Semi-Daily Journal

Heh - this one turned up in the Economist a few years ago, and I'm afraid it took me ages to be convinced.

Eventually I think I knocked up a perl script to simulate the payoffs to switchers vs non-switchers over 1000 runs of the game.

It did cheer me up, however, to find that putting the puzzle to a bunch of graduate students in basically quantitative disciplines provoked not the response I had half expected (namely cries of 'duh - obviously you switch, dummy'), but a good hour of furious head-scratching, scribbling and argument.

http://www.crummy.com/articles/monty

This started a crossposting orgy that seemed to run for thousands of posts on usenet a few years ago.

Among other things it's an example of psychology and the importance of the effect of the point of view one takes towards a question.

Almost everyone assumes the point of view of the player, finds it confusing, and then it's off to formulas and matrices and simulations.

But if one looks at it from the point of view of Monty manipulating the doors (or a dealer manipulating cards in an equivalent card game) it's simple and people usually can see it right away.

So one really shouldn't judge a situation without standing in the other person's shoes.

The assumption articulated in the last sentence of the answer is absolutely crucial -- and is not in fact stated in the question as you pose it, nor in many other versions. It is sometimes not made clear even in supposed answers. Omission of, or confusion about, that key premise is at the heart of many of the sulphurous debates that have arisen about this problem.

Dear Ken Doran

I think it is clear that Monte wouldn't open the door with the big prize and ask the contestant if they want to chose that one. To contestants and viewers it was clear that he always showed a little prize. I watched the show (let's make a deal) and didn't figure out that one should switch.

When it was finally explained to me (by Larry Katz) I covered my tracks in a dishonest way. I claimed I didn't know that there was a big prize and two little prizes (or didn't correct him when he suggested that explanation for my lame brained ness). Sorry Larry but I'm confessing in public.

I think there is something deep here. One guess was on my blog (thanks for the link Brad). Another follows. If you are dealing with someone and you say A and he asks if you don't want C instead, you can usually be sure he is trying to trick you.

Monty Hall didn't pay for the prizes and got higher ratings if contestants won the big one. However, the general rule is if someone offers you a chance to switch you don't want might be deeply engrained (even inate).

This effort is related to the cheater detector hypothesis of Leda Cosmides & John Tooby which might be in www.psych.ucsb.edu/research/cep/primer.html
(which I haven't read). The idea is we are very good at some things involving probability like catching cheaters and very bad at other things which are mathematically the same.

If the host has the option of whether or not to offer a switch after the first choice, he might choose to offer a switch only, or primarily, when the guest has first chosen the valuable prize. If he wanted to save the show money that month, that is exactly what he would do. And if that is what is going on, switching is a losing move, turning a one-third chance into something significantly less. As the answer given above correctly but belatedly states, the switching-is-good result is reached only when the contestant knows that an offer of a switch is required, or close to certain for some other reason, independent of the value of the first choice. As to the actual "Lets Make a Deal Rules", Monty apparently sometimes offered a choice, but he was not required to do so and often did not.

Without getting into the probability, you can get an intuition for the correct answer by changing the problem slightly:

Monty gives you 1000 doors. You pick one. He then reveals 998 doors as having nothing behind them, leaving the one you picked and one other door. Should you switch?

Now, if you justify your answer for that problem, you just have to convince yourself it's still valid for the original.

yeah, i remember getting this during my first microsoft interview, before i had encountered a single stats class at my university. this was my evil interview question. it took a lot of head scratching to convince myself it wasn't just a 50-50 chance.

yeah, i remember getting this during my first microsoft interview, before i had encountered my first stats class. this was my evil interview question. it took a lot of head scratching to convince myself it wasn't just a 50-50 chance. the best way to think about it was imaging 1000 doors, with 999 being unveiled by monty. it made it pretty clear you should switch.

this is one of my all time favorite interview questions. the other one is the lion that prowls around the arena. that one goes as follows: there's a lion outside the arena you're trapped in that runs 4 times faster than you. assuming the lion only runs on the perimeter (meaning if you get past the arena border, you're safe), how do you get out? you start off in the center of the arena.

joel spolsky maintains a database of all kinds of good questions like this and a friend from college, noah, wrote a good book on programming interviews that contained a lot of these.

