, do you agree the 52-card version (shuffle, draw one, look at the remaining 51 and discard 50 that are not the ace) is equivalent? In both of these, after the action of discarding there are only two cards left. Are the chances 50/50 in both of these? If you truly believe this, I strongly encourage you to actually try it (say, 10 times) and let us know how it works out. -- [[user:Rick Block|Rick Block]] <small>([[user talk:Rick Block|talk]])</small> 05:12, 15 December 2011 (UTC)
:: Yep, Rick, I do understand your logic, as I stated above. I have read the article and many others. Have you got any comments on my last paragraph, where I say that the MHP "paradox has been created because stats and probabilities have a meaning on events that have many occurrences, like going to the casino and playing the game many times. When there is only one occurrence, like when you can only be once in your life a contestant on the Monty Hall game and you can make that choice only ones, all (equal) options available have exactly the same weight. This would explain why people using common sense believe that it doesn’t matter if you change, while mathematicians, who look at repetitive applications of the paradox, think that it is better to switch"
:: I haven't seen anyone not acknowledged the maths. I have seen many with different points of view and different calculations. My last assumption is that stats and probabilities are irrelevant on a single application. They only have meaning if they are applied multiple times, which is not what the MHP is about. What do you think?
::[[Special:Contributions/121.44.184.159|121.44.184.159]] ([[User talk:121.44.184.159|talk]]) 07:58, 15 December 2011 (UTC)
:The difference between [[frequency probability]] and [[Bayesian probability]] is what you're talking about. In a pure sense, a single, non-repeated (or non-repeatable) event cannot be reasoned about with frequency probability - for such an event a frequentist can't say anything about the odds, because by naming any odds what you're saying is that if this event were repeated the proportion of various outcomes will (in the limit as the number of repetitions approaches infinity) have a particular value. Indeed, by asking a question whose answer is presumably based on the probabilities of various outcomes the ability for any frequentist to answer at all presumes repeatability. The only "correct" answer for a strict frequentist who is assuming the MHP is a non-repeatable, one-time only offer is "My notion of probability has no relevance to this question, so I can't say whether you should switch or not". Note that this is completely different from saying it's a 50/50 choice because there are two doors left (saying it's 50/50 is saying "if this were repeated over and over, half the time the car will be behind door 1 and half the time the car will be behind door 2"). However, this is not a view of the MHP I've seen published anywhere (I've read much of the available literature).
:On the other hand, Bayesian probability allows reasoning about probabilities of single events based on pure logic. As it turns out, these two seemingly wildly different interpretations of probability are much closer than you might imagine. In particular, the rules for computing conditional probabilities (for repeatable sequences of events for frequentists, and using Bayesian logic) are identical. In a sense, a Bayesian analysis provides a "prediction" that will be born out by experiment - i.e. if your understanding of the rules governing a particular situation is correct the Bayesian prediction will match the experimentally observed (frequentist) probability. Regarding the MHP, vos Savant's nationwide experiment is an example of this. In response to persistent claims that switching does not matter in the face of a logical explanation, she described an experiment (similar to the three card version) which many school classrooms did, which showed switching wins twice as often as staying with the original choice. Perhaps unfortunately, neither her explanation or experiment address the specific case the question asks about (player who has picked door 1 and has then seen the host open door 3) - but many people completely overlook this (by assuming the results must be the same for either door the host opens).
:Is this last point perhaps the actual source of your concern, i.e. many solutions do not address the situation ''after'' the host opens a particular door? -- [[user:Rick Block|Rick Block]] <small>([[user talk:Rick Block|talk]])</small> 17:13, 15 December 2011 (UTC)
:: Thanks for the links. I was not aware of those theories; I will have a look at them a bit more.
::It seems to me that there is a reason why this is a contested issue. The arguments and the variable as not well defined, too many assumptions and that is why not all agree. After all, every game and riddle is a construction of illusion in order to make the answer unclear and thus fun to solve. Is it possible that the reason for this is that MHP appears, not in a philosophical paradox form, but as a well visualised practical situation, a TV quiz, where people perceive it as a one off scenario? Are people presented with a situation that is an example of the frequency probability theory which makes everyone become a frequentist? Nowhere is MHP does it say that the problem will be repeated, that you will have the option to choose again from the start.
