I think the Monty Hall problem is a "broken" question and would like your guys input on my thoughts.

Instead of saying there is a car and two goats, let's get abstract and say there is an A, a B, and a C.  You always pick door 1 and C is always revealed.  You want an A to win.  1/3rd of the time you get an A, 1/3rd of the time you get a B, and 1/3rd of the time you get a C. 

So it would look like this:

1 2 3
A B C
B C A
C A B

Here's why I think that monty hall is broken.  There's a right answer, a wrong answer, and a false answer (is never be chosen).  In this case A, B, and C.  Basically what the problem does is it throws out the false answer from being initially chosen.  If you could choose the false answer and were forced to switch between two unknowns:

1 2 3
C A B
C B A

Then half the time, you would get it right and half the time you would get it wrong.  A more accurate version should look like this:

1 2 3
A B C
A C B
B A C
B C A
C A B
C B A

Remember you are always choosing door 1 initially.  In the first scenario (1/3rd of the time), you get A and switching would make you lose.  In the second scenario (1/3rd of the time), you get B and switching would always win.  In the last scenario (1/3rd of the time and normally excluded from the problem) it would be revealed to you that you've made a bad choice initially and you'd have a 50-50 shot of getting things right with your switch.

Note I'm not arguing that in the original version switching should give you a 50-50; logically it shouldn't.  What I do think though is that it is improperly created as it disallows you from initially choosing the false choice.  Here's why: let's say that in a realistic situation people will only choose the 1st door 1/3rd of the time.  Now lets say you preselect which goat is to be revealed.  In the original version, you've thrown out the possibility of that goat from being chosen, which should happen 1/3rd of the time.  Thus you've skewed the results. 

In the original "broken" problem, for the 3rd scenario (you choose C) C and B are switched beforehand and you are forced to pick B twice as often as you normally would be.

So what do you think and how would you create the problem?   Would you disallow the person from choosing the false choice initially (ie the goat that is to be shown) and give him a 2/3rds chance of winning if he switches, or would allow him to know he's picked wrong and have a 50-50 shot of getting it right on the switch?  Does the false choice really exist or not?

In case you're wondering, I've posted it just because I've been thinking about this lately and think it's interesting being able to see how a change in perspective can change results.

A couple of years ago I posed this conundrum to my son, to no avail. It went a bit like this:

ME: So, you're trying to win a car...
HIM: I don't want a car.
ME: You don't want a car?
HIM: No.
ME: Ok, what would you rather have?
HIM: Abi Titmuss.
ME: Ok, so there's Abi Titmuss in a box...
HIM: Cool
ME: ... and two goats in the other boxes
HIM: I want a goat
ME: Uh? I though you wanted Abi Titmuss?
HIM: I'd rather have a goat.
ME: Why?
HIM: You can do more things with a goat.

I gave up.

You're analysis is wrong right from the start.

You said:

Quote - Instead of saying there is a car and two goats, let's get abstract and say there is an A, a B, and a C.  You always pick door 1 and C is always revealed. 

You left out the case where door 1 is C. In that case it is not revealed, because Monty never shows you what's behind the door you picked.. He reveals B instead. You said that that case was excluded from the problem, but it can't be excluded because that is the third case. This is the crux of the decision that Monty makes, and that you are not aware of.

The ABC thing is really leading you down the wrong path.

The way that most people finally understand this is to extend the problem. Restated, there are N doors. Exactly one has the prize. The rest are losers. Try listing all the scenarios for 1000 doors. That's beyond what anybody could do in a lifetime. You have to look at it differently.

So in the general situation where there are N doors...

You pick a door. The chance that you have the right door is 1 out of N.

Monty carefully opens N-2 doors, leaving only two closed doors. While doing this he makes sure that since he knows where the prize is, he does not choose that one. This is the key to the problem.

He then asks you to stay or switch.

With this restatement, it should be obvious what is going on.

When Monty picked all those doors to open, he was avoiding the prize. One out of N times, your choice was the right door and it didn't matter what he opened. But (N-1)/N times the door he didn't open is the prize.

Let's say there are 1000 doors.