I don't think it's necessary to get into Monty Hall's head to get an intuition for this puzzle. Just change the game a little: take 10 doors, and choose one. Now have Monty open 8 of the 9 remaining doors, carefully not opening the one with the prize. Do you want your original choice, or the one not opened by Monty?

If 10 doesn't do it, try 100.

Oops, didn't read the last comment or I wouldn't have bothered posting the same. To atone, here's one of my favorite puzzles:

4 people are trying to cross a bridge, but they all have different rates: one takes 10 minutes to cross the bridge, another takes 5 minutes, another takes 2 minutes, and the last takes 1 minute. The bridge can only support the weight of 2 people at a time. It's dark and they have a single flashlight, so they have to travel together at the rate of the slowest person. The river is too wide for them to toss the flashlight back, so someone has to carry it back whenever there are still people waiting to cross (so, 2 go across, 1 comes back, 2 more go across...)

What is the minimum time required to get all 4 across the bridge? (Hint: it's less than 19 minutes).

I remember I found this in The Economist a long time ago. And then I found that people were in fact argueing about whether or not you should change the door. I remember I've always thought that *that* aspect of the problem is the one I fail to understand. How can you seriously argue about what the best strategy is when even a simple sketch on paper will show you the best strategy? I guess that's the difference between being an economist and a physicist. ;-)

It took me a long time to get a common-sense (intuitive)presentation of the answer. If you initially choose the right door (1/3), then Hall can open either of the other doors. You're best off staying in that one case. But if you initially chose a wrong door, then Hall has to open the other wrong door; he has no choice according to these rules. So the door remaining is the right door (2/3). You don't go back to zero and start off at 50/50 with two doors, because the door Hall opened was opened for a reason, based on the original 3-door situation; you're still in the three-door game.

There's initially a 2/3 chance that the prize is behind one of the two doors you didn't choose; once one of the two doors you didn't choose is opened, there's a 2/3 chance that it's behind the other door.

My initial intuition was completely wrong on this, and I never would have figured out the answer without being assured that changing is the right choice.

This is a lot like card-counting games such as poker or blackjack, where the odds change with each deal if the cards aren't shuffled. But it's such a simple case that people who immediately understand the principle in card-counting (me) miss the point here.

Ken Doran is right, though, that the solution is only good of it's really a one-player game, and that Hall is a dummy whose moves are all prescribed. If Hall had a motive (either for you to win or to lose), and if he had free choice whether to open a door or not, the solution would not be good; Hall would try to get inside your head. It also assumes that the valuable prize is instantly recognizable; wouldn't work for a real diamond plus two good fakes. If the puzzle were presented formally with x's and o's, people would probably solve it more often.

After Monty shows you there's nothing behind one of the other (not chosen) doors, your odds of winning automatically go from one in three to one in two -- WHETHER YOU SWITCH OR NOT!

Once Monty shows you that one of the three doors hides nothing, you know that the prize is behind one of the other two. There is a 50% chance it is behind the one you've already chosen and there is a 50% chance it is behind the door you haven't chosen.

Changing doors gives you a one in two chance of winning. Not changing doors gives you a one in two chance of winning.

Change if you want, but it doesn't matter.

The confusion comes from comparing the odds of the initial problem (choose one of three doors, one in three chance of winning) with those of the second separate problem (choose one of two doors, one in two chance of winning). Your odds of winning the second problem are greater, but you don't have to change doors to get those odds.

In the game against nature, obviously you switch.

In the game against a player (Monty, who gets to take home whatever you don't), it seems neither player should adopt a pure strategy (Monty should sometimes offer you the choice even when you've already "lost", to chum you into switching when you've "won" on your original pick, and you should sometimes decline to switch, frustrating Monty's attempted manipulation). Somewbody in all those thousands of posts must have worked out the mixed-strategy equilibrium. Cites?

The Monty hall problem is, or should be, easily understood by bridge players. In bridge, the key to the solution goes by the name of "The principle of restricted choice."