:: Of course, this takes away all the fun, but strictly speaking, it could be correct. After all, the MHP was not proposing that you are in a casino playing this game multiple times, what will your strategy be for the night? It explicitly (and intentionally confusingly) describes a one-off problem. The answer to the MHP could be "I can't say whether you should switch or not", so .. whatever!! It doesn't matter if you switch or not, which is same as 50-50. I don’t think this is an extreme theory. This is how dice and roulette works: The shape of the dice and the way our hand throws it is so random that makes it practically a one-off event. Same with the ball on the roulette. That is why stats on those two games are more superstition than anything else.
::[[Special:Contributions/121.44.184.159|121.44.184.159]] ([[User talk:121.44.184.159|talk]]) 20:29, 15 December 2011 (UTC)
:Yes, in its usual form the MHP is not well defined. However, most people who read it assume the same things making it well defined and repeatable (see the Krauss & Wang paper referenced in the article). In particular, the "game show" context provides a strong implication of repeatability and most people assume (even if these aren't explicitly stated)
:1) the initial ___location of the car is (uniformly) random
:2) the host always opens a door showing a goat (never opens the player's door)
:3) if the host has a choice between two doors to open (the player happened to initially select the car), the host chooses which door to open (uniformly) randomly
:4) the host always makes the offer to switch (the offer is not made more or less often depending on whether the player initially selected the car)
:With these assumptions (what the article refers to as the "standard problem") the game is immanently repeatable resulting in a probability of winning by switching of 2/3 regardless of whether you're talking about the overall outcome across all players (regardless of which door they initially picked and which door the host opens) or you're talking about the probability in a specific case (such as player picks door 1 and host opens door 3), and regardless of whether you choose to interpret probability in the frequentist or Bayesian sense. The psychologists who have studied this kind of problem (for example, Ruma Falk or Fox and Levav) say the issue people have with it is not that it is ill-defined but that the result goes against a very strongly held probabilistic intuition (two unknowns, therefor the probability must be 50/50). Piatelli-Palmarini's quotes are very apt: "no other statistical puzzle comes so close to fooling all the people all the time" and "that even Nobel physicists systematically give the wrong answer, and that they insist on it, and they are ready to berate in print those who propose the right answer." They're not misunderstanding it because it's sloppily or ambiguously presented - they're simply getting the wrong answer. -- [[user:Rick Block|Rick Block]] <small>([[user talk:Rick Block|talk]])</small> 00:24, 16 December 2011 (UTC)
:: Well, These are the assumptions I was talking about. We assume that most people who read it assume the same things making and you said that the "game show" context provides a strong implication of repeatability. Maybe those who don't assume these are those who do not agree with these answers. In any way, shouldn't this be stated in the article?
:: In “Statistical methods in experimental physics”, by W. T. Eadie, Frederick James, page 11, it states that: “ Frequentists probability can only be applied to repeatable experiments. This means, for example, that one cannot define the frequentist probability that it will rain tomorrow, since tomorrow will only happen once and other days are no identical to tomorrow.”
:: Also, professor of Computer Science at Olin College, Allen Downey, analysing a variation of the problem, he said “We can't reject the null hypothesis, so if we play by the rules of conventional hypothesis testing, I guess that means we can't take advantage of Monty's tell. If you are a committed frequentist, you should stop reading now.” http://allendowney.blogspot.com/2011/10/blinky-monty-problem.html
:: This should be enough doubt to grant a comment in the MHP wiki page that in order for the problem to have a meaning, we should assume that this is not one off, and that the contestant, unlike real life, can have multiple goes. That could ease many doubters, like me.