You pick door #1. You have only 1 chance in 1000 that you're right. Most of the time, you're wrong. He opens 998 doors no matter what, all the while carefully avoiding the prize. There are two doors left - yours and the one he decided not to open, which more than likely is the prize. In fact, the chance that the remaining door that neither you nor Monty chose is the prize is 999 out of 1000.

So back to the original with 3 doors.

You pick door #1. You have only 1 chance in 3 that you're right. Monty opens one door, carefully avoiding the prize in the situation where you're wrong. That is 2/3 of the time. So 2/3 of the time the remaining door that neither you nor Monty chose is the prize.

The key is to understand that you know nothing and are choosing randomly. Monty knows everything and is not choosing randomly.

Quote - In the last scenario (1/3rd of the time and normally excluded from the problem) it would be revealed to you that you've made a bad choice initially and you'd have a 50-50 shot of getting things right with your switch.

This is the wrong part. Monty opens B, not C. He never opens your door, and he never opens the prize - that's the rule. In the case where you have C, he opens B, leaving C and A. Switching wins the prize in that case.

In the case where you have A, switching loses. This is 1/3 of the time.
In the case where you have B, he shows C, and switching wins. This is 1/3 of the time.
In the case where you have C, he shows B, and switching wins. This is 1/3 of the time.

Switching wins 2/3 of the time.

Winterclaw,

  Let me go back to the two goats (called G1 and G2), a car (C) and three doors, because that's the specific case.  To simplify things, you always pick door 1, and Monte always opens door 3, unless it has the car, in which case, he has to open door 2. (The math works the same whether you simplify things or not.)

The prizes can be distributed 6 ways:

1   2     3
C  G1  G2
C  G2  G1
G1 C   G2
G2 C   G1
G1 G2 C
G2 G1 C

After Monte opens the door (Door 3, unless that's the car), here are the possibilities

1   2     3
C  G1 
C  G2 
G1 C  
G2 C  
G1       C
G2       C

In a third of the cases, Goat 1 is behind Door 1, Goat 2 is behind it in a third, and the car is behind  Door 1 in a third of the cases.

I didn't believe it, either, until I sat down and worked the problem for myself.  It's one of those counter-intuitive things that make some areas of math, like orbital mechanics, fun.

Quote - You're analysis is wrong right from the start.

You said:

Quote - Instead of saying there is a car and two goats, let's get abstract and say there is an A, a B, and a C.  You always pick door 1 and C is always revealed. 

You left out the case where door 1 is C. In that case it is not revealed, because Monty never shows you what's behind the door you picked.. He reveals B instead. You said that that case was excluded from the problem, but it can't be excluded because that is the third case. This is the crux of the decision that Monty makes, and that you are not aware of.

The ABC thing is really leading you down the wrong path.

Thank you.  While I was reading the initial post, this same thought struck me and I was left scratching my head.  But I didn't want to say anything because math isn't exactly my strong point, but I did kick butt with probabilities.  In fact I used the probability of dice rolls to my benefit when playing Backgammon in my 20's.  Such fun!!!  ;)

Quote - You left out the case where door 1 is C. In that case it is not revealed, because Monty never shows you what's behind the door you picked.  He reveals B instead.

Bill, you've essentially restated why I think the problem is wrong: he can never reveal C if you pick it.  That's why switching is the better option as the original stands.  I'm not arguing that.  Switching wins 2/3rds of the time in the original problem.  That was never what I was arguing against.

In the case of N-2 doors being opened if he opens the door you've picked then behind those other two you have one right and one wrong answer.  So if N was 5, you'd have an A, a B, and 3 Cs.   If N = 1,000,000 then you'd have 999,998 Cs.  Now if he is allowed to show you that you picked a C, then behind the last two remaining doors you have one A and one B.  There are no other logical possibilities.  Again the Cs are the false options (the things being revealed that you will never pick).

Almostfm: yours has come out wrong because you've changed the conditions back to the original form, not the new form.  Redo that under the assumption monty always shows you G2, even if you pick it and you'll see where I'm trying to go with this.

Let's restate the monty hall problem this way.  There are three doors.  One has a honda Accord, one has a blue goat, and the other has a chartreuse goat.  After you've made your selection, Monty always reveals the location of the chartreuse goat (which it is possible for you to pick before hand).  Then he gives you the opportunity to switch, if you've picked the chartreuse goat you can switch to either remaining door.  What are your odds of getting the car if you switch?

