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Deal Or No Deal/Monty Hall Problem
Hello,
Long time lurker, first time poster here. I was arguing with some friends over whether Deal or No Deal is essentially the same probabilities issue as the Monty Hall problem. Although there are some fundamental differences, our basic question was: "When it is down to your case and only one other case, with $1 million and $0.01 on the board, are you better off switching your case?" We were also talking about it for when there's 3 cases left, but that part doesn't really matter. I continually switch off in my own thinking, completely convincing myself back and forth that it is, or isn't, the same. Also, don't let this thread turn into a Monty Hall problem argument. It's generally known that you are better off switching if you disagree, I ask you to do some other research. Here's my thinking both ways $1 million is the car, $0.01 is the goat: Yes, it's the same: At the beginning, you pick one case, hoping for $1 million. You have a 1/26 chance of getting it. Just like in Monty Hall, other cases (that you KNOW aren't $1 million because of the ground rules I've laid out) get eliminated, and in the end, you're offered the switch. As far as I can tell, this is exactly the same as Monty Hall. No, it's different: In the sakeofargumentstyle rules of Monty Hall, Monty always has knowledge of which door has the car, and purposely chooses a goat. This knowledge is a fundamental difference in Deal or No Deal and Monty Hall, and affects the probabilities. So, tell me dopers: You've rejected all the deals, and are down to $0.01 and $1 million. You are offered the chance to switch. Should you? Thanks, Matt 




In Deal or No Deal, you first choose one door and then choose 24 doors to open. So you've essentially chosen both of the doors yourself. In Monty Hall, you choose one door, and the host chooses one of the two remaining doors.





To make it more explicit, imagine a setup with 26 doors. You choose one, and then 24 more doors are opened and their contents revealed. In the Monty Hall scenario, Monty chooses which door remains unopened. In the Deal/No Deal scenario. you make the choice. Does this help?







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You've already illustrated why. The assumption in the Monty Hall Problem is that Monty knows where the goats are and deliberately reveals a goat, which is what makes it logical to switch. But if you were to consider a scenario where Monty DOESN'T know where the goats are, switching would be pointless; if Monty pulls back Door #2 to reveal a goat, he's simply hit his 66.6% change of hitting a goat and the odds are now 50/50 for either remaining door. Assuming no knowledge is involved in the reveal, there's no advnatage to switching. In "Deal or No Deal," there's no knowledge involved; the host, Howie Mandel, is not picking any cases or you and anyway he doesn't know where the million dollars is, either. The "I might have a million bucks in my briefcase" idea in "Deal of No Deal" is purely illusory. It would make no difference at all in the way the game is played if you didn't choose your briefcase to start the game, but instead just picked 25 straight cases to eliminate and were left with the last one. Personally, I'd be the worst Deal or No Deal contestant ever. "Okay, Rick, pick six brief..." "Gimme one through six." "Uh, okay. (Reveals cases) Now, the banker is calling..." "Tell him to go to hell, no deal. Now open cases 7 through 11." 




It doesn't make one bit of difference. Both cases which are unknown have a 50/50 chance of being the $1million case.
If anyone could explain why it would be a good strategy to switch or not switch besides some supersticious reason I'd like to hear it, 




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This thread appears quite dead but I feel compelled to offer an alternative explanation to the author's question for anyone who stumbles on this as I did. I would argue that deal or no deal and Montey hall are essentially the same. People seem caught up on the idea that the host's knowledge of which door the car is behind fundamentally differentiates this game from the Monty Hall problem and I have yet to see a good explanation here of why. For me the most important thing regarding the last 3 briefcases/doors is that one is revealed and that that one is not the million dollars/car. It does not matter that the revelation of the goat or nonmollion dollar briefcase was revealed by chance or because of the host's knowledge. What is most important is simply that it is revealed. Regarding the last three brief cases, you know you have a 1/3 chance of holding the million dollar briefcase, and a 2/3 chance of holding a briefcase that does not contain the million dollars. 2/3 of the time, the revelation of a nonmillion dollar third briefcase will simultaneously reveal the million dollars. There is only a 1/3 chance that of the remaining three you hold the million dollars, and thus 1/3 of the time it is in your best interest not to switch after the 3rd briefcase. 2/3 of the time, however, you will be better off switching, simply because of the greater chance of having chosen a nonmillion dollar briefcase in the first place. Perhaps this helps explain it. Just think that if in the montey hall game the goat WERE revealed by chance, it would still be a good idea to switch.







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But the whole point of the Monty Hall game is that the goat ISN'T revealed by chance. It's purposefully giving you the "wrong" choice of your 2/3 chance when the odds are still 1 or 2 out of the three.
In the Deal or No Deal situation there is always a chance the "goat" reveal may be the $1million box. In the Monty situation that chance does not exist. 




If a goat is revealed by chance and two doors remain, each has an equal chance of being the winner. What asymmetry can exist here that would make one more likely than the other?





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If the host is Monty, you pick one box to hold on to, Monty opens 998 boxes that he knows are goats, and gives you the option of switching. Obviously, you're going to switch. The remaining box *has* to be the one with a million dollars, because there's only a slim chance that the first box you opened was the correct choice. In fact, you have a 99.9% chance of getting the million dollars if you switch. On the other hand, if the host is Howie, you pick one box to hold on to, and Howie makes *you* open 998 boxes. Most of the time, you're going to open one with the million dollars some time during that 998 run, and the game will be over you're going to go home with a goat no matter what. But in the unlikely (0.2% chance) event that you managed to open 998 boxes at random and every one happened to be a goat, it doesn't matter if you switch or not. What's the difference between the two boxes? Both of them *you* decided to hold off from opening, for whatever reason. What difference does the order make? To sum up, with Monty as the host, you always have a 99.9% chance of winning as long as you switch. With Howie, you have a 99.8% chance of *losing* during the initial opening phase, and then 50/50 chance of winning if you manage to get past that phase through sheer luck. Overall, you only have a 0.1% chance of winning no matter what you do with Howie. I'd much rather be playing with Monty. 






I didn't know how Deal or No Deal worked, so I looked up the Wikipedia article on it, and when I saw the picture they chose to illustrate the article, I was pretty sure that the editor of that article is a marketing major who has studied the art of website promotion.
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Add your photo to the SDMB Portrait Gallery! Last edited by Arnold Winkelried; 02162011 at 06:32 PM. 




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1) Does Monty EVER open the door with the prize? Or does he ONLY open doors with goats? 2) On Deal or no Deal, does the contestant EVER open a case with $1 million? Or ONLY cases with lesser amounts? So, the difference is simple. In one game, the contestant can randomly reveal the big prize (thus eliminating it from play). In the other game, the host deliberately avoids revealing the big prize. Of course, I like me some goat meat, so the Monty Hall problem is a no lose proposition. Last edited by Great Antibob; 02162011 at 06:42 PM. 




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So yeah, amongst the last three briefcases (and definitely yamatotwinkie's 999 boxes also), Monty's knowledge is significant because it prevents the contestant from eliminating the prize. Definitely, without a doubt. Perhaps though the revelation of the goat by chance still gives one sufficient reason to switch from their 1/3 chance of winning the prize to a 2/3 chance. Someone try understand what I'm saying so I don't feel so completely insane . I'm damn sure I'm right, though . 




or, better yet, perhaps someone could try to explain why Monty's knowledge versus his ignorance of the what lies behind the doors alters the original Monty Hall probability after the third door has been shown to conceal a goat. How does the "asymmetry" disappear? I am referring only to the last three doors or briefcases and not to the monumental feat of probability that you would have to achieve to make it to the last three briefcases with Howie as the host.
Last edited by Camdgoodman27; 02172011 at 12:54 AM. 




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"I'm going to set aside box #12 first, then set aside #752 as my second box" or you could have decided: "I'm going to set aside box #752 first, then set aside #12 as my second box" Either way, the game proceeds identically. In both events you result in two unopened boxes that you could have chosen randomly from the getgo. The order makes no difference. In the unlikely event that you didn't reveal the million dollars in the 998 other boxes (which seems to be the part you're still stuck on), it doesn't matter which of the two boxes you set aside "first". Why would it? You could have chosen the two at the exact same time! 






IMHO I always thought that if it came down to the $1 million case and the $0.01 case you were better off switching. My reasoning was like the OP’s first thought; that the odds that you picked "correctly” the first time was 1/26 so the odds that the $1 million case was still on stage were 25/26.
I think the key is that if through chance it comes down to the 1cent and 1m cases you still know what the odds originally were. If Monty didn’t know where the prize was but got lucky and revealed a goat, the odds are still in your favor to switch, right? The OP set the condition that chance played a major part and got you to this point; you were lucky and opened 24 “wrong” cases. So what are the odds that you picked the right case at the start? Still 1/26. Of course I only took one class in statistics and I’ve only had 1 cup of coffee so far today. 




What is possibly being missed in the random reveal scenario is that as goats are randomly removed from the "stage pool", the probability that your original choice is correct increases. This is usually the intuitive thought  it's often harder to grasp that when one of the stage pool goats is revealed intentionally, your odds of winning don't improve.
Another way of looking at it is this: If Monty intentionally reveals a goat, he is selecting from all goats excluding your choice if it is a goat. That skews the results toward the winner being on stage once more and more goats are revealed. Once the number of goats left gets down to 2, it's more likely than not you've got one. On the other hand, a random selection that reveals a goat excludes your case as well. This time, your case is as likely as any unopened case to contain a winner. Since only goats are revealed in this scenario, the likelihood that any unopened case (including yours) contains a winner is equal. Once it gets down to two, there's no reason to switch as you can't do any better. 




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Imagine you're playing the lottery. You and your friend have the exact same ticket, except the last number is different (you picked 0, he picked 1, and there were only two options). The radio station starts slowly reading out the numbers, and both of you have matched the first set of numbers with only one number left to be read. Do you switch tickets with your friend before the last number is read? According to your logic, since you only had a 1 out of ten million chance of winning the lottery when you first bought the ticket, your friend's ticket should have a 99.999% chance of winning. But then wouldn't your friend think the exact same thing about your ticket? You both can't be right. 




The entire point of the Monty Hall problem is that Monty has told you something you didn't know at the start (or, another way, he is constrained in his actions during the reveal  your initial choice dictates his choice).
If Howie chooses the cases for you, always avoiding the $1 mill, then you should switch at the end. But since you choose the cases yourself, each time risking the 1/26 chance of opining the $1 mill, it makes no difference if you switch or not. 






And, like RickJay pointed out five years ago, in the game of Let's Make a Deal, if Monty didn't know which door had the prize and which had the goats, but just got lucky on this one event, you have no advantage by switching. The "2/3 probability of winning if you switch" answer applies only in the case where Monty knows where the prize is and avoids opening that door.





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1/3 You have picked the prize. 1/3 Monty opens the prize door. 1/3 The remaining door has the prize. If Monty opens a door that reveals a goat, one of these possibilities has been excluded, leaving 2 possibilities with equal probibility. He could have just as easily opened the prize door, ending the game. It's this new information that changes the odds of your door winning. If Monty knows what is behind each door and only opens a goat door, the chances are: 1/3 You have picked the prize door. 2/3 One of the other doors is the prize door. When Monty opens a goat door in this situation, you have learned nothing new. He always opens a goat door, so you have no new information. The odds of your door winning are still 1/3, and the odds of the other door(s) winning is 2/3. 




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Let the doors be numbered 1, 2, 3. The prize is behind one of them, call it A, and the others B and C. You can choose A, B, or C. Then, a second one is choosen (by you in Deal or No Deal; by Monty in Let's Make a Deal), and the last is left. So, our options are ABC, ACB, BAC, BCA, CAB, CBA, which exhausts the possibilities, and all of which are equally probable. In Deal or No Deal, you win by staying with the two cases ABC or ACB, but you can also win by switching in the two cases BCA or CBA. Your chances are the same. However, in Let's Make A Deal, Monty would never choose the winning door, so of the six possibilities, what's left is A(B/C), A(C/B), BA, BA, CA, CA, after Monty's door is removed. You can see that your chances are improved by switching. 




RM Mentock, among others states it clearly. Monty Hall will never reveal the grand prize, only a lesser prize. I don't know much about statistics or probability, so to resolve this for my own understanding, I wrote a program to simulate the circumstances. It was immediatly obvious that the selection of the door to reveal was not random, it had to involve knowledge of the prize behind the remaining two doors. On Deal or no Deal, that isn't a factor. First of all, the choice is not made when there are three cases left, just two, when you have a choice between your initial pick and the only remaining case still held by a model. Secondly, on Deal or No Deal, the choice to switch between the final two cases isn't based on any knowledge of the contents of either case.
In layman's terms it's easy. On Let's Make a Deal, you have a one in three chance of picking the right door. One out of three times you will pick the winner, just by chance. When Monty reveals one of the other doors has a goat behind it, he is telling you that there is a one in two chance that the door you didn't pick contains the grand prize. Why? Because two out of three times the door you didn't pick the right door contains the grand prize. Here are your choices: 1. You picked the right door, and the other two don't contain the grand prize. 2. You picked the wrong door, and the other door A that Monty has not revealed contains the grand prize. 3. You picked the wrong door, and the other door B that Monty has not revealed contains the grand prize. So two out of three times, the other door that did you didn't pick contains the grand prize. One out of three times the door you picked contains the grand prize. So if you switch, you change from a one out of three chance, to a one out of two chance, to win the big one. On Deal or No Deal, you get the opportunity to switch only when there are two cases left. So here are the choices: 1. You picked the right case, and the other one is the wrong case. 2. You picked the wrong case, and the other one is the right case. It doesn't matter whether you switch or not, the chances of winning are the same. One out of two. Keep in mind, on Deal or No Deal you are only given the choice to switch when two cases are left. On Let's Make a Deal, there are still three choices when you are given a chance to switch, and Monty only reveals the door which is not the grand prize. It's not a random selection. Monty isn't going to show you the door with grand prize, because then the show would be over, because everyone would know you lost, and there's no point in revealing what's behind the door you picked. On Deal or No Deal, if they did something similar when there were three cases left, what would they do? If they reveal the contents of case you didn't pick, and it's the big prize, the show would be over, nobody would care what was in the case you picked. So they would have to reveal a case based on the knowledge that it didn't contain the big prize. 




If Monty Hall is too easy for you, here's a generalization that may be a good exercise in applying symmetry principles to find optimal mixed strategies.
Montgomery and Sally play a game with K closed boxes. Montgomery hides a milliondollar jewel in one box (the other boxes are empty) and proceeds to open the empty boxes one by one as Sally plays (see below). When there are only two boxes left, Sally opens either box and wins if it contains the jewel. Prior to each of the K2 box openings Sally chooses one box and locks it, preventing Montgomery from opening it next. That box is unlocked after the opening and cannot be so locked twice in a row. What are the optimal strategies for Montgomery and Sally and what is the fair price for Sally to pay to play the game? 






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It should read: "So if you switch, you change from a one out of three chance, to a two out of three chance, to win the big one." Note that the MH problem is exactly equivalent to this: Behind 3 doors are 2 goats and one big prize. You have the choice of choosing one door and keeping what's behind it, or two doors, keeping your choice of what they conceal. 




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Dead thread, but...
It's been a while and Deal or no Deal isn't making new episodes, but I wanted to try my hand at this.
1. A lot of people here seem to be saying the whole POINT of the Monty Hall problem is that the host KNOWS where the goat is and picks FOR the contestant. That it's NOT done at random. No. This is not at all the point. The host knowing ONLY simplifies things and gets us to the ACTUAL point of the MH problem. Imagine if he didn't know. Or you choose. Or your overweight nearsighted autistic swimming instructor chooses. Or a butterfly sneezed in the studio and accidentally knocked down a door revealing a goat. The host knowing is just a vehicle to quickly and simply get us down to the scenario where a goat is shown and you have the option of sticking with your door or switching to he other. How we get there really does NOT matter. 3 doors 1 car 2 goats You are down to a prize and a lesser prize 2. Yes, technically the $1,000,000 is the car and every lesser amount would be a goat. We're using "goat" here as a variable. Now THIS is where the randomness and NOT KNOWING DOES MATTER. It determines what amounts are goats and what amount will end up being the "car". This is the source of the confusion. At the end of DEAL or NO DEAL you will always have two cases. One with a higher amount and one with less. ALL the OTHER cases have been opened and revealed an amount that is not the prize. The prize is the highest amount you can get. Say it's $.01 vs $50,000 or $200 vs $100,000. We'll go with the latter two amounts. It does not matter what amounts were actually in all the other cases. We are standing on stage with $200 and $100,000 still on the board. All the other cases are our third door. Many of those cases revealed higher amounts than our $100,000 and many revealed less. But $100,000 is NOW what we stand to win or miss. Since you didn't pick any of the other cases you couldn't have won them... they may as well all have contained goats at this point. Or they may as well all have been one door containing a goat (or in this case $200) What are the odds that you chose the $100,000 at the beginning? What are the odds that it's still on stage and you have another $200 in your case? Monty Hall problem. 2/3 odds again if you switch. 3. THE ONLY THING THE RANDOMNESS of the game does is gradually determine by chance just how much the "car" is worth when we get to the MH Problem at the end. 4. This thread was frustrating to read 




I should change the wording at the end bit. It's not what the odds are that you chose $100,000 at the "beginning" (beginning of the game). It's as if you only had two rounds to "begin with" knowing the options were two "not $100,000" (or in our hypothetical scenario two goats... or $200) and one "$100,000" (car). The entire game thus far has revealed all the cases that are "not $100,000" and can be consolidated into opening one door with a goat behind it. You're left with two doors and the option to switch.