:: [[Special:Contributions/121.44.184.159|121.44.184.159]] ([[User talk:121.44.184.159|talk]]) 09:41, 16 December 2011 (UTC)
:That most people make the standard assumptions is what the reliable sources that discuss this say (i.e. it is not "our" assumption). See, in particular, the papers by Falk, or Fox and Levav. Mueser and Granberg explicitly tested responses to different versions of the problem and found no statistically significant difference between versions where 1) the "standard" assumptions were explicit, 2) no assumptions were provided (like vos Savant's version), or 3) the host explicitly opens a random door, and only happens to reveal a goat (in which case the probability is 50/50 for the remaining two doors). Anyone can change the article however they'd like, but if you want to make a change suggesting that there are a significant number of people who get the wrong answer because they're strict frequentists and view the problem as a non-repeatable event I'd encourage you to find a reliable source that says this (as I read it, Downey's blog is not saying this). -- [[user:Rick Block|Rick Block]] <small>([[user talk:Rick Block|talk]])</small> 17:11, 16 December 2011 (UTC)
::To 121.44.184.159: First of all, this is not "Let's make a deal", so nothing is known about tapes of previous shows that - if available - could eventually give additional hints on the actual constellation, resp. on the actual ___location of the car behind those three doors. And nothing is said about this actual show to already having been repeated or will ever be repeated. You can take it as a singular show if you like, or to be repeated in future, if you like.<p>But just think of this one variant: '''There are one hundred doors''', with just only one single car behind anyone of those one hundred doors. Just one car. Each door with equal chance of 1/100. And the contestant may choose just one door. Only one. Then the host asks the contestant "do you want to stick to your one chosen door and open your only one door, or would you like to swap to the residual 99 doors, you may open the rest of all of them at once and they all will be yours, if you like." Irrespective of the host having opened one or more empty doors, or not having opened any door in this variant, or having opened none or one door in the original MHP, the question is: Should the contestant stick to her '''one''' chosen door or should she prefer to take all the residual doors altogether, instead? Hard to believe, but I am sure there will be some saying that there is no difference, and it doesn't matter whether only 1 door – out of 100 – or 99 doors – out of 100 – have been chosen. Or in the original MHP one door or two doors have been chosen.<p>Please consider carefully that basic interpretation of the [http://en.citizendium.org/wiki/Monty_Hall_problem#Explicit_computations Monty Hall problem] in Citizendium, do you feel that this concise report on odds of 2/3 : 1/3 eventually could help to guess and to recognize the underlying basics of the MHP? Regards, [[User:Gerhardvalentin|Gerhardvalentin]] ([[User talk:Gerhardvalentin|talk]]) 20:32, 16 December 2011 (UTC)
:::: It is an one off for the contestant of the show: Monty is not picking doors.
:::::121.44.184.159: apart from eventually expected "additional hints", for the basic proportion of "99 : 1" in the aforementioned variant and for the proportion of "2/3 : 1/3" in the MHP, it does not matter whether the host has shown a goat or not. Read it again, please. [[User:Gerhardvalentin|Gerhardvalentin]] ([[User talk:Gerhardvalentin|talk]]) 14:12, 17 December 2011 (UTC)
:: Rick, you are seriously thinking I can find a reliable source that says "that there are a significant number of people who get the wrong answer because they're strict frequentists and view the problem as a non-repeatable event"?. I wonder if you can find the opposite. Anyway, the souce from the "reliable" Eadie is here: http://books.google.com.au/books?id=QbBm2VhV5TQC&lpg=PA11&dq=Frequentist%20probability%20can%20only%20be%20applied&pg=PA11#v=onepage&q=Frequentist%20probability%20can%20only%20be%20applied&f=false and I think his claim is applicable to the MHP.
:: After 5500+ words on our discussion, I would have thought that we had come to an understanding that there is a position/opinion/variation which does not appear in the article, a frequentists view on this problem that practically will occur only once, but for argument sake we assume that it will happen multiple times.
:: You appear more knowledgeable than me and as you have assume the position of the protector of this article, maybe that gives you the power to say NO to any change that is not your cup of tea, but maybe it also gives you the responsibility to spend some time searching on an extra +5500 words to find a source that would be good enough to grant a mention in the original article of a position that you yourself brought up. I am just passing by and by no means do I want to spend more time on “the ongoing farce that Monty Hall problem has become”. I think W. T. Eadie's source is good enough, apparently it is not.