The simplest illustration deals with the case where, with four cards of a suit outstanding, a player , required to follow suit, plays the queen or jack, and is known to have begun with either: (a) only the card played or (b) with both the queen and jack but no other card in the suit. It is now roughly 2-1 that the original holding was (a) the singleton, by the Monty hall logic. With a singleton the player's choice of card was restricted to that card. With both cards he might have played either one.

Note that in terms of scoring there is no reason to prefer the queen to the jack, or vice versa, when holding both. The odds are only roughly 2-1 because the a priori probabilities are not quite the same.

The principle applies to a large number of card play situations, and often leads to surprising conclusions.

What is most interesting, from the point of view of this thread, is that it took some decades before this principle was widely accepted, and that new applications still pop up occasionally.

Amazing! zizka and I actually think alike about something. And we were both wrong initially. As is James Erlandson now.

In effect Monty Hall is giving the player a chance to exchange his initial one door for two doors. Exposing one "loser" door before offering the chance to switch, obfuscates that.

Obviously there are no bridge players among the commenters. This is a variant of what's known in bridge as "the principle of restricted choice." That is, when an opponent drops an honour under yours, you assume he had to and play the other opponent for the cards that (therefore) the one that dropped the honour dooesn't have.

Hey, hey. I got it as well. What I had to do was decide whether to settle for a 1 of 3 choice or opt for 2 of 3 choice with a switch. Of course, this goes against all sorts of superstitions. Funny, even after I knew I might like my old door. Oh dear.

This is one of those questions (and answers) that irritates me to death, and completly turns me off all math-based sciences to a degree. The two ways you can take the question either lead to a lack of social engineering, or too much information.

If it is Monty's goal to minimize winnings, then it depends on if he tends to give you the reduced options if you have the first one or not. If Monty tries to confuse you by reducing the options if you already have picked the right one, then given the reduced choice, you are better off staying. (Mind you this is all psychological and based off tendencies).

If it is an automatic part of the game that Monty reduces the options, then it is effectivly only a 2-option game. Easy as that. Anything past that is making things way too complicated.

Sorry. Bernard beat me to it. There is another bridge player reads Brad DeLong.

Let me add that even in the general case (where Monte doesn't have to open a door) there's still enormous power to restricted choice analysis, without trying to get into Monte's head.

In bridge, one's opponents can, sometimes do, "false-card." That is, play a card higher than the card they're forced to play. In practice, at the card-table, one ignores that possibility. The 2-1 odds that the analysis gives you are better than your opponent's feeble attempts at deception can overcome.

Similarly, here. If Monte opens a door at least 2/3 of the time, one should switch if given the opportunity (pace Glenn). Even if Monte is perfectly malevolent--that is, he opens a door every time you've picked the good door and randomly chooses not to give you the option if you've picked a bad one--if more than half the time you've picked a bad one you get the opportunity to switch, you should. And if Monte is not perfectly malevolent, your gain from switching is even greater. Math beats psychologizing.

Well put, Jam. Except that your analysis depends on the contestant having extensive empirical evidence about, and carefully analyzing, what the host chooses to do and why. This quickly gets into something more akin to reading bluffs in poker than simple logic analysis. That is simply not part of the problem/answer as set forth above. Most formulations of this problem, including the one here, are sufficiently defective to give Glenn good grounds for his annoyance.

"After Monty shows you there's nothing behind one of the other (not chosen) doors, your odds of winning automatically go from one in three to one in two -- WHETHER YOU SWITCH OR NOT!"

False, in spite of all emphasis. This belief *assumes* there is a "fair chance" that when Monty opens one of the doors it may be the winning door, thus ending the game -- which never ever happens.

This sort of assumption is what makes all kinds of elemental magicians' tricks and street scams befuddling to observers, while being simplicity itself to the magicians and scam artists. Try looking at it from the other point of view...

Imagine a game of "high card" in which you are the dealer, and in which from a fair deal you get two cards while the "player" only gets one. What are the odds that you hold the winning card?

Now add an extra step to the game -- as dealer you always select, show, and throw away the lowest of your two cards, the one that has zero chance of winning. Now what are the odds that you still hold the high card?