If you'd like to do it with cards, do it this way: take out an Ace, a 2, and a 3 of the same suit.  Pick a card, any card.  The other person always reveals the location of the 3 and then gives you the opportunity to switch.  He is never allowed to show you the location of the Ace or the 2.  What are your odds of getting the Ace if you switch?  Now this should give you different results because the conditions are different than the original monty hall problem.

So my question for everyone is should monty be allowed to let the player know he's picked wrong (the chartreuse goat or the 3) to begin with?

 OK.. Being a foreinger and not familiar with the quiz show.. Can someone, in small, easy words, explain to me why "switch" should be a better solution than to stay with the door chosen? For all I can see .. if door 1 is chosen, and door 3 is then revealed as hosting a goat, all we know is that the car is in EITHER Door 1 or Door 2 - giving it a 50/50 chance of each.

OR do I get the idea of the show wrong? IF the car is behind Door 1, is it then instantly revealed? The Wikipedia article didn't make that bit clear, only that the game ost - Monty Hall - would open a door that did NOT contain the car. 

In the original, switching is a better option because:
1.  You are wrong on your initial pick 2/3rds of the time.
2.  You never know when you are wrong initially.

So if you have a 2/3rds chance of being wrong and no way to eliminate the door you picked, then 2/3rds of the time the other hidden door will be the one with the car.

EDIT:  Let me put it like this, if you always choose Door 1 then two out of three times you are getting a goat on your pick.  Since it is impossible for the revealed goat to be a car, then the other door has a car.  The only time switching loses is when you picked the car to begin with.  That only happens one out of three times.

Ah. I didn't understand what you're getting at because I'm dense. Sorry.

So what you're really saying is:

1. You've invented a new game, the Winterclaw game. It has some similary to the Monty Hall game, but the rules are different. I'm OK with that.

2. Playing the best strategy in the Winterclaw game results in a different average probability of success than in the Monty Hall game. That's true, too. In the Winterclaw game, the best average probablity of success is indeed 1/2. I'll explore that a bit more in a second.

3. The Monty Hall game is broken. That I disagree with. It's not at all broken.

What is unclear is what you mean by "broken". First of all, the Monty Hall problem is not the same as the Monty Hall game. You said the problem is broken. But the game is not the problem. The game is the premise of the problem, and the problem is to answer what is the best strategy and why.

We're not talking about whether the game is good or bad, apparently, but rather whether the problem is interesting or not. (Side note - a broken game is one in which you always win or you always lose - not very interesting. Broken games automatically produce no interesting problem of choosing a strategy.)

Now we all agree that in the Monty Hall game as well as the Winterclaw game, you win or lose are both possible, so the game is interesting. The question then: is the problem of choosing a strategy interesting?

In order for a problem of this type to be interesting, there has to be at least two strategies, and one has to be better than the other. If they have the same outcome, then it isn't an interesting problem even if the game is interesting, since your decision to follow one or another would not affect the game outcome.

For example, consider the coin toss game within the game of football. You can choose head or tail. The coin toss game is interesting. I care whether I win or not, and I can lose. But is the problem interesing? is there a best strategy, and if so, which is it. There are two strategies - choose heads or choose tails. The value of both strategies is 1/2. Therefore, while the game is interesting, the coin toss problem is dull and not worth discussing. Do whatever, it won't matter.

Now in that sense, then the Monty Hall problem, which is a question about the Monty Hall game, is interesting. Many smart and/or educated people give the wrong answer as to the best strategy. Either they choose "stay", which is not the best strategy because it has a value of 1/3, or they deny that the "switch" strategy has a value of 2/3. This makes it interesting - the choice of strategy matters, and many people perversely get it wrong.

So, I do not think the Monty Hall problem is broken at all. Quite the opposite. It provokes a lot of thought and discussion, and is surprising and interesting to a lot of people.

Now let's consider the Winterclaw game and the Winterclaw problem.

In the Winterclaw game, there are two strategies as well, stay or switch. And as you've discovered, in the Winterclaw game, the value of both strategies is 1/2.