:: Nevertheless, thank you for helping me prove (to myself) that there is a problem with Monty.
:: [[Special:Contributions/121.44.231.208|121.44.231.208]] ([[User talk:121.44.231.208|talk]]) 20:51, 16 December 2011 (UTC)
:::“There is no doubt that with respect to a large enough sample of Monty Hall games the player should switch. But what if we look at a single game (cf. Moser and Mulder 1994; Horgan 1995)? I want to argue that we run into serious problems if we apply probabilistic notions and arguments like the one above to a single Monty Hall game. The application of such notions and arguments to a single case (a single game) does not make sense; hence, there is no answer to the question what the rational player should do in an isolated case, at least no probabilistic answer.”
::: Peter Baumann: Single-case probabilities and the case of Monty Hall: Levy’s view: SYNTHESE, Volume 162, Number 2, 265-273,
::: http://www.springerlink.com/content/652828555l629711/fulltext.pdf
::: [[Special:Contributions/121.44.231.208|121.44.231.208]] ([[User talk:121.44.231.208|talk]]) 21:22, 16 December 2011 (UTC)
::::Paul K Moser and D Hudson Mulder of Loyola University of Chicago, PROBABILITY IN RATIONAL DECISION-MAKING : Philosophical Papers Vol. XXIII (1994). No. 2
::::Page 112 : “ Why should we hold that the results of switching over a hypothesized long run of trials automatically bear, in any relevant way, on the advisability of switching in an isolated single case, or even in an isolated short run of consecutive trials? The view that certain hypothesized long-run results bear on an isolated single case needs explanation and argument, as that view is not self-evident. Typical results in a hypothesized long run are not automatically matched by results in an isolated single case or even in an isolated short run of consecutive trials. A typical long-run result does not influence the propensity of outcome in an isolated single case or in an isolated short run of trials. In particular, a typical long-run result has no causal influence on a single case, although sometimes long-run results reflect the causal powers at work in a single case.”
::::Page 126: “The rational advisability of switching in an appropriate long run of Monty Hall games depends on a presumable statistical correlation that will hold only in runs of games where the rate of success on the initial choice is about 1/3. This statistical correlation is irrelevant to an isolated individual case, because it makes no sense to talk of a rate of success x, where 0<x<1 , for the initial choice in an isolated individual case. It is impossible for one to have a 1/3 rate of success on the initial choice in an individual play of the game; one’s rate of success here can be only 1 or 0. The rational advisability of switching depends essentially, however, on a presumable 1/3 rate of success for the initial choice. Such a rate of success is not only impossible in an individual case but also highly unlikely in a short run of cases. It becomes progressively more likely only as the length of the run gets progressively greater (and where there is no decisive contravening evidence, of course). Switching, then, is not rationally advisable in an isolated single case, and it becomes only gradually more advisable, under certain conditions, as the length of the run of games one considers increases. If one is offered 10 games, switching is less strongly rationally advisable than if one is offered 100 games. If one is offered only one game (as one is in the Monty Hall game of section I), switching is not at all advisable, given the bonus gained by staying.”
:::::Kahneman and Tversky 1972, pp. 434-36 : “People expect that a sequence of events generated by a random process will represent the essential characteristics of that process even when the sequence is short . . . Thus, people expect that the essential characteristics of the process will be represented, not only globally in the entire sequence, but also locally in each of its parts. A locally representative sequence, however, deviates systematically from chance expectation: it contains too many alternations and too few runs. “
:::::[[Special:Contributions/121.44.231.208|121.44.231.208]] ([[User talk:121.44.231.208|talk]]) 21:50, 16 December 2011 (UTC)
:: And there is where my argument stands. MHP is not presented as a PHD thesis in statistics, but as a simple, visual problem that anyone with a TV can relate. MHP is placing us exactly in a one case scenario, a “what would you do if you were so lucky to appear in a game show” ? Everyone thinks that the prize could be behind any door, so it does not matter if you change or not. And the logic makes sense because everyone’s head is focusing in this one case scenario. MHP has no suggestions for multiple applications; it a question for YOU to answer for this one event. For the sake of the argument we should indeed take it to another place, where multiple events take place, like a casino etc. But wiki should point this out and should have a paragraph for this theory, the one off event. It is not that weird or too unique: Dice and roulettes are exactly the same, they are one off events.