It's as simple as that. No truth tables or 1,000-hand simulations necessary.

Ken Doran is right. The question as posed doesn't have enough information to determine an answer. You need the caveat from the answer ( Monty is guaranteed to show you a bad door every time after you choose a door).

If solving the problem requires knowledge of the show (ie. to contestants and viewers it was clear that he always showed a little prize) then that disqualfies many readers from being able to solve the problem.

In probability the exact procedure for selecting something is called a protocol and this is an example of a problem that can't be solved without knowing the protocol. See Stirzaker's book "Elementary Probability" Sec. 2.12 Part B, the exercises after and his "Remark" on the following page which is worth quoting: "... the correct answer is, there is no unique answer...There seems to be no way of preventing the futile, acrimonious and incorrect discussions accompanying " the regular appearance of such problems

Jim:Again, it depends on if the revealing of a door is automatic or part of the competition. James, I belive was running on the assumption that it was automatic. If part of the competition, again, it comes down to bluffing, poker style.

As for the card game analogy, that does not fix exactly, as two cards give a more distinct chance of beating the one card. A more apt example is..

3 Player game, the player with the lowest card is removed by an independent judge. To a blind observor the chances of each of the remaining two players winning is 50/50. If one bet on one of the three players before cards are drawn, your odds have just increased from 3:1 to 2:1.

Given an perfectly random situation, the odds are even on which door you take.

As Prof. Delong outlines it, the needed information is provided. What I found was amusing was that my initial intuition was to switch, but then I talked myself out of it, without writing out the table provided in the solution. Jim Glass is correct in that the answer lies in viewing the situation as the house, instead of the mark.

I think some people are not reading the setup, the crucial part of which is:

" Monty then counters by
showing you what is behind one of the other doors (which is a bogus prize)...."

Thus, Monty is offering the contestant two doors for one.

The idea that Monty could be a player in this game didn't occur to me, I thought it was clear that his role was passive like that of the Black-Jack dealer.

After reading these comments, it seems obvious that the natural assumption that Monty plays, and does so with superior information, is forming a strong intuition for not changing doors: don't let him fool you!

A simple parallel is the quite classic question of which of two fairly long sequences of results from tossing a coin is the most likely: one with only heads, or a given sequence involving both heads and tails.

It is assumed in the question that the coin is perfect, but as we know that not all coins are perfect, and that some gamblers sometimes cheat, we just dismiss that unnatural assumption, and answer:

"The heads only sequence is from a cheater's coin, the mixed sequence is from a fair coin." Hastily rephrased to be an answer to the posed question this becomes: "the mixed sequence is the most likely".

The math.stat. teacher calls this a variant of "gambler's fallacy".

The gambler could equally well call the question and its right answer the "mathematicians fallacy".

"Ken Doran is right. The question as posed doesn't have enough information to determine an answer."

Sure it does.

"You need the caveat from the answer (Monty is guaranteed to show you a bad door every time after you choose a door).

That is explicitly specified in the problem. Look at it again.

"Jim: Again, it depends on if the revealing of a door is automatic or part of the competition."

That is specified.

"As for the card game analogy, that does not fix exactly, as two cards give a more distinct chance of beating the one card."

Exactly as the Monty's two doors gives a more distinct chance of beating the contestant's one door.

"A more apt example is .... the player with the lowest card is removed by an independent judge."

Nope, note the specification in the problem...

~~~
... Monty then counters by showing you what is behind one of the other doors (which is a bogus prize)...
~~~

In every game it is guaranteed that Monty will show a door, and also that it will be a bad "bogus prize" door (of which of course he must have at least one). By the terms of the problem there is no independent judge, no blind observer, and no three players -- just Monty and one player who share all information from beginning to end, just as in the high-card example.

It's all specified in the problem right there, and the answer is obvious if you look at it from Monty's point of view. He starts with a 2/3 advantage, he always opens a door, and it is always a bogus prize door that retains his 2/3 advantage. Which clearly indicates to the player what he should do, if he thinks about it for a moment from Monty's point of view.

"Given an perfectly random situation, the odds are even on which door you take."

