Monty on a Mission (Monty the Traveling Cat Book 5)
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Do you believe that this item violates a copyright? Amazon Music Stream millions of songs. Suppose there are three doors, A,B and C and you originally chose door A.
If you stay with your original door, then the only way that you win is if originally the prize was behind that door A, which has a chance of 1 in 3. If the prize was originally behind door B on the other hand which has a chance of 1 in 3 , then when you pick door A, door C will be removed. Hence, if you switch you will be switching to door B, and therefore you will win. Finally, if the prize was originally behind door C which again has a chance of 1 in 3 then door B will be removed, and if you switch you will be switching to door C and therefore will win.
Hence, if you stay with your original door, you win if and only if the prize was originally behind door A. If you switch though, you win if it was originally behind either door B or door C. Since the chance the prize being behind door A from the get go is 1 in 3, whereas the chance of it being behind either B or C from the get go is 2 in 3, you are better off switching!
Let me post 2 cases to challenge those that have assumed 2 things. Bear with me on the slight differences to the original example. Host decides to reveal 0 doors: Host decides to reveal ALL of the remaining doors. Many have stated that the initial probability of the Door choice is to remain unchanged despite additional info provided by the host. In this case, 2 possibilities can happen, host reveals the prize by opening all the remaining doors or reveals all the goats. I have given a counter case that suggests otherwise. I believe in some cases, it is not apparent that you should switch like in Case A, switching or not makes no difference if no information is revealed.
Even though it would be an act of switching to a higher probability. When a host reveals the location of some of the goats, the total number of possibilities of the actual location of the prize is directly reduced, how can the probability if Door 1 is a prize be unaffected?
Sorry for the presentation if it confuses anyone, I am rather unconvinced at what the majority is accepting which motivates me to find out if I am wrong in my understanding. The remaining doors regardless of position will be considered as per stated in my previous example. Which in my opinion is a rather true reflection of real life choices. Much has argued that remaining doors collectively contain a higher chance, then follow through their calculations while assuming the probability calculated on the first chosen door should remain the same throughout regardless of additional information revealed on the remaining doors, which i find rather weird.
In the game, whatever your choices are at the start, in the end you will have two doors. By eliminating the rest of the bad doors you are not helped at all, in the end there will always be just 2 doors with equal probabilities. One with the car which might be your first choice and one with the goat.
Nobody corrected this in 8 years, lol! The fundamental nature of this problem is that Monty is asking you to bet on whether your initial choice was right or wrong. I love when this question comes up, sooooo many people never get it. The way I get most people to see it is by looking at Deal or No Deal. Say you have selected your case and there are 24 or is it 25? Then what is the difference if you open them dramatically. Not the same exactly as the Monty Hall, but sometimes helps people figure out what is going on.
Thanks for the great site! Frank is right comment Therefore, your initial pick has no relevance to what he is going to do and is a completely independent event. Of the remaining two doors, exactly one contains a car. Monty is essentially asking you to flip a fair coin and then showing you a picture of a goat. That game is a demonstration of conditional probability.
You already know this. That is the very definition of an independent event. I choose door 1: Monty says show the goat. I choose door 2: I choose door 3: Have a great day and thanks for an awesome site. There being 3 doors does matter. Think of it this way: Therefore the car is behind the doors you did not choose with 2 out of 3 times. This does not change no matter what Monty does. Randomness is the issue, which is what confuses everyone. In a random choice if he doesnt reveal a car then the odds go to 50 However, because he will always reveal a goat, he in essence claims 2 doors to your one and the odds change.
Kalid has caused me to spend the day in primal scream therapy with this Monty Hall problem. Fortunately my head did not explode notwithstanding the impassioned efforts of several posters here to make that happen. Monty is a showman, yes, with the costumes and goats and tricky switchbacks. I read out there in cyberspace that presented with this problem, Monty himself said that yes, if the host is always going to show a goat and give the contestant a chance to switch, then switching makes sense.
Here are the possibilities of the other 2 doors with the probability after each. Note that all I did was remove the first column to concentrate on our next choice:. Monty MUST open a door. Here Monty picks a door randomly. I also initially refused to accept that switching improves the odds though perhaps not as vociferously as you. Every argument including the one I just made felt completely wrong to me.
The only thing that convinced me was me writing a program simulating the game myself, and running it a million times. Since I was where you are now, I offer no argument to you, as I know arguments led me nowhere. I only ask for your own curiosity that you take some time to write the program yourself with the game exactly as it is stated, and to run it a ridiculous number of times.
You disagree with the results of other scientists, but thanks to the reproducibility property of the scientific method, you get to check for yourself! Monty Hall Problem Think of it this way. You have a choice of one door or two doors which has been collapsed to one by elimination. First, either contestants are always given a second choice, or they are not.