::[[Special:Contributions/121.44.231.208|121.44.231.208]] ([[User talk:121.44.231.208|talk]]) 22:07, 16 December 2011 (UTC)
:First of all I am in no sense the "protector of this article". I said above "Anyone can change the article however they'd like". Not only did I actually mean this, but it's literally true as well. If you want to change the article, do it. If you want to propose a change, propose it on the [[Talk:Monty Hall problem|article's talk page]] (not here). Second, please look at the info box at the top of this page. This page is for discussing mathematical aspects of the MHP, not for suggesting or discussing changes to the article. I thought we've been discussing the math behind the problem. You seemed (still seem) to be somewhat confused. I've been trying to help. If your stance is it's a 50/50 choice and you're going to ignore anything anyone has to say that contradicts this, that's certainly your choice. However, I can assure you there is absolutely no academic controversy about this - including the references you've quoted above (which are not saying that it's a 50/50 choice). What these sources are saying (and, by Wikipedia's standards they'd be considered [[wp:fringe|fringe]]) is that if the MHP is viewed as a singular event the (frequentist) probability of winning is indeterminate - not 50/50, not 1/3:2/3, not anything. I'm somewhat surprised [[User:Gill110951]] hasn't commented on this thread yet. He's a professor of mathematics with an interest in the MHP. If you're finding this conversation with me to be not to your liking I'm sure he'd be willing to talk about it with you (on his talk page at [[User talk:Gill110951]]). In any event, I apologize if I haven't been helpful. -- [[user:Rick Block|Rick Block]] <small>([[user talk:Rick Block|talk]])</small> 08:45, 17 December 2011 (UTC)
:: Hi, sorry I have not been watching these discussions. I have been taking a break from MHP and got deeply into the [[Two envelopes problem]] instead. But sure I am still very interested in good old Monty Hall. <p> I see that the discussion is about frequentist versus Bayesian approaches to MHP. I agree that this is an interesting issue. I believe that the typical assumptions of a Bayesian and of a frequentist for MHP would be different. The meaning of probability is different, so this is no surprise. Because their assumptions are different, their solutions are different too. They both decide to switch, but for different reasons. I have written about this issue in a published article which you can also find on my university home page, together with some other MHP writings; the link is http://www.math.leidenuniv.nl/~gill/#MHP. <p> And for what it's worth, here's my conclusion. The frequentist will realise that we know nothing about how the prize is hidden, nor how the host chooses a door to open when he has a choice. So he will take fate into his own hands by choosing his own door, in advance, uniformly at random, and then proceed to switch, whatever door is opened by the host. For him, the chance that he'll get the car, given his initial choice and given the door opened by the host, is at best unknown, at worst undefined. But he doesn't care. He does know that he has a 2/3 unconditional chance of getting the car, since his initial choice is correct exactly 1/3 of the times; he also knows that one cannot do better (by the minimax theorem from game theory). This solution is the unique minimax solution: it minimizes (by the player's strategy of combined door choices) the maximal unconditional chance of not getting the car (maximal with respect to possible car-hiding and door-opening strategies of the host). <p> On the other hand, for the subjectivist, all doors are initially equally likely to hide the car since he has no information on whether to prefer one door to another. Similarly, for him the host is equally likely to open either door when he had a choice, because he (the subjectivist player) has equal reason to believe a possible bias of the host would work in one direction or the other. Consequently the subjectivist will initially choose door 1, say, because 1 is his lucky number, and switch because after the host has opened, say, door 3, the other door has a 2/3 chance of hiding the car. The reasoning behind 2/3 is because initially door 1 has chance 1/3 of hiding the car. The host is going to show a goat behind another door anyway, so when he does this, the chance that door 1 hides the car doesn't change. Finally, *which* door is opened by the host does not give any new information to the player, by symmetry. So given the door chosen by the player and the door opened by the host, the subjectivist gives 2/3 chance to the car being behind the other door, and switches. <p> Two meanings of probability, two quite different solutions, both perfectly acceptable I think, and all this is written in my paper if anyone finds it notable enough to use in the article. [[User:Gill110951|Richard Gill]] ([[User talk:Gill110951|talk]]) 17:28, 6 January 2012 (UTC)
===The doubting continues===
Initially there are 3 possible outcomes. Each door has a 1/3 likelihood of having the car.
Once a door has been opened, there are only 2 possible outcomes.
The car is behind door 1 or the car is behind door 2. The odds of the car being behind 1 or 2 were the same at the beginning, 1/3 each. That relationship hasn't changed because door 3 had a goat. The odds are now 1/2 for each of doors 1 and 2. Switching does not make a difference.
The vos Savant chart is flawed. Once door 3 has been opened, there are only 2 possible outcomes. Outcome A has the car behind door 1 and outcome B has the car behind door 2. Door 3 is no longer relevant. The odds of the car being behind either #1 or #2 have changed, from 1/3 to 1/2, but the relationship of the relative likelihood is still the same as it was at the outset.
Similarly, if door 2 is opened, there are only 2 possible outcomes -- car behind #1 or car behind #3. [[User:Kwinzenried|Kwinzenried]] ([[User talk:Kwinzenried|talk]]) 01:04, 7 January 2012 (UTC)
:Are you interested in talking about this, or are you absolutely convinced you're right and nothing anyone says can change your mind? -- [[user:Rick Block|Rick Block]] <small>([[user talk:Rick Block|talk]])</small> 01:56, 7 January 2012 (UTC)
This is an interesting problem. I am certainly open to obtaining a clear presentation of the correct answer, whatever that might be. Per my comments above, I don't see that switching makes any difference. Additionally, the problem of symmetry hasn't been dismissed if switching is anyways the preferred answer -- if switching were always preferable, then the initial choice must matter, but since the initial choice is random, symmetry can't apply if switching is the best answer. <small><span class="autosigned">— Preceding [[Wikipedia:Signatures|unsigned]] comment added by [[User:Kwinzenried|Kwinzenried]] ([[User talk:Kwinzenried|talk]] • [[Special:Contributions/Kwinzenried|contribs]]) 03:31, 7 January 2012 (UTC)</span></small><!-- Template:Unsigned --> <!--Autosigned by SineBot-->
:OK. First - I'll agree with you that vos Savant's solution is flawed. It's definitely part of the story - but she changes the question from "is it better to decide to switch (having picked door 1) ''after'' seeing the host open (say) door 3" to "is it better to decide to switch (having picked door 1) ''before'' seeing which door the host opens". Do you see these as two different questions, and do you agree vos Savant's solution addresses the second one? -- [[user:Rick Block|Rick Block]] <small>([[user talk:Rick Block|talk]])</small> 04:34, 7 January 2012 (UTC)
:: I do not agree that vos Savant's solution is flawed, but that's another matter. <p> Anyway let's agree on what we mean by probability, so we know what we are talking about, and what we can assume, and what we can't assume. Let's take probability to have its subjective meaning. Let's agree that a priori, as far as we are concerned, the car is equally likely to be behind any of the three doors. Let's also agree that whether or not the host has any bias in opening doors when he has a choice, for us the door numbers are mere labels, completely uninformative. So in particular, if we happened to choose the door hiding the car, so that the host had a choice, we find it equally likely that he chooses either door. OK so far? <p>
:: Now here follows a solution which ends with the 2/3 '''subjective''' probability of the car being behind the other door, '''given''' our choice (Door 1), and given the door opened by the host (Door 3). <p> Go back to the moment when we have chosen Door 1, but the host hasn't opened a door yet. At that moment we believe the chance is 1 in 3 that the car is behind our door. We know that the host is going to open Door 2 or Door 3 and show us a goat. Keep your eyes shut, let him open Door 2 or Door 3. We hear a goat bleating, but we don't know yet which door was opened. Nothing has changed. This was going to happen anyway. The chance is still 1 in 3 that the car is behind our door, Door 1. Finally, we open our eyes. We see that Door 3 is the one which has been opened. '''Nothing changes regarding Door 1'''. Because, '''whether or not''' the car was behind Door 