Which means that the Winterclaw problem is broken. It doesn't matter if you analyze it incorrectly. There is no advantage to you knowing the strategies and the probabilities. If two people compete at the Winterclaw game, one playing with an understanding of the strategies, and the other totally clueless and choosing at random, and we measure the outcomes of these two people over a large number of trials, we wil not be able to tell which one is clueless and which one is wise to the game. This meets my definition of an uninteresting problem.

[Edit]

On the other hand after further study, there is a slight wrinkle on the Winterclaw game and problem.

I suggested the Winterclaw problem is dull because a clueless random player will not do worse or better than a wise player, and that watching these guys, I can't tell which is which.

But there is another kind of player. A player who chooses to always stay, even after being shown the third case, where he knows he has the wrong door.

In that case, the probability of winning is only 1/3.  So the stay strategy has value 1/3, and the switch strategy has value 1/2. Strange isn't it?

Given three players:

DumbGuy - always stays
SmartGuy - totally understand the game and always switches
CoinTosser - if his door remains closed, he tosses a coin to decide to stay or swtich. If his door is opened and reveals that staying is stupid, he switches, but tosses a coin to pick one of the other two doors.

DumbGuy wins 1/3 of the time.
SmartGuy wins 1/2 of the time.
CoinTosser wins 1/2 of the time.

:)

A respectable answer.

However, for the sake of the argument, let's say that part of the fun of the game is not knowing the outcome and the greater the risk or uncertainty the greater the reward.  In the monty hall problem you are afforded a simple strategy that has a much more predictable outcome.  Basically you know that you are going to win 2/3rds of the time if you switch and if everyone knows the trick you as a viewer can predict the outcome rather often.

That's pretty boring if you ask me.  Low risk and only one viable strategy really.  Not much of a game if you ask me because once you know how it play it all the fun is gone.  Plus it'd be pretty boring to watch as well.

The ante has to be upped somehow.  Basically for it to be fun, the more likely to win strategy has to have a lower payout or else it isn't a strategy at all, it becomes who knows the game and who doesn't.  So going back to the monty hall game, let's say instead of a car, we have a "prize" instead.  If you pick the prize on your first guess, you get a car, $50K, whatever.  If you switch and get the prize, you only get $5K because your odds are better.  Now we have something of a game.  You understand it, but there are now two real strategies.  You can try to play it safe for a higher probability of making money or you can go for it all.  

Now things are interesting.

Or in the Winterclaw game, there is a good prize, an okay prize, and a consolation prize because every good game show has three prizes.  The okay prize is always revealed first and if you picked it to begin with, you can keep it or switch.  However you can't pick it after it is shown.  If you get the okay prize, do you switch or not?  Again, there are varying outcomes and strategies.  

And again, more interesting.

Bill, I think you are right when you say that in order to be interesting the problem needs to have some strategy but I think there needs to be a little more to it than that. 

What you describe is captured in the probability concept called "expected value".

The expected value is just the probability of an outcome multiplied with its value.

And as you note, there are two kinds of players. Those that:

Type 1) Attempt to get the maximum expected value, and are willing to end up with 0.
Type 2) Are unwilling to end up with 0 and avoid it in exchange for a lower chance at the maximum prize.

This is the heart of the game "Deal or no deal". It's fascinating to see where the player switches from type 1 to type 2, or watch the permanent type 1's go down in flames.

In that game, there is no strategy. You have no idea which suitcase to pick at all times.

The entire strategy has to do with managing expected value, given where you are, versus where you want to be.

In the Deal or no deal game, I often calculate the expected value as the game goes along. The amount that the banker offers to buy the player out is roughly proportional to the expected value, but the actual proportion changes. At times it is close to the EV, other times he offers as little as half the EV.

I cringe when I see somebody with, for example, $200K, $100K, $1, and $5 left. The EV here is about $75K. If the banker offers $70K and the guy doesn't take it, he's an idiot.

Here's why.