If you picked the door with a goat, the other doors have a car and a goat. The probability that you first picked the car door is one out of three, so there is a two-thirds chance that you first picked the one of the two goat doors. That means the other door now has a two-thirds chance of having the car, and your initial door selection has only a one-third chance. Take a pizza divide it into 3 pieces. Make 2 plain and 1 pepperoni. You want the pepperoni.
Choose 2 of the 3 pieces randomly. Pretend those slices are instead doors. We know where the car pepperoni is, so that no longer a mystery. Pretend the 1 pizza slice was the 1 door pick by the Monty Hall contestant. Him removing a door is a kin to you dropping a piece on the ground when exchanging the 1 piece for the 2. If you choose 2 doors aka switch your choice after the first pick Monty is willing to take one donkey off your hands.
Another way of explaining — Let there be 3 doors — A,B,C. Valid — very well done. I admit I get a weird pleasure from seeing the logical knots the non-switchers tie themselves into.
One correction, in comment 32 you said to Frank that Monty is revealing new information with the pick. He can always show you a goat. I have not read all of the above, so if this is already stated then … sorry you have to read it again. Some bozo tells you that at least one of the other cards is a 9 if you accept this like it means something special then you are a bozo as well … because every time at least one of the other cards is a 9. It has been twenty years since I thought of this scenario … thanks Kalid, for the opportunity to keep thinking about stuff. Actually I find learning most enjoyable when you come back to it, vs.
Monty just does you a favor by throwing one away early. Actually Monty does you a confusing to most people disservice by throwing one away early. If he left the unpicked two doors closed and said you could open them simultaneously, and if you see a car … you win. Almost all would then trade for the two doors. When initially trying to understand this problem, I was so focused on the red herring of Monty throwing away door that I failed to see the basic logic as you so clearly explained. This article was so long and drawn out and seemed to apply a somewhat unrelated rationale for conclusion more information allows you to make a more informed choice, hence increasing probability of decision.
This is a rather peculiar argument. Suppose the game worked like this: John and Sarah are playing as a team. They have to agree on which door to pick. Then, John is blindfolded and Monty opens one of the other doors, showing Sarah that it has a goat. What is their probability of winning?
Yet if they play repeatedly, their win rate has to converge on one or the other of those. A more relevant statistic is the simple total of wins, and for trials, that total should be much closer to than to This is assuming that you made sure to always switch, of course! On these occasions, the other goat is eliminated and the car is in the other door, giving a certainty of a car if you change doors with your second choice. Monty revealing a goat does nothing to increase your original choice from Who will win the car more often? Your friend, of course. Which is more likely to have the car?
Misters Groove and Smith Jr should take special note. You better not, because one of his doors HAD to be a goat. Please, someone, expound upon that. Groove 53 You are correct that Monty will always reveal a goat, there is no doubt about that. The question is not whether or not Monty will reveal a goat, but rather is Monty forced to reveal the particular goat that he did, is his choice of goat restricted?
You select Door B. Monty must reveal Door C. In this situation Monty is restricted to choosing Door C. You select Door C. Monty must reveal Door B. In this situation Monty is restricted to Door B. You select Door A. Note that the probability of picking Door B and of picking Door C add to the one third. If this experiment was ran times with the car behind Door A everytime and the player selecting each door times the results would be as follows:. Never opened because the car is behind it. Opened times, times because it had to be as in Scenario 2 and 50 times by the choice of Monty as in Scenario 3.
Opened times, because it had to be, 50 by the choice of Monty. If Monty was forced to choose a door, that means that there was only one available goat for him to choose from. If there is only one available goat to choose from, the player must have chosen the goat to begin with. Yes, Monty will always pick a goat, but it is whether or not he is forced to pick a goat that matters. If I were playing the game, knowing the rules, I would decide in advance always to switch.
Then the situation would be equivalent to the following. I walk on to the stage and select a door. Or would you like to open both the other doors? If a car is behind either of them, you win the car. As you can see, it has a goat behind it. There are 3 doors and you know only 1 has a prize.
You pick a door. Then, you are given the option to pick the 1 door you already chose, or the 2 other doors. Obviously, you would choose the other 2 doors. You already know that at least one of the other two doors are empty— the fact that Monty opens one of the doors is irrelevant. In reality, the information is exactly the same; Monty simply reinforces what you already know that at least one of the other doors is empty. Eliminating the psychological trickery makes it intuitive.
This is an interesting probability problem. We can use tree diagram to work up the chance of getting the right choice. If you originally chose a goat without knowing it , then if you switch you will get the car. If you originally chose the car, and switch, you will get a goat.