The situation is NOT random, as clearly specified in the problem -- and realzing that is the key to finding the best strategy for the player. It's the solution to the problem.

You are here confirming my prior observation that people who look at the problem from the player's point of view for some reason seem hypnotically drawn to treating the situation *as if* the door opened by Monty is randomly selected, when the problem specification makes it very clear that this is not so -- I find that quite curious. They are also drawn to thinking that information available at the end point (one of two doors holds the prize) is all that matters, while the info available at the start of play (Monty has a 2/3 chance of possessing the winning door) somehow is irrelevant.

But again, if one looks at it from the point view of Monty or a card dealer, the truth about those naive mistakes becomes readily apparent.

Patrick -

"Thus, Monty is offering the contestant two doors for one." Before I continue to congratulate myself, should this read two doors for three? I thought Monty was making the choice simpler. Rather than 1 of 3, 1 of 2. Do I understand?

Anne

I think that "those naive mistakes" Jim Glass is referring to is about psychology or sociology. Polite behaviour is perhaps best based on the assumption that people you haven't met before are in some respect like you, that there is some sense of equality involved. Assuming that people were trying to cheat you based on their superior information would perhaps shine through as "cynical".

Therefore, you assume that you and Monty have the same information set, that he randomly opens a door that just "happens" to have the bogus prize behind it. After playing the game for some time, you become a gambler, then it's OK to show that you understand the true distribution of information, and you learn show some calculation and cynism, and you switch the doors.

anne, Patrick, am I correct to assume that accepting to change you get what is behind two doors (Monty's and the other you didn't choose at first), rejecting you get what is behind one door (the one you choose first)

Mats:Years ago back in high school I did a paper on that very concept. Depending on which coin you use, one side is for sure heavier than the other. This will change the flight dynamics of the coin, making the results not exactly 50/50. (But close). It also depends on how you flip the coin, and if you let it drop, or catch and flip. (on the back of the hand). It IS indistinguishable from randomness, as it is nigh impossible to keep track of the varibles and do a "perfect" flip, (especially on a drop flip, NFL style, as the bounce will provide almost infinite variations), but it is still not 50/50.

In any case, you are right, I would call this a sort of "mathmaticians fallacy", allowing raw numbers to get in the way of reality.

PS. If I was not such a gracious guest to our good host, I would call it an "economist's fallacy". ;)

Ken,

No. My analysis doesn't depend "on the contestant having extensive empirical evidence about, and carefully analyzing, what the host chooses to do and why." It depends on the contestant having cursory empirical evidence about the host's habits. The show was (is?) shown on weekdays; the door choice occurred once a show. I need only the information as to whether Monte pulled the door opening trick less than three times a week or more than three times a week. Given a months run of the show, five minutes work with a TIVO.

Given extensive evidence and careful analysis, I can estimate the likely malevolence of the host. Define a malevolence index as the proportion of times a switch is offered when the contestant has chosen the right door less the proportion of times it's offered when the contestant iis wrong, suitably normalized (this information is always available: it's required the grand prize be shown and described by the voice-over announcer, since "a promotional fee has been paid"). If over a run of the show, a moving average of the malevolence index converges, that value, plus the likelihood of the switch being offered can tell me fairly precisely whether I should switch or not. That's the analogue of choosing to call a possible bluff by an habitual opponent in poker.

But now we're out of theory and into application.

We agree with the setup: Three doors, one hiding a prize, two hiding nothing. We also agree that the probability that the prize is behind any one door is one in three.

You choose a door.

Monty opens one of the doors (one of the two you didn't choose) and behind it is the goat.

At this point, all you know is that there are two closed doors and behind one of them is a valuable prize. You know nothing about Monty's motives, Monty's history, whether Monty is having a bad day or if he has to put up the prize money himself. You don't know if Monty plays bridge or poker or runs around with loose women.

You know there are two doors and one prize. The odds of picking the right door with this information is one in two, no matter which door you choose.

Glenn

Did you see those dice that always end up showing "6"? And I did find an "economist's" fallacy, or at least a "stock market's" fallacy in a forthcoming issue of Journal of Finance, www.afajof.org. The authors claim that the stock market overperforms with Dems in the White House, and underperforms with GOP. Significantly so, they say.

