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Please explain how my chances of picking the one door in three with a car behind it increase from 1 in 3 to 1 in 2 because one of the doors I did not picked is opened first? No, it is the one and only problem, ie the same. It is new problem if someone enters the game who does not know which door the contestant chose. To you and Steve? I do not find it confusing, nor do those who understand and appreciate that the solution is counter-intuitive.
Not to mention that there are hundreds of sites explaining the MHP with logical and mathematical proofs , it is a standard text for students in maths textbooks worldwide, and the literally millions of computer trials demonstrating the But these, and maths experts everywhere, are all wrong, Phil and Steve however know the correct answer.
This is a problem I have always been stumped by and I do understand probability very well because of the apparent conflict between the model and the reality. At the purported start of the game, I have no door and so no chance of having the car. So mathematically — if I switch it reduces my likelihood to end up with the car from 0.
However, it never works out that way in the simulations. I guess I need to spend the time to program it the way I conceptualize it and see what the results are. Ableger, I do not mean to be rude, but just about every significant premise in your post is incorrect. Firstly, the game actually starts when you make your initial 1 in 3 pick. No matter what happens thereafter, you will win 33 times for every time you play this game, no matter what Monty does, if you stick with your first choice.
You will win 50 times in You will win virtually every time. This is the difference between Choosing the car, and Having the car. As I said — this is a conflict between the statistical calculation and the reality of the situation. It is an actual choice. That your first choice is irrelevant is a false premise as your first choice tells you two things:.
Valuable information if you know how to use it. There is no conflict between statistical I take it you mean probability calculation and reality, they are both in agreement: You start the game when you pick a door. You now have 1 card left in your hand, and the card you picked to begin with face down on the table. Which one is the Ace? Repeat as many times as necessary for you to realise that the card left in your hand is 51 times more likely to be the Ace than the one face down on the table.
Kalid, this is really great work and I think your explanation is very clear. The outcome of the puzzle is highly counter-intuitive at first, but once explained clearly as you have done I think it can be readily grasped by most people. Palmer — so you are now saying keeping the card is more likely to get you the win? I thought switching was the winning strategy?
Damn Monty for opening that door and turning Goats into Cars! The show would go broke! And while this may not be an intelligence test, it is certainly a test of your logic and reasoning ability and ability to see that a logical conclusion will survive all manner of irrelevant events, like doors blowing open, or Monty opening a door with a goat. The conclusion will survive intact to because it is an inevitable conclusion.
That means the correct answer goes against instincts, not maths. Secondly, there are millions of people who thought like you and realised they were wrong. There is simply no need to get emotional when the answer is logical and clear underpinned by maths rather than emotion or intuition and yelling at people acting on instinct simply does not work.
We simply use logic. Apologies if someone already pointed this out. My boss gave me an explanation that makes it clearer to me than the or 1 million boxes even. Same scenario, but instead of Monty taking one box that always has a goat away, he says, you can either stick with your choice, or you can have whatever is under BOTH the other two boxes. You are not a machine and you have information from the first selection.
You are just more likely to have a goat. Forget what is removed. You choose a door — The fact that a door is removed is immaterial — it does not reset your odds. All you can say is you are Forget that he removed the door or that there are now two doors. Nothing will change the initial odds of you having a goat which are He can paint the doors blue or dance the Macarena.. On that you must agree.
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So of course you switch..! There is never a When the problem was originally posed it was poorly worded and open to different interpretations. I invite you to a friendly game! We will do the monty hall problem with a full deck of cards. The ace of spades always wins the hand.
You draw one card randomly from the deck, our friend Call him M. MH will then lay out the remaining 50 cards between us on the table to show that the ace of spades is missing. Ok guys, Groove almost got me with his argument, but only things that actually conviced my is this short clip by BBC ,. This is the first time I have read these posts, but I am surprised at the number of ways folks have of complicating what is at heart a simple issue.
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This is clearly true at the moment you make the choice, and absolutely nothing that happens later can affect it in any way. Incidentally, of course, Monty does not make a random choice — he cheats in the interest of the game, but nothing he could possibly do affects the a priori probabilities in any way. You choose one door, and Monty gets the other two doors. He always gets a goat, since he gets two doors, so every game, he will remove one door, which will always be a goat.
Think of it this way! Another way to look at it. The door you choose is a random person, and when you choose your random person, Monty is given a world-renowned chef. They will do battle in the kitchen, making spaghetti! Then, Monty reveals a goat at random. Now, Monty opens a door at random. When Monty reveals a goat door at random then it makes no difference whether you stick or switch doors. That was my point: Patently wrong, whether when you are referring to when you make your first pick or after Monty has opened the door.