1, '''we considered it equally likely, that Door 2 or Door 3 would be opened''', by symmetry. <p> Consequently, the probability the car is behind Door 1 still remains unchanged at 1 in 3. But we do know now that the car certainly isn't behind Door 3. So we would bet 2 against 1 that the car is behind Door 2, we will accept the offer to switch. <p> If you want to go through this argument with full mathematical details, I recommend you use Bayes' rule for updating probabilities on obtaining new information in the form: posterior odds equals prior odds time likelihood ratio; where "likelihood ratio" means the ratio of the probabilities of the new information, under each of the two alternatives considered. Take as alternatives: Car is behind Door 1, Car is not behind Door 1. Given that we initially pick Door 1, the initial odds that the car is behind Door 1 are 2:1 against. In either situation, the host is equally likely to open Door 2 or Door 3, by the symmetry of our prior beliefs. No need to do any calculations. So the likelihood ratio, for and against the hypothesis "car is behind Door 1", coming from the information "Door 3 got opened" is 1:1. So the posterior odds remain 2:1 against. <p> Vos Savant's argument can be considered to be completely correct. In essence she is ignoring the specific door numbers from the start, by symmetry. This is a completely intuitive step of many people thinking about the problem, and is also suggested by the problem statement's use of the words '''say''' Door 1, '''say''' Door 3. The words are used to help us visualise the problem, but also to trick us into '''seeing''' the problem at the stage after we have chosen a specific door and a specific door has been opened, forgetting the procedure whereby one particular door got opened, which does depend in fact both on our own choice and on the ___location of the car. The host does not have a free choice, and the probabilities of what he does, depends on the situation at that moment. We need to keep in mind the whole procedure. We are mislead into discarding the relevant information ('''how''' we got to this situation), and remembering only the final situation. And then we intuitively come to the wrong decision. The right information to discard are the actual numbers written on the doors in the case at hand! <p> Here is another, rigorous, complete, solution, on the lines of vos Savant: Because the numbers are completely arbitrary, one can discard them in advance. The only relevant things are the '''functions''' or '''roles''' of the doors. Forget the numbers painted on them! There are three doors: a door chosen by us, a door opened by the host, and a door remaining closed. One of the three hides a car. The question is, what are the probabilities that the car is behind each of these three doors. '''These''' subjective probabilities are determined in advance, and they never change, till the very end of the game. The probabilities are 1/3, 0, 2/3. That's obvious, isn't it? From this point of view Vos Savant's solution is completely correct, and there are reliable authorities enough who support this point of view. Forget the door numbers, they are irrelevant! <p> MHP is a brain teaser, a trick, a joke, a surprise; the formulation is cleverly chosen so as to entice you into the wrong line of thought. Just like a good joke. One needs to step back, rethink, look at the whole problem from a new point of view, and only then will intuition give us the right answer, and you see the joke. As the Joker said in the film "[[The Dark Knight (film)|The Dark Knight]]", "not so serious!". The good intuition is: '''the door numbers are irrelevant. The initial choice has 1/3 chance to hit the car. Hence switching gives the car with probability 2/3. End of story''' (at least, that's my opinion, but also the opinion of quite a few reliable sources, including many professional mathematicians). See http://www.math.leidenuniv.nl/~gill#MHP for further contributions by me. [[User:Gill110951|Richard Gill]] ([[User talk:Gill110951|talk]]) 09:44, 7 January 2012 (UTC)
I said I could change my mind and I have.