This would be something for Brad I guess...

Don't know about some of you guys, but it seems straightforward to me: the initial probability distribution is 1/3 per door, so the other 2 doors give a 2/3 chance of holding the prize.

Short of physically changing the objects behind the door, nothing the host does alters these odds, so showing that one door doesn't hold the big prize essentially "collapses" the 2/3 probability onto the other door, ergo, it pays to switch. In a way, it reminds me of quantum mechanics ...

Mats -

"Am I correct to assume that accepting the change you get what is behind two doors (Monty's and the other you didn't choose at first), rejecting you get what is behind one door (the one you choose first)?"

Makes sense - thanks thanks. Nice logic exercises.

Anne

Mats

The line goes a bit like vote Democrat if you wish to live like a Republican. Whew, does that ever seem true.

James Erlandson

"You know there are two doors and one prize" - but you know more, which I think is Jim Glass' point. The door you picked first is randomly chosen out of three. The door you are offered to swich to is what remains of two doors randomly choosen out of three, after one door hiding a bogus prize is thrown away.

Monty is "enriching" the set of doors you didn't pick. Try the exercise proposed by at least two others above. Pick one door out of ten. Let Monty open eight other doors, all with bogus prizes, of the remaining nine. Then intuition should work.

Mats -

Yes. What I did was change the puzzle to have Monty opening a raft of doors, and decided I had to change. Notice how strong out tie is to the door we in itially chose. Superstitution is that, but it is compelling. The more I think of this, the more helpful the exercise.

Anne

Mats:I've seen how loaded dice work..and they work to a similar principle to the coin flipping example just more extreme.

It really does seem to be the case that the economy seems to go topsy-turvy when it comes to ideology, at least in the short-term. My interest in all this is the transfer of memes..my theory on that is that it has absolutly nothing to do with ideology, and everything to do with competence. A competent leader is able to take the economy through that narrow channel that ensures prosperity. Someone fixated on their ideology tends to pull the boat into the shoals. The Clinton boom, I believe had everything to do with this, yet competency was not a major factor for most voters. I believe in 2004 it will be, as this whole war debate is quickly evolving (IMO) into a judgement of the competency of the Bush Administration. This is a good thing, I think.

Aboila:Actually, all you need to compare is the second choice. In reality, if an option is taken away each time, you only have n-1 options. In this case, the only real option is to keep or switch doors. 2:1 odds.

I want to join with Patrick and seize this rare opportunity to say that I am in complete agreement with Zizka.

I also got this wrong on my first go (back in 1986!). Since then, I have met two people who got it right the first time, and too many to count who have not.

In thinking about how critical it is to know that Monty is guaranteed to show you a bad door every time after you choose a door, it is helpful to consider two separate strategies that Monty can follow if he does not follow this rule. It will follow from these two strategies that, if Monty is not guaranteed to show you a bad door every time after you make an initial choice of doors, your chance of winning if you are given the opportunity to switch doors and you do switch could be any probability between 0 and 1, inclusive.

Case 1: Evil Monty.

In this situtation, if your initial choice is one of the two bad doors, Monty gives you the bogus prize behind the bad door you chose and the game is over. You lose. If your initial choice is the one good door, Monty opens one of the bad doors and gives you the chance to switch to the other bad door. Thus, with Evil Monty, if you are given the chance to switch, and if you do switch, you will always lose.

Case 2: Good Monty.

In this situtation, if your initial choice is the good door, Monty gives you the good prize behind the door and the game is over. You win. If your initial choice is the one the two bad doors, Monty opens the other bad door and gives you the chance to switch to good door. Thus, with Good Monty, if you are given the chance to switch and you do switch, you will always win.

Finally, note that Monty could choose among these two strategies based on some probability p, with 0 <= p <= 1, such that the probability that he plays "Good Monty" is p. In this circumstance, if a player is given the opportunity to switch, and he switches, his chance of winning is p.

Whoops, I seem to have written my conclusion too quickly about the chance of winning when Monty chooses between Good Monty and Evil Monty. Because the player is given the opportunity to switch 2/3 of the time when Monty follows Good Monty strategy, and 1/3 of the time when Monty follows Bad Monty strategy, I think I should have said:

Note that Monty could choose among these two strategies based on some probability p, with 0 <= p <= 1, such that the probability that he plays "Good Monty" is p. In this circumstance, if a player is given the opportunity to switch, and he switches, his chance of winning is 2p/(p+1).

Ken Doran is right. The question as posed doesn't have enough information to determine an answer. You need the caveat from the answer ( Monty is guaranteed to show you a bad door every time after you choose a door, something that was not assured in the original game show.
).

In probability the exact procedure for selecting something is called a protocol and this is an example of a problem that can't be solved without knowing the protocol. See Stirzaker's book "Elementary Probability" Sec. 2.12 Part B, the exercises after and his "Remark" on the following page which is worth quoting: "... the correct answer is, there is no unique answer...There seems to be no way of preventing the futile, acrimonious and incorrect discussions accompanying " the regular appearance of such problems.

Writing a program doesn't prove anything as the program will embody a protocol even if none is given in the problem statement

Mr. Colpitts, Prof. Delong specifies what is needed, when he writes:

"Monty then counters by
showing you what is behind one of the other doors (which is a bogus prize), and...."

Thus, it is specified that Monty will reveal, every time, one of the bogus prizes. One knows before the game begins that Monty will toss out one of the bad choices, so one only need eliminate the other bad choice.

Will Allen's reading of the problem as stated is well short of compelling. I interpret the problem as stating, in effect, "You find yourself on a game show, and here is what happens to you." I find nothing to specify coherently what if any constraints the host is under, or what strategy he might be following.

Yes, Ken Doran is right. The problem, as stated,
does not provide enough information.

You have to know not only what Monty did
in your case, but what he had determined to
do in any case, before you made your
choice.

It is not enough to know that Monty will
always show you a booby prize. If you
you choose the valuable prize, both of
the remaining choices will be booby
prizes. Suppose Monty will always show
you the lowered number door in this
case. And suppose Monty does indeed show
you the lower of the two remaining doors.
What does this variant imply?

I also disagree that one must immediately
discuss Monty's psychology if the protocol
is not fully specified. If we postulate
something about Monty's goal, it becomes
a problem in game theory. vN and M, for
example, discuss bluffing, while assuming
"rational" actors. Same here.

Or course, we know people are not
"rational" in this sense.

Mats:

You are selected to be a contestant and you choose door A. We agree that there is a one in three chance of the million dollars being behind door A (and one in three for door B and one in three for door C).

Monty shows you the goat behind door C and offers you the chance to stick with A or change to B. You are so excited that you have a heart attack and are taken to the hospital. Monty is too distraught to go on. Fortunately, your wife (who was in the bathroom and hasn't seen anything) returns and agrees to finish the game for you. Bob Barker is brough in from the next studio to stand in for Monty. Your wife is presented with two closed doors (A and B) and one open door (C) with a goat.

Being a sensible person, she says, "Two doors. One prize. I'd say the chances are 50% for door A and 50% for door B."

Clearly, we have an independent event. New people. No history. Two doors. One prize. A one in two chance that A is a winner and a one in two chance door B is a winner.

Monty is "enriching" the set of doors remaining in the game, increasing the odds for each from one in three to one in two.

"You know there are two doors and one prize" - but you know more, which I think is Jim Glass' point."

Exactly. You do know more than that.

"Don't know about some of you guys, but it seems straightforward to me: the initial probability distribution is 1/3 per door, so the other 2 doors give a 2/3 chance of holding the prize.
Short of physically changing the objects behind the door, nothing the host does alters these odds... "

Exactly. The odds at the start are only 1/3 that the prize is behind the door that you initially picked. And you know from the specification of the problem -- from Monty's mandatory behavior --that these odds do not change after Monty opens a door.

(I.e., in each and every one of the 67% of all cases where Monty has the winning door, he opens that door that is *not* the winning door. Thus, even after opening a door in each case so he only has one door left, he still has the winning door in 67% of all cases. This is information you have!)

Ergo, you remain with a 33% chance of having the winning door, while Monty's one door has a 67% chance of being the winner.

Monty's opening the (*always*) bogus-prize door is just a bit of flim-flam misdirection, like magicians and real three-card monte dealers use, to distract one from what is indeed the straightforward logic of the situation. And it works!

My final post ...

Several have suggested trying the case of ten doors, nine goats and one prize. It has also been written that in the three door problem, the odds assigned to the "not-chosen" doors (2/3) "collapse" to the remaining "not-chosen" door when one of the goats is revealed.

Following that logic, a ten door game would pose a one in ten chance of choosing the right door, leaving a 9/10 probability for the nine "not-chosen" doors.

You choose door A. Monty opens eight of the nine "not-chosen" doors revealing eight goats.

Applying the logic that the odds of all the "not-chosen" collapse to the one remaining "not-chosen" means that changing doors gives you a 9/10 chance of winning the big prize by changing doors. I don't think anyone would look at the two doors and say that one has a 10% chance of hiding a winner and the other has a 90% chance.

A reverse raffle works this way. They sell a million tickets for $1 each and put the $1,000,000 into the pot. Then, they pull tickets out of the hopper. The last one pulled wins the million bucks. After 500,000 tickets have been pulled (and discarded) how much are the remaining 500,000 live tickets worth? Each has a one in 500,000 chance of winning a million bucks, so each ticket is "worth" $2. After 999,998 tickets have been pulled (and discarded), how much are the two remaining live tickets worth? Is one worth more than the other? If you had one of the two remaing live tickets and your neighbor had the other, how much would you pay to trade tickets?

James Erlandson:
Try thinking of the first choice as being one to discard. There is a one third chance that you got it wrong. However if you did get it right and discard one of the doors hiding a bogus prize the Monty will help you out by letting you know which remaining door not to choose and you will get the right door for sure. Otherwise his door opening does not make any difference. Effectively you get two doors for the price of one.

The only problem would be if Monty(e?) could move the prize after you have chosen or can decide whether or not to ask you. Then things would indeed be fifty fifty in the first case or, assuming an evil Monty, probability one that you had already got the right door in the second.

If you are still sure about things being fifty fifty, can we do this for money?

I can hear Chris Auld laughing all the way from Calgary at anonymous's:

" You have to know not only what Monty did
in your case, but what he had determined to
do in any case, before you made your
choice."

Which is, of course, known. It's told to us, and why so many are ignoring it is not a matter of game theory.

anon, what would Sraffa do if Monty told him that he could select one of three doors, but that after he'd made his choice he could swap it for BOTH of the other two?

Jack:
I'll put you back stage so you can make sure Monty doesn't move the goats.

When I pick a door (A), there is a one in three chance I got it right and a two in three chance I got it wrong.

If Monty shows me a goat behind door C, that door is no longer in the game. What is left are two closed doors (A and B), one goat and one million dollars.

At the beginning of the game, I got one door for the price of three. After Monty shows me where one of the goats is (C), I get to choose one door (A or B) for the price of two.

I will be happy to play this game for money as long as we play a statistically significant number of times. I'll be Monty.

If as a contestant you are sufficiently confident that the host will offer you a switch, it is a mistake to think about "making a decision" after you are offered the switch. You should decide on a strategy on the outset, and then simply follow it, the way a skilled chess player has a pre-planned second move for any plausible response to her first move. The sound strategy, given the above, is to plan to switch and then do it. You will win if the good prize was behind either of the doors you didn't choose initially. James Erlandson goes astray by thinking in terms of making a new decision after being offered the switch. That makes sense only if you weren't confident that you would be offered it, and think you can "read" and outfox the host -- but, as discussed, you are then out of a strict logic exercise and into poker.

I think the (other) bridge players on this comment thread have it backwards: I don't know who first formulated the law of restricted choice, but it only became common knowledge among bridge experts as a consequence of various popularizations of the Monte Hall problem.

Mr. Erlandson, your reverse raffle model is all WRONG, because the winning ticket is NOT pre-specified. However, we can turn it into a valid model: after 999,998 tickets are pulled out, you are told that one of the two remaining tickets matches a number that was