The offer is like money in his pocket. It's already his - he can walk away with it. So here we have a person who is essentially BETTING $70 THOUSAND DOLLARS of his own money. The next pick has a 1/4 chance of being the $200K. That would change his EV to $33K, and the dealer would probably offer $25K next time. So the player is accepting a 1/4 chance that he will lose $45 THOUSAND DOLLARS, while trying (hoping) for the maximum of 1/4 chance of to gain him $130K. (Because he already has $170K)

That's a stupid bet.

Note - I edited the previous post multiple times as I was distracted (am at work) and wrote wrong numbers repeatedly. It may still be wrong - but you get the idea.

Yeah, I was reading about that game the other day.  I think the banker offers a lower value early on and it gets better the longer people stay in the game (or until they shoot themselves in the foot).

For a game to be fun and interesting, I think there has to be risk, strategy, and the possibility of managing expected values as things go on.  That's probably why games like chess and go have been around forever.

Risk and Strategy have no "surrender" option, so managing expected value is difficult. There is still only "win" or "lose", and there's no point in giving up, except to save yourself some time, or to have more fun by starting over with a clean slate.

There is a popular game with an expected value option, which makes it more interesting. Ironically, for the very reason I like it, most of the people I play with don't. The game is Backgammon, and the EV feature is the doubling cube. When you think you're doing well and highly likely to win, and/or you know your opponent is risk averse, you double the value of the game. The other person has to choose to keep playing at the doubled value, or they have to surrender the current value without any more play. I love that.

Ugh. I just read what I wrote and I said the wrong words.

I meant to say that chess and go have no "surrender" option, not "Risk and Strategy". Certainly in chess you can resign, but there's no value in it in terms of outcome. If you resign, there is no consolation prize. You lose.

Sorry - I'm very woozy from working. I did two all-nighters in the last three nights. 

Quote -  OK.. Being a foreinger and not familiar with the quiz show.. Can someone, in small, easy words, explain to me why "switch" should be a better solution than to stay with the door chosen? For all I can see .. if door 1 is chosen, and door 3 is then revealed as hosting a goat, all we know is that the car is in EITHER Door 1 or Door 2 - giving it a 50/50 chance of each.

OR do I get the idea of the show wrong? IF the car is behind Door 1, is it then instantly revealed? The Wikipedia article didn't make that bit clear, only that the game ost - Monty Hall - would open a door that did NOT contain the car. 

There is one car, and two goats, each hidden behind a door. You would have to choose one of the three doors. The host then open one of the doors you did not choose, showing you a goat. After that you will be asked if you want to switch to the third door or if you keep your door which you've choosen in the first place. 

The show, in the US aired as "Let's make a deal" was running as "Geh' auf's Ganze" in Germany for a while. We did not have goats, we had the "Zonk", a big red furry puppet. 

But as everyone knows, the "Law of Averages" is a fallacy according to probability theory.   :)

The law of averages is the common term for the gambler's fallacy. 

Given time and a large enough sample, switching will approach but not reach (or maintain) exactly 2/3rds.  However, as they say, the die doesn't know that it's due.

Bill, its no problem being tired, we've all been there.  Chess and go as the whole game don't have surrender options for the game itself but there are things you can do, be challenged on, and then give up to save time, material, or there's a better opportunity to gain space.  So in the small picture, you can realize there's not much value to be gain in continuing a certain course of actions.  In fact in Go you are supposed to stop playing once both people hit the point of no more territory you can expect to be gained.

Quote - The law of averages is the common term for the gambler's fallacy. 

Given time and a large enough sample, switching will approach but not reach (or maintain) exactly 2/3rds.  However, as they say, the die doesn't know that it's due.

I read a 19th century book (whose title and writer I have forgotten about over the intervening 30 years) about mathematical crackpots (and other subjects), which described how the author and George Boole conducted experiments in probability. Back in the early 19th century, it was not at all clear whether our current notion (that successive flips of a fair coin, or rolls of a fair die, are statistically unrelated) was correct. They assumed this as the null hypothosis, but also wanted to investigate whether what we now know as the "gambler's fallacy" was true (that an imbalance of heads would be counteracted by an imbalance of tails), or whether the notion of streaks was true (i.e., that after several heads in a row, the next flip would be more likely heads). So you had these mathemticians rolling dice and flipping coins and writing down the results. Pretty cool.