Therefore the probability of being right after switching is the same as the probability of being wrong with your first choice i. And the probability of being wrong after switching is the same as the probability of being right with your first choice, i. I get it now. You want to TRY and pick a goat the first time. Now that Monty has removed the other goat, you switch.
Monty has removed the other goat, so you should pick the last one. First, we need to know our basics in adding fractions. Try and pick a goat you pick door 1. Door 2 is removed, so it IS a goat. Which one will you choose now? Enjoy your new car: I have spent countless hours online trying to explain this to people, usually using the or, when really desperate, 1 billion door, analogy, often to no avail. Revealing which of the two is the Goat does not reduce the probability of one of them being the car.
However, today, I came up with what may be the best way of having naysayers challenge their thinking. This should cause the naysayer to concede that, all things being equal, in shows, there will be 33 winners off a straight pick with no one switching after the first goat is revealed. It should be a short leap of logic then for them to see that if 33 people are destined to win, then 67 people are destined to lose, and get only a goat.
The next step is to ask what would happen if each contestant was ordered, under pain of death, to change their initial selection when given the opportunity by Monty. The inescapable conclusion should then be that the 33 people who won a car when sticking with their first choice would become losers and the 67 people who originally picked a goat would become winners, simply by switching to the only other closed door. So another distinction needs to be drawn between the probability of a contestant picking correctly after being given the option to switch which is virtually impossible to compute and the probability of one door or another holding the prize.
Evidently two different things as how people use the information they have is unpredictable. A Vulcan, of course, would always switch. As the whole Monty Hall 3-doors game is mired in preconceived ideas, it may be simpler to ask the person to consider 3 playing cards and ask them how many times in a hundred they would expect to pick a lone Ace amongst the three cards. Editor, could you please change that last number from 67 losses to 33 losses and delete this post, please?
If this were so, 50 contestants would each win a car if no contestant switched and, ipso facto, 50 contestants would each win a car if all contestants switched. So how can the number of contestants who pick the correct door in a one in three guess suddenly increase from 33 to 50? Many people have shown this when they point out the analogous scenario of a game comprising or one million then, usually in desperation, one billion or perhaps, a Googol doors to show that the probability of the first door being a winner very much depends upon the number of doors from which it is chosen, as with any random guess.
You choose door A and Monty can open either door B or C and show a goat. In both cases it will be wrong to change door 2 misses in total. If you choose door B, Monty can only open door C to show a goat, and it will be correct to change door. If you choose door C, Monty can only open door B, and it will be correct to change door. In total 2 good and 2 bad changes The same will repeat with the car behind door B and door C. In total 12 possible solutions, 6 of them will give the car if you change the door, 6 will make you go away from the car.
Be careful with the counting: Door A has the car: Door B has the car: Door C has the car: Try writing out all the scenarios explicitly to be sure. When you pick the right door A , Monty can reveal either B or C. A miss is a miss, and you only lose once when you switch away from the winner. The point is that Monty looks at the two doors you have not picked and he has to take away one goat. I have to open the goat door. As I say, this happens two out of three times. Therefore switching makes much more sense as the chance is higher. Trust me, it works! Even apparently so called maths experts conned by computor programs.
Having revealed a goat which he always does theres two options remaining a goat and a car and the contestant has to choose one of those two. Those odds of winning do not change with the order in which the doors are opened. Try it at home with a deck of cards. Hardly surprising is it?
Consider also a sporting contest between two contestants. However knowledge of form in that analogy is like knowledge of which door was chosen first in the Monty Hall Problem. There are 10 doors and you randomly pick 1. Monty closes 8 of them. So there are 2 doors left: You increase your winning chances by switching. So there are 2 doors left. One with the car and one with a goat. So switching doors increases your winning chances. Switch and win the car. Regards, Snehal Masne http: You have have no idea which of these answers is correct so you choose one at random lets say answer A.
You use your You are left with answers A and D. Should you switch to answer D? My understanding is that because answer A was not protected from the Like Steve, I am amazed at all those people who do not understand probabilities. For the first part, the probability is 1 in 3 of picking the car. After the host opens a goat door the probability of picking the car is now 1 in 2.
Once a door is opened, you are looking at a new problem. The old situation no longer applies. The way the Monty Hall problem is phrased is intentionally confusing. I see websites that wrongly say there are only 3 scenarios for the switching, but there are actually 4. Work through it yourself. Phil, rather than be amazed, like Steve, you should take some more time to consider what you know is counter-intuitive.
That means, your first instinct is most probably wrong. 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. 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.
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.
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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?
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.
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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.
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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: