## Monty Hall Problem [solution discussion]

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Theres a really neat diagram explaining the monty hall problem in "The Curious incident of the Dog in the Nighttime" by Mark Haddon
In a forum dedicated to logic puzzles most ppl here have probably read it hehe
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Yeah, I read some of it. I think the whole concept is a bit bollocks though, so I'm not in a hurry to read any more of it. Just didn't feel like it was real

Hey, it would be cool if there was a section for talkin about books. I really want to talk about Philip K Dick - Wait, some of his books are a logic puzzle!
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A book forum here would actually be an excellent idea, I agree whole-heartedly = D
ph33l_da_l0v3

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### Re: Monty Hall Problem [solution discussion]

Since the problem thread is back on the main page I might as well post the reason that "lay people" (in this case, correct people) have so much trouble with this solution:

Firstly: the stats don't lie, on a number of trails, you are better off to switch. No question.

The problem is that the stats are based on a change in the problem without accepting that it is a new problem (ie. that the stats are no longer relevant *for that trial*).

Here we go:

I pick a door, I am either correct or not. At this point all discussion of the 66%/33% chances goes out the window; I've picked and I am either right or wrong, not 33% right and 66% wrong.

If I was right, switching makes me lose

If I was wrong, switching makes me win. (my door + monty's door leaves only the correct option if switching)

There are a couple of problems loosely related to this, that I'll now give:

"The population box": You're told that a box has either 1 black and 9 white marbles, or 1 black and 99 (or 999, whatever) white marbles. You hit the button to drop a ball and (gasp) it's black. Was it 1/10 or 1/100(0000000).

Problem: You don't know the odds, because you've already picked... The odds have no sway over an individual, singular event; only multiple trials.

"The scratch lotto" (this is a good one to piss off stats majors): You have a 99.9(99999999)% chance of losing the scratch lotto the instant you walk into a store. Despite the "odds of winning" of 1 in 100000(00000000), there are only 12 or so tickets available to purchase, meaning that you have VERY slim odds of winning at all.

The bottom line is this: *once you have chosen* you are no longer a statistic to be, and the odds of you winning are not dictated by statistics to be. Whether or not switching is better for *you* is up to whether or not you guessed right or wrong to start with.

On the other hand, if you got a whole bunch of shots at it, the best strategy to *maximize overall winnings* would be to switch. Within each individual trial though, the pre-guess statistics would still be meaningless.

That is why "lay people" think that the Monty Hall problem is bogus (and they're right).

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### Re: Monty Hall Problem [solution discussion]

adriankemp: OK, your post was an illogical rambling wreck, but I just have one question:

Would you switch doors?
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### Re: Monty Hall Problem [solution discussion]

adriankemp: You may want to look into Bayesian probability. In that interpretation, you can be in the situation where you're either right or wrong, there's no more randomness... but which state is true is unknown to you. And when that happens, you can still talk about the probability of you being right.

Like, say, with your scratch lotto ticket, which is a great analogy - after you've bought the ticket, you've either won or lost, but you don't know which. However, you can say that you have a 1 in however many million chance of having already won, based on the information you have so far. Then, as you scratch the ticket, you'll get more information, and your probability will go up or down based on that. Until finally you've scratched the whole ticket, and the probability will hit 1 or 0, as you have full information.

The alternative to Bayes is frequentist probability, and it would phrase the numbers as not "what are the chances I've won" (as, by that interpretation, as you said, that is either a 1 or a 0, we just don't know which)... but rather "of all the situations where I would know what I do know now, how many would turn out to be winners?". Both this and the Bayesian thing end up boiling down to the same conditional probability equation, P(I won | stuff I know) = P(I won && stuff I know)/P(stuff I know).
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### Re: Monty Hall Problem [solution discussion]

adriankemp wrote:I pick a door, I am either correct or not. At this point all discussion of the 66%/33% chances goes out the window; I've picked and I am either right or wrong, not 33% right and 66% wrong.

The idea that the 66/33 business "goes out the window" is incredibly misleading.

If you repeat the Monty Hall scenario a large number of times, your first choice will be correct about 33% of the time, and your first choice will be incorrect about 66% of the time.
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### Re: Monty Hall Problem [solution discussion]

On the scratch-lotto analogy, suppose there are 1,000,000 tickets, of which exactly 1 is a winner.

You buy one ticket. But then, before you scratch it off, someone else buys all the other 999,999 tickets.

Now either you hold the winning ticket, or that other person does.

But then the other person says, “Hey, tell you what, if you want to, I’ll switch with you. I get the ticket you have, and you get the best ticket I have. But you don’t have to decide now. First, I’ll scratch off all my tickets, so I will know where the winning ticket is. Then I’ll show you 999,998 losing tickets, since I definitely have at least that many, and I’ll throw them out. Then we’ll each have one ticket. That’s when I’ll ask you if you want to switch.”
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### Re: Monty Hall Problem [solution discussion]

phlip wrote:adriankemp: You may want to look into Bayesian probability. In that interpretation, you can be in the situation where you're either right or wrong, there's no more randomness... but which state is true is unknown to you. And when that happens, you can still talk about the probability of you being right.

I don't disagree with what you're saying, and in fact you are very correct. But it also doesn't actually contradict any of what I said.

Every method of statistical analysis will point to the same answer: 66% of the time you will win if you switch. 33% of the time you will win if you don't. This is correct and I am not arguing it.

Ultimately, the problem is that Bayesian probability is (pretty much by definition) a sort of fiction; it's a bit like imaginary numbers... They can be a part of the underlying mechanics, but go ahead and try to give me 5i pencils. My chances don't *actually* change as I'm scratching the ticket... surely you recognize that the ticket is not re-ordering and changing as I'm scratching?

This is the heart of the problem: Much like a lotto ticket isn't changing as I'm scratching it; the car isn't jumping around behind the doors and I'm not magically half-guessing a different door. Once a choice has been made *that trial* is subject to the choice that was made and cannot be changed. Ergo, if you pick correctly; switching simply is not the best option for you (no matter how good a strategy it is for maximization).

Trust me, I've spent many hours debating this with many very smart people; eventually they all realize how silly they were to even question it in the first place. I don't mind more debate on the issue as long as everyone is staying friendly. Individual trials simply are not governed by the statistics they generate (which really, that shouldn't seem like an outrageous statement... if it does you may need to go back to basics for a minute or two).

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### Re: Monty Hall Problem [solution discussion]

The bottom line is this: *once you have chosen* you are no longer a statistic to be, and the odds of you winning are not dictated by statistics to be. Whether or not switching is better for *you* is up to whether or not you guessed right or wrong to start with.

Think about choosing your strategy ahead of time, and let's compare the two strategies a) pick a door at random and stick with it, or b) pick a door at random and switch. Then we can ignore all this nonsense about how once you have picked, you are either right or wrong (which, while true, is uninformative and misleading). From this prospective viewpoint, it is clear that the chance of winning with strategy a) is 1/3, and with strategy b) is 2/3, so a wise player will adopt strategy b) over strategy a).
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### Re: Monty Hall Problem [solution discussion]

Just because it's a fiction doesn't mean it's useless, though. Perhaps a better term is "abstraction". Having a way to quantify how your limited knowledge affects the situation is useful. And if you have a choice between a small-Bayesian-probability-of-winning situation and a larger-Bayesian-probability-of-winning situation, then it's rational to choose the latter. Even if the former would be the right option in a particular case - you can't tell you're in that case ahead of time, so worrying about that will gain you nothing. And the latter would be right more often.

Ultimately all this "but the probability is really either 0 or 1, you just don't know which" is just so much semantic wankery... sure, there's some philosophical points to be made about the basis for it all (be it frequentist or Bayesian) but ultimately all that changes is that you change a few words in how you have to phrase the question... the calculations, the answer, and the general applicability of the result to the situation are all identical. This is why they're merely interpretations of probability, rather than different entire theories of probability. Regardless of what you think about the philosophical side, a player who switches will win 2/3 of the time, while a player who stays will win 1/3. So if you're playing, you should plan to switch.

Also, a friendly tip... this self-aggrandising stuff:
adriankemp wrote:Trust me, I've spent many hours debating this with many very smart people; eventually they all realize how silly they were to even question it in the first place.
is rather at odds with this goal:
adriankemp wrote:as long as everyone is staying friendly.
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### Re: Monty Hall Problem [solution discussion]

adriankemp wrote:Ultimately, the problem is that Bayesian probability is (pretty much by definition) a sort of fiction; it's a bit like imaginary numbers... They can be a part of the underlying mechanics, but go ahead and try to give me 5i pencils.

Can you give me 1.5 pencils?

I mean really 1.5 pencils. If you just snap a pencil in half with your fingers, do you know you're giving me 0.5 of a pencil as opposed to 0.48?

Imaginary numbers aren't any more fictional than other numbers.
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### Re: Monty Hall Problem [solution discussion]

phlip wrote:A well reasoned post

Okay firstly I think you need to go back to the basics of statistics for a second. Stats is the information that arises out of data points, it does not control or dictate them. Thats actually vitally important here.

You have not refuted anything I've said except that talking of absolutes is pointless. Since we are talking about a real game show that you only get a single shot at, absolutes are in fact the only thing there is *any* point in discussing.

The fact that my single shot may well be destroyed by applying a maximization strategy is very relevant to the problem at hand. The question is not the best strategy to get the most cars out of ten goes.

To answer the question from someone else re: would I switch? I honestly can't say. The maximization strategy is sound, and for lack of a better option it's a reasonable way to go. But I also understand that a coin coming up heads four times does not mean the next flip will be tails. Just because the probabilities say I'm more likely wrong doesn't mean I actually am. Besides I like my current car.

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### Re: Monty Hall Problem [solution discussion]

Well, let's consider another situation... trim away the non-disputed parts of the puzzle, and ramp up the situation to extremes.

I uniformly pick a random positive integer less than a million. If that number happens to be 314159, then I write the letter "A" on a piece of paper, otherwise I write the letter "B". I then put the piece of paper in a box, and call you in. You have to guess what's on the piece of paper. If you get it right, you win some kind of prize. One shot, no repeats. What wouldst thou deau?

Now, there's a right answer and a wrong answer - I've already written it down, after all. But you have no information one way or the other, except for the description above. Intuition would say that "B" is about a million times as likely as "A", but intuition is unreliable. Frequentists would claim that guessing "B" would have you win about a million times as often as guessing "A", but you claim that doesn't count because you only get one shot. Bayesians would claim that the suggestion that "B" is right has about a million times as much credence as the suggestion that "A" is right - or, in other terms, it agrees with intuition that "B" is about a million times as likely as "A"... but you say that these figures are purely fictional. So would you truly say that it's a fallacy to claim that guessing "B" is the superior choice here? I mean, you certainly can't rule out the suggestion that the right answer is "A"... given what you know, it is a possibility...
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### Re: Monty Hall Problem [solution discussion]

phlip wrote:A well reasoned post

Okay firstly I think you need to go back to the basics of statistics for a second. Stats is the information that arises out of data points, it does not control or dictate them. Thats actually vitally important here.

You have not refuted anything I've said except that talking of absolutes is pointless. Since we are talking about a real game show that you only get a single shot at, absolutes are in fact the only thing there is *any* point in discussing.

The fact that my single shot may well be destroyed by applying a maximization strategy is very relevant to the problem at hand. The question is not the best strategy to get the most cars out of ten goes.

To answer the question from someone else re: would I switch? I honestly can't say. The maximization strategy is sound, and for lack of a better option it's a reasonable way to go. But I also understand that a coin coming up heads four times does not mean the next flip will be tails. Just because the probabilities say I'm more likely wrong doesn't mean I actually am. Besides I like my current car.

You appear to be making the argument that statistics only applies when you have a sample size greater than one. Why is that the case?

Sure, applying the maximization strategy to the Monty Hall problem could result in you getting the goat. However, it could also result in you getting more goats than cars (or even all goats) if you have ten tries, or a million. While the odds of that are progressively less likely the more tries you have, it's still possible for any finite number of tries, from one to Graham's number or more.

Your argument about absolutes is equally valid or invalid no matter how many tries you get. Just as there's a correct answer for one try regardless of what statistics tells you, there's a correct sequence of answers for any number of tries regardless of what statistics tells you. Thus I don't understand why you accept statistics for n>1 but not for n=1.
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### Re: Monty Hall Problem [solution discussion]

phlip wrote:one in a million

This is an entirely different problem, for which a basic probability is very representative. You're asking me whether one number in a million is more likely to come up than all the rest: yes, obviously.

With Monty Hall, You offer me a choice (am I more likely to guess 2/3 than 1/3, this is equivalent to your last question and I've repeatedly agreed that I am). The scenario is now irrevocably changed and has two completely different branches of statistics depending on whether or not I chose correctly.

If the game show worked such that Monty said "I'm offering you a choice to pick a door and stick with it even after I open another door that doesn't have the car, or a choice to pick a door and then switch after I eliminate a door" then the only sensible course of action is to pick and switch. That is *not* the scenario, and it makes a very real difference. It's a multiple stage problem that (for individual trails) does not always follow the maximization strategy.

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### Re: Monty Hall Problem [solution discussion]

adriankemp wrote:This is an entirely different problem

What's different about it? In the second half of the Monty Hall puzzle, you're asked a question: switch or no? One answer is right, one is wrong. But one has a 2/3 chance of being correct, the other 1/3.
In my analogy, you're asked a question: guess A or B? One answer is right, one is wrong. But one has a 999999/1000000 chance of being correct, the other 1/1000000.
They're identical except for the probability.

adriankemp wrote:If the game show worked such that Monty said "I'm offering you a choice to pick a door and stick with it even after I open another door that doesn't have the car, or a choice to pick a door and then switch after I eliminate a door" then the only sensible course of action is to pick and switch. That is *not* the scenario, and it makes a very real difference. It's a multiple stage problem that (for individual trails) does not always follow the maximization strategy.

Why is the situation between the two different, when the only change is when you make the decision? You get exactly zero extra information in the meantime... Choosing "pick and stay" in this variant is identical in all respects to picking, and then later staying, in the original; choosing "pick and switch" in this variant is identical in all respects to picking, and then later switching, in the original. And you have exactly as much information when making the delayed decision as you do when making the early decision. The two puzzles are strictly isomorphic. So how are they getting different results?
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### Re: Monty Hall Problem [solution discussion]

adriankemp wrote:Trust me, I've spent many hours debating this with many very smart people; eventually they all realize how silly they were to even question it in the first place.

Pity they aren't good at math It's OK, I'm not either.

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### Re: Monty Hall Problem [solution discussion]

This isn't really looking from a mathematical perspective, but the way I look at the problem is this:

Keeping your first choice makes the question "What door do you think the prize is behind?" and switching makes the question "Which door do you think the prize is not behind?"

Another way to look at it is this:

Pick a door 1-3, you have a 1/3 probability of getting the right door, 2/3 probability of being wrong.

Then the second question is "Do you think you were right or wrong on your first guess?" Well, you had a 2/3 probability that you were wrong, so saying you were wrong wins with a probability of 2/3.

This is all assuming, of course, Monty Hall knows what door the prize is behind and only shows what's behind a non-winning door. If he doesn't know, then it's 1/3 and then 50/50.
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### Re: Monty Hall Problem [solution discussion]

adriankemp wrote:If the game show worked such that Monty said "I'm offering you a choice to pick a door and stick with it even after I open another door that doesn't have the car, or a choice to pick a door and then switch after I eliminate a door" then the only sensible course of action is to pick and switch. That is *not* the scenario, and it makes a very real difference. It's a multiple stage problem that (for individual trails) does not always follow the maximization strategy.

So you are saying that if you are given the choice of the switch strategy or the stick strategy by Monty before anything happens, you should choose the switch strategy. Right? Doesn't it also follow that you should choose the switch strategy over the stick strategy just knowing the rules of the game, even if Monty doesn't give a choice of strategies at the start, but instead gives a choice of switching or sticking after opening a door?

If you disagree, please explain the difference to me, because as far as I can tell, the two situations are completely analogous.
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### Re: Monty Hall Problem [solution discussion]

adriankemp wrote:This is the heart of the problem: Much like a lotto ticket isn't changing as I'm scratching it; the car isn't jumping around behind the doors and I'm not magically half-guessing a different door. Once a choice has been made *that trial* is subject to the choice that was made and cannot be changed. Ergo, if you pick correctly; switching simply is not the best option for you (no matter how good a strategy it is for maximization).

Trust me, I've spent many hours debating this with many very smart people; eventually they all realize how silly they were to even question it in the first place. I don't mind more debate on the issue as long as everyone is staying friendly. Individual trials simply are not governed by the statistics they generate (which really, that shouldn't seem like an outrageous statement... if it does you may need to go back to basics for a minute or two).

The issue you seem to have is the same one I had with stats in high school. I dismissed it because, ultimately the car is either behind one door or the other, not 33% behind one and 67% behind the other. However I changed my mind about statistics when I can to see it not as telling me what was (i.e the car is behind that door but also perhaps behind the other door), but instead telling me how much I know.

You're knowledge of the situation allows you to pick a door and win with 67%. The game show host has the knowledge to pick a door and win with 100%. A random person who enters the room and sees two doors has the knowledge to pick a door and win with 50%. Three different percentages, but the car is still behind the same door. Hence we see that what changes with the probabilities is not the whereabouts of the car (the reality we live in), but the information each person has about that reality.

The lottery ticket is either a win or a loss, whether you scratch it or not. But the information (or probability) you have about it being a winning or a losing ticket, jumps dramatically after you've scratched the ticket.

If we view probability as information, then the Monty Hall problem is easier to understand. You're probability of winning the car increases since you get more information from the host (who has all the info) when he opens up the empty door. This is the key. If he had opened up a door at random, the probability would've been 50%. But since he knows where the car is, and since he uses this knowledge when he opens a door (he avoids the door with the car), this information is passed on to the contestant who's chances to win thereby change (increase).
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### Re: Monty Hall Problem [solution discussion]

adriankemp wrote:To answer the question from someone else re: would I switch? I honestly can't say. The maximization strategy is sound, and for lack of a better option it's a reasonable way to go. But I also understand that a coin coming up heads four times does not mean the next flip will be tails. Just because the probabilities say I'm more likely wrong doesn't mean I actually am. Besides I like my current car.
You honestly CAN'T say? Or you honestly WON'T say? You are deluding yourself - if you wanted the car, you would switch doors. Period. UNLESS you had additional information - like Monty gave a wink to his pretty assistant when he opened the door, or you have x-ray vision - but in the original problem we have no additional information.

How does the coin coming up heads four times in a row apply to this scenario? Wouldn't a better analogy be a weighted coin with 66% chance of coming up heads? Surely you would pick heads instead of tails, right?
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### Re: Monty Hall Problem [solution discussion]

I like to think of it in the context of http://xkcd.com/525/

If the odds of winning if I switch are 50-50, than I loose nothing by switching, if the odds of winning if I switch are 2/3 than I am better off switching.

Thus the optimal strategy is to switch because in the worst case it is as good a choice as not switching.
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### Re: Monty Hall Problem [solution discussion]

If you think the odds are 50/50 then please do the following:

Get a friend and get three cards, one of which is a queen. Your friend will try to pick the queen by playing the game. You shuffle the cards, they pick one, you look at the other two and reveal one that is not a queen, then they choose to switch or not.

Play until you know you are wrong.
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### Re: Monty Hall Problem [solution discussion]

Not sure if a similar post has been made (if so please remove) but this reframing is what I have seen convince most people that you should always switch (given the information you are presented with)

You want to choose the ace of spades. I lay out an entire deck of cards face down, you choose one then I will flip over every card except for your card and one other making sure I do not flip over the ace of spades. You can stick with your card or switch to the other face down card. Should you switch?

Still not convinced? I will secrectly write a number on a peice of paper (any number, complex, real, rational, irrational whatever). Then you guess a number. I will tell you "Either the number you guessed or [X], is the number I wrote down" where X is either the number I worte down (if your first guess was wrong) or some other random number (if your first guess was right). Should you switch or stay with your original guess.

In both cases as with the original monty hall probablem you are asked to choose from a large set (3, 52, or infinite) and are then reduced to two otpions (what you picked and one other).

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### Re: Monty Hall Problem [solution discussion]

phlip wrote:one in a million

This is an entirely different problem, for which a basic probability is very representative.

You are saying that at some point between one in three and one in a million the problem suddenly becomes "an entirely different problem, for which a basic probability is very representative." What is the exact probability at which this switch occurs, and why does it do so?
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### Re: Monty Hall Problem [solution discussion]

There are some philosophical subtleties when articulating what probability is. Like other people, I have sometimes been troubled by statements like "with probability 1/3, there is a car behind the first door" when we know there either is a car or there isn't.

And I appreciate that subjectively, probabilities like 1/3 or 2/3 might "feel close enough" to 1/2 that we don't "notice" the difference in a very small number of trials.

Here's something that may be relevant. One not-very-serious hobby of mine is sports betting. Assigning a probability to a future baseball or football game is not a pure math problem, and involves an element of personal judgement. In practice, for many sports and many leagues, a reasonably strong "favorite" may sometimes be assigned a probability around 55-60%, say, and a probability as high as 66% might be rarely seen.

Very informally, one might describe this by saying "No one game is really that far from a 50-50 proposition." Which, as informal advice for gamblers, could perhaps have some value. People sometimes say things like "Any given team can win on any given night." What this does NOT mean, however, is that probability somehow "just doesn't apply" to single events.

If I were willing to bet you a dollar against a dollar for any team against any other team in any MLB or NFL game, you would do well to take my bet, and just adopt a simpleminded strategy such as "Always pick the team with the better record." In the long run, you will win money. In practice, you might win approximately, say, 52-53% of the time, so you would win in a rather gradual and "fluctuatey" way which you might find frustrating. But it would still be the right thing for you to do.
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### Re: Monty Hall Problem [solution discussion]

rigwarl wrote:
adriankemp wrote:Trust me, I've spent many hours debating this with many very smart people; eventually they all realize how silly they were to even question it in the first place.

Pity they aren't good at math It's OK, I'm not either.

I'm not sure that's how I would interpret their reaction from that post <.<

But anyway, math is always about modelling what we know about reality. Why should we care whether or not "the math is the reality" or whatever? The car is behind a specific door, but we don't know that, so we use statistics to MODEL the situation we're in, to determine that we should probably switch. And if you're looking for a definite strategy, switching is the best one. If you're not looking for a definite strategy, then I suppose you could whip out your calculator and use its RNG to figure out what to do...
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### Re: Monty Hall Problem [solution discussion]

skullturf wrote:If I were willing to bet you a dollar against a dollar for any team against any other team in any MLB or NFL game, you would do well to take my bet, and just adopt a simpleminded strategy such as "Always pick the team with the better record." In the long run, you will win money. In practice, you might win approximately, say, 52-53% of the time, so you would win in a rather gradual and "fluctuatey" way which you might find frustrating. But it would still be the right thing for you to do.

Don't use that strategy in finance though.
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