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If Monty doesn't absolutely know where the goat is, he opens the door with the car some fraction of the time (actually 1/3 of the time). But this never happens, because Monty KNOWS where the car is and NEVER opens that door. So, yes, the point IS that Monty knows where the car is, and that affects the probability. 




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If Monty is forced to always show a goat before offering the choice, then the answer to the problem is that you should switch. In the real TV show, Monty was capricious and there is not a clear answer about what's best to do. 






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The times he does, the game is over and the problem doesn't come into play. The participants chances were always 1/3 The 2/3 of the time he doesn't would be physically identical to the problem where he intentionally opened a door with a goat behind it. In these 2/3 of the time the problem is actually allowed to occur and becomes a subcategory upon which we base the likelihood of the location of the car are the same since we're dealing with same results. This is where things get confusing for me and everybody else You're all right in saying I don't see how this subcategory could be any different than th formal version of the problem. I don't grasp how his KNOWING and opening the wrong door intentionally somehow confers a different probability. By breaking down into a subcategory he still opens the wrong door every single time within this category, the only time the full game is played. The options and possible permutations are still the same where one possibility is he switches and gets a goat and the other two possibilities he switches and gets a caras broken down graphically in the "simple solutions" portion on the Wiki page for the problem. It would still appear to me that including the times where the game is over early (and we could therefore consider the times it's not as a new and separate game) is to also consider how often a contestant wearing a blue shirt wins as statistically relevant. I don't see it bearing any influence on the new game any more than... what's that logical fallacy that relates to flipping a coin? No matter how many times you flip it, though odds are it should turn up say "heads" eventually, each flip is it's own and the odds, possibility remains only 1/2. Different logic problem, unfortunately brings up the 1/2 stat again which could lead to confusion, but I hope it illustrates better where I'm coming from. 




My apologies
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3 doors, A,B,C. Assume A is the winning door. Pick A: Monty Picks B, switch loses Pick A: Monty Picks C, switch loses Pick B: Monty Picks A, game over Pick B: Monty Picks C, switch wins Pick C: Monty Picks A, game over Pick C: Monty Picks B, switch wins Half the time when we pick wrong, the game ends early. But every time we pick correctly, the game does not end early. In this case, if we are given the option to switch, there is a 1 in 2 chance that we started off by choosing the winning option to begin with. Consider a variant where we have a goat, an ass, and a car. If we are still in the game, the probability that we picked the goat, ass, and car are not the same. 






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What purpose is served by changing the game so it is entirely different? As originally presented, Monty knowing what's behind the doors isn't the important thing. The important thing is that he NEVER opens a door with the car. Whether that's accomplished because he has foreknowledge or because he gets signaled by the producers or through magic, it doesn't matter. It's NOT the same game if he does sometimes open a door with the car and the game ends there, because that doesn't ever happen in the traditional "Monty Hall Problem". Monty is guaranteed, under the terms of the game, never to do that. Again, it doesn't matter "how" this is accomplished, only that it does. Basically, you want a variation on the game where the key factor (Monty never opening the door with the car) is no longer part of the game but also doesn't count towards the probability. That's a radical change that is fundamentally different from the original version. 




No, this scenario NEVER happens. That is the whole point of the set up, MH knows where the car is and never opens that door. If you ignore that point then you've set up a different problem that may be interesting, but isn't the Monty Hall problem.





Here's an example that should make this difference crystal clear. Imagine that you're playing against "Evil Monty," who wants you to lose, and only offers you the choice of whether to switch if you've picked correctly with your first guess. In this case, the fact that he has even offered the choice tells you that you had better stick with your first guess.





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The game goes like this: Three doors. Contestant selects a door. Monty reveals one of the other two doors, that does not have a prize behind it. Contestant can either have what is behind the door they selected, or what is behind the third door which was not selected or revealed. Two out of three times the prize is behind one of the doors the contestant did not select initially. In every game Monty reveals one of those two doors that does not have a prize. That means two out of three times the third door has the prize. That's why it's better to switch. Another way to look at it is this. At least one of the two doors not selected initially has no prize behind it. Therefore switching is like getting both of the prizes that are behind both of the doors not initially selected. One out of three times neither of those doors has a prize behind it, but two out of three times the prize is behind one of those doors. Last edited by TriPolar; 01072013 at 02:39 PM. 






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In your bolded part above, did Monty open a door without a prize because a) he has to, those are the rules that he always has to open a nonwinning door, or b) because he picked a door at random and in this particular case that happened to reveal a nonwinning prize? To state the puzzle properly, you have to be specific that Monty is constrained that he must open the nonwinning door because he knows which door is the winning one and he has to open another one. 