And that if you play that game times you are likely to have picked the Ace of Spades 50 times when statistically you would be lucky to pick it twice? The dealer shuffles the deck, and asks you to pick a card at random, without looking at it. You place this card face down on the table. Next, the dealer looks through all 51 remaining cards, and selects one. Then, he places that one card face down next to yours.
He throws the remaining 50 unused cards away. Now, should you bet that the card you picked originally is the Ace Of Spades or the card the dealer picked is the Ace Of Spades, which is more likely? Purposeful Summary — Vinny Colantuoni Counterintuitive Monty deliberately removes a losing outcome door C. Same thing with the cards. The dealer turns over a 2 of diamonds. He then puts down one card face down. You put your card face down. The dealer turns over the 2 of Diamonds I assume he deliberately avoids the Ace of Spades if he has it.
The dealer however did not choose his card at random — he chose the Ace of Spades if it was available. Now when the two cards are placed side by side, in one out of 52 games, the card you chose will be Ace of Spaces, and in the remaining games, 51 times out of 52, you will have failed to choose the Ace of Spades initially, thus the dealer MUST have it.
The dealer then manually looks through the remaining cards and picks a card. You know that he always picks Ace of Spades if possible i. The dealer then places his card face down in front of him. You can then swap cards if you want, or just reveal both cards as they are. Further to my previous comment: Nobody is claiming that if you play 3 games you will always win 1 of those 3. Swap offered, but rejected or ignored. But what is also clear is that if you lose, Monty wins the car for himself. How can that be true? Consider a random set of numbers: At best you can only make an approximation.
But if you DO know the odds per door, you can win that often. The dealer thus gets the Ace of Spades in any game where the player failed. Consider a Player 2 that saw the draw and know which card is which. See, the skewed chances of either card being right do not chance the nature of random choice, but if you know the relative odds on each card, you can win proportional to those odds. Gee, Jason, you take the prize, if not the cake, for the most complicated attempts at explaining the odds of the random player scenario.
The fact is, it is typically the simple explanations that are the correct ones and those positing incorrect hypotheses rely on complex faulty formulae to support reasoning that itself is wrong. Put simply, Jason, if the player with no knowledge plays times he will, on average, win 50 times. Just like you would win, on average, 50 times, if you picked heads or tails in a game coin toss.
It is the same as if sporting contests were set up between the top players and complete novices. A sports fan will pick the winner nearly every time whereas a guy from Mars who has no knowledge of sport will pick the winner, on average, 50 times. Now, these 4 possible situations add up to 1 unity , with half the outcomes being wins and half the outcomes being losses. This is like year 8 mathematics. Total probabilities add up to 1. It was in response to Zenasdad, who wrote:.
My posts were in response to that conversation. Probability trees show how that works. Jason, I do understand probability trees and, as someone else will chime in shortly and explain to you, they are for calculating the probability of outcomes dependent upon multiple sequential decisions ie more than 1 not single decisions in once-off scenarios like picking from one of two doors. A decision tree has no place in explaining a situation where there is only a single, random decision to pick either door A or door B. There is an equal probability of an outsider picking the door with the car from the two remaining doors each time they guess.
So yes, I do understand decision trees, what they are, what that are for and how to use them. I should use them more often, eg chances you will read this post:. The question is, if you understand decision trees, why are you using one for a problem involving a single decision when they are designed solely for use in situations involving multiple sequential decisions? As for your second set of even more flawed calculations that now show that it does add up to one:.
It only now adds up to one because you created 4 possible outcomes when there are only two! Definitely the most imaginative yet completely misguided post on this forum so far. PS — sorry, that should have read: Chance you will respond if you realise you were wrong: That is as clear as mud. A probability tree proves that:. Obviously, a player who knows which is door A and which is Door B gets better odds, but Zenasdad disputed that.
Mapping it out as two independent decisions shows why that is flawed. The doors were selected via the MH system, thus they carry the MH odds for a smart player. Therefore it is correct to think: You are stating as a fact, things that are suppositions: Then, add to that the dealer offering you the Monty-Hall type swap.
It breaks down like this:. Jason I meant to write: Before you pick a door, math, knowledge, truth is: GAME 2 C is revealed to be a goat. You are choosing between two. There are only 2 choices! The car IS behind either A or B. There are three possible outcomes, all equally likely:. Say you picked door1 and Monty opened door3 knowing it contained a goat. Now the car is obviously behind either door1 or door2, and both were equally likely at the outset.