The misdirection in the wording of the problem (and several of the explanations of the answer) is the focus on the participant rather than the host. The host captures the benefit of the 2/3 odds of having the car in the first stage since he gets two of the three choices. Once he has seen the first stage outcome, he always discards a loser and then, in the 2 of 3 first stage outcomes where he has the winner, he keeps it. So in 2 out of 3 first stage outcomes, the host has kept the winner. The participant, not knowing the first stage outcome, can elect to swap his 1 in 3 chance of having won the first stage for the host's 2 in 3 chance of having the first stage winner.[[User:Kwinzenried|Kwinzenried]] ([[User talk:Kwinzenried|talk]]) 16:19, 7 January 2012 (UTC)
: That's right. Under the minimal assumption that the player's first choice is right 1 times in 3, he should always switch: that way he gets the car 2 times out of 3. Moreover, if *each* door has a 1 in 3 chance of hiding the car, the player can't do better. That is because whatever strategy he uses (two stages of door choices) there is always a prize ___location for which his strategy will lead him to lose. <p> One does not have to do conditional probability calculations in order to solve MHP adequately. Careful considerations of strategy do the job, too. [[User:Gill110951|Richard Gill]] ([[User talk:Gill110951|talk]]) 17:12, 7 January 2012 (UTC)
:Yes, you can change your focus from the situation ''after'' the host has opened a door (and you can see which door the host has opened) to the situation ''before'' the host opens a door (or, as Richard does above, to the situation ''after'' but keeping your eyes closed so you don't know yet which door the host has opened) - this is indeed what vos Savant does. If the show is on, say, 300 times and the players all picked door 1, if all of these players keep their original choice about 1/3 will win the car and if all of these players switch (to whichever of door 2 or door 3 the host does not open) 2/3 will win the car. The flaw is that this doesn't actually answer the question about a player who sees the host open door 3 without the additional argument that the cases where the host opens door 2 or door 3 must be the same. This argument is where the trick actually is! You can choose to completely gloss over this and beat people on the head to change their focus, but it's completely unnecessary to do this. Just fill in the blanks in the following (imagining 300 shows, i.e. each group of answers sums to 300):
:1) ___ Number of times the car is behind door 1
:2) ___ Number of times the car is behind door 2
:3) ___ Number of times the car is behind door 3
:4) ___ Number of times the host opens door 2
:5) ___ Number of times the host opens door 3
:6) ___ Number of times the car is behind door 1 when the host opens door 2
:7) ___ Number of times the car is behind door 1 when the host opens door 3
:8) ___ Number of times the car is behind door 2 when the host opens door 3
:9) ___ Number of times the car is behind door 3 when the host opens door 2
:The answer for #5 must be the sum of #7 and #8 (if the host opens door 3 we know the car must be behind door 1 or door 2), and the answer for #1 must be the sum of #6 and #7 (if the car is behind door 1 the host must open either door 2 or door 3). With a little bit of thought, it's quite obvious the answer to #2 and #9 (and #3 and #8) must be the same (any time the car is behind door 2 the host must open door 3, and vice versa).
:From these answers we can then look at the proportion of times the car is behind door 2 when the host opens door 3 (#8 out of #5) and the proportion of times the car is behind door 1 when the host opens door 3 (#7 out of #5).
:There is no trickery involved whatsoever. The problem is people leap to the wrong conclusion based on faulty intuition. As Carton (referenced in the article) says: ''We often cringe when our students, or members of the general public, make rudimentary mistakes in probability. But even qualified scientists and mathematicians can make such mistakes, although theirs are sometimes less obvious. The misapplication of conditional probability, the haphazard use of “equally likely” outcomes, and the non-use of Bayes’ Rule can lead to all manner of incorrect answers.''
:The point is that people's intuition is what screws them up here. Appealing to a different intuition, even one that happens to produce the correct answer in this case, simply isn't helpful. A very good test of this is to ask the followup version - is it better to switch if the host blindly opens door 3 (without knowing where the car is) and happens to reveal a goat? Fill out the same table above. Think very carefully about the answers to #8 and #9. -- [[user:Rick Block|Rick Block]] <small>([[user talk:Rick Block|talk]])</small> 18:01, 7 January 2012 (UTC)
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