So you switch to double your chances of winning. A mathematical solution using Bayes Theory produces the same answer. Say there are doors, and one prize is initially placed at random. Imagine that all the doors are opened at once, revealing where the prize is. Next, lets ask what happens when the doors are opened one at a time?
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This should give the same result as opening them all at once, right? Imagine, the host opens 98 of the doors but he never opens the prize door and he leaves one closed door per player. There are now two doors left, one of player 1 and one of player 2. Nobody has switched or made another choice. What are the odds of it being behind either remaining door?
According to Jason and the rest, this is not true. This can be shown mathematically very easily, and in fact I already laid it out. Here we go again. Adding all the probabilities together gives 1. The maths works out the same no matter what the different odds are.
The actual odds of the two answers being correct are not related to it. I ask you to stick with the 3 door problem. If you pick a door at random then you have a: So many of us agree so we must be right!! The number of doors. There are two non-dependant individual choices being made. I am not a denier, I am unconvinced. Again, refer to comment I will be in the field for 2 days but would love to hear back from you. Car is Door C, Monty reveals goat in door B, swapping wins. So it only works if you assume the car was twice as likely to be behind the exact door that the player chose than the other doors.
I am looking for hoping someone can find flaws in the data. I am trying to explore all possible outcomes; If we can agree on the data-set that will be a good start. I remain unconvinced by maths tainted with intuition. I realize you guys know far more math s than I. Jason has already explained this, but anyway….. Take the example where the car is behind Door1. The only possible outcomes and the probability of each outcome occurring are: There is no mathematical rule that guarantees that. ZenasDad if you want to look at all possible outcomes in an exhaustive way, we have to look at every step of the game.
Next, we add a third letter, which is the door Monty opens revealing one of the goats. The ones in brackets are also the ones where switching loses. This is clearly impossible, because it would mean the players choice was dependent on where the car was, which contradicts how the game was set up. Those are ALL possible combinations when you flip three coins one time. This yields 8 equally probable outcomes. If you plot all possible outcomes: We should NOT see equal totals for staying and switching. If you really want to test it, I just wrote this bit of. Chuck it in a file called monty.
Please can you post that to pastebin? This thread removes extra whitespace and tabs. There is a clear error here. But if Door B has the car, they are more likely to have chosen Door B at the start than the other doors. VBS uses a variable with the same name of the function to store the return value from the function. The table should be unbiased: You can still stay or switch. For every possible car location in our 3-door problem, — there are eight 8 possible outcomes.
Consider this version with coins: Show me a version where the player is equally likely to pick any door regardless of where the car is, and we can talk. Here is the crux of your error, ZenasDad. You seem to simultaneously hold two contradictory views, yet have avoided confronting the contradiction. Let me make an uncontroversial statement. The following three possibilities are all equally likely:. You also claim that these four outcomes are equally likely.
This is in direct contradiction to the first outcome set. This error has been repeatedly pointed out by others. I am actually trying to prove myself wrong. So the conventional majority? Show me math s that explain 24 possible [Begin][Choice][Switch] endstates. Therein lies a revelation. So far, none do. You have not answered any of my questions.
Please tell me how you reconcile your competing ideas about likelihood. You are assuming that all outcomes are equally likely, which is prima facie absurd, something my simple example above demonstrates. It is akin to saying — well, if I buy a lottery ticket, there are two outcomes: Just because two outcomes exist does not mean they are equally likely, a fact you have yet to come to terms with.
What exactly is wrong with the truth table I generated? There is no unrepresented path through the game. If you accept that, which I think you should… All paths are equally likely. Are the following outcomes equally likely? You have not provided any maths that backs that up. I have shown you every shown every possibility. Janus if every possibility is not equally as likely, how do you explain your idea that my initial likelihood is one of three. Plus, another contradiction is the chance of the car being behind each door changes based on what the player chooses, e.
A clear failure of probabilities. This is why you are wrong. Ok, at this point we have 9 outcomes, all outcomes are equally likely. I think my table could be flawed — hence request for critique. The chances of the player choosing the different things in your table are not equal, they are dependent on the location of the car, this biases the data. The same for Door B and Door C. Mathematically, this is identical to the doors example. Did you even read my post ? If so where do you disagree with the analysis that showed that staying and switching are not equally likely outcomes?
This is a consequence of two game rules: IF you choose the car, monty has goats 2 and 3. Anyway, add up the probabilities in YOUR previous comment: ZenasDad might is wrong, but he did try and he put out every possibility. Palmer say zenasdad matrix show correctly all possibility. So i see why zenasdad thnks your math is wrong.
Someone can explain why he is wrong to show all possibility. There no game he did not show. Interesting to note that in the simulation atop this page, computerised Monty always chooses the left most door when he has the option of doing so ie if you have chosen the door with the car car or if the car is behind the door to the right. It is mistake to say zensdad is wrong because of maths we are taught. He say there are two choices, after first choice, every tim, we can stay or switch. No mistake in his program http: If you modify line 1 of my code to:.
Computerized Monty in my code follows two rules: I thought I did.. My code should put 2 blank line between car locations a,b,c 2. My code should put 1 blank line between stay, switch. Change source code line 18 from: Should clarify my point. Because of rule 2 cannot reveal car , computerised Monty only opens the right-most door of his two options when the car is definitely behind the other non-contestant door. That is a mistaken inference. Monty has two goats he can only reveal a goat..
Other possibilities with Car behind Door A: It should be obvious why, when Choice! Guys, my boss istampering with the thread I left my machine logged-in while I went to the bathroom. Editors, pls erase comment snd There are not 4 outcomes. Purpose of model is to represent reality.
Code is at pastebin. So you ran the program 24 times: Not only that you randomly picked Door A exactly 8 times and precisely 4 times when the car was behind Door A , Door B 8 times 4 times when the car was behind Door B , and Door C 8 times 4 times when the car was behind Door C. If you look at only the branches where the player choose Door A i.
I cant run the VBS. What the data output implies is: Any sane person would have warning bells going off that they made a mistake at this point. The overall statistics have me puzzled. When I have time I will modify the code. The fix is very simple to explain. Since I am insane [a blasphemer? Simple to under stand right? What I did what multiply the size of each choice array together as the calculation goes on.
Here is the output:. Use view page source to see the formatting: Listing the outcomes is trivial see post and can be done manually in less than 10 minutes, the hard part is getting you to understand that not all outcomes are equally likely. Apart from being pointless, your program fails to take the probability of each outcome into consideration, and is therefore WRONG. The maths works out the same, either way. That is how probability trees work, the number of splits determines the likeliness of an outcome, and you multiply the number of splits at each level. Chooses the leftmost door when he has two goat doors from which to choose or the car is behind the rightmost door; and.
I achieved seven correct guesses in a row and a I asked a professor. The answer is, and has always been, Two independent and one dependent probabilities: This has already been explained a gazillion times. Look at the tree structure in your post for a visual representation. My results do not reflect the true probabilities. PalmerEldritch, my above average wins were achieved by always choosing the door that last won the car after initial guess. I think I also elected to not switch on some occasions when the leftmost of the two remaining doors was opened by Monty.
Is ignoring the fact that Monty has TWO choices of doors to open when the contestant chooses the right door giving misleading logic? Hi Pamela, great question. Hrm… something is fishy here! Originally, you pick one door out of 3. Yes you are right. I thought it through and realised that because Monty is constrained not to choose either the door with the car behind it nor the door already chosen by the contestant then it makes it more likely that the remaining door has the car behind it.
I now feel intuitively that this is true. A bit annoyed with myself for not seeing it earlier but thanks for helping me to think further. Thanks also to Jason. A player then chooses 1 door out of For the sake of argument they always choose Door 1. Casual Essay — Vinny Colantuoni Counterintuitive I disagree with your above statement. The probability of picking the car by not switching is different than the probability of picking the car by switching. Consider switching after your first choice. After you pick your initial door, the player does not know whether their initial door held the car or not, but Monty already knows whether you picked correctly, because he knows where the car was placed for that specific round of the game.
This also shows how different observers can know different probabilities for the same choice, dependent on their level of knowledge:. I said pick the car meaning pick the door with the car behind it. Picking the car at the first step does not mean win the car. I still think my math works. How do you reconcile the math of the probabilities for the example I gave in my first post?
This is mathematically incoherent. You need to enumerate ALL branches on your tree, not just one, and add up the probabilities of the outcomes. That will show you the problems in your analysis. That means that in every other situation, switching MUST get you the car, even if you chose the car in your first choice.
The upshot is that improperly calculated probabilities lead to paradoxes and contradictory statements. Palmer, I noticed that you asked the author why this was so but he never responded. I was thinking of what this shows but admittedly it came out jumbled. In that blog http: Research Argument — Vinny Colantuoni Counterintuitive The question is really: Would you rather have one door or two doors?
Monty is letting you have two doors with your second guess. Who would not take this deal? Yes, one of the two doors will be a goat, but Monty will tell you which of your two doors it is. It is not clear! Andrew, the player picks first. Out of 3 doors. Now You Know Absolutely Everything: Absolutely every Now You Know book in a single ebook Dec 06, Provide feedback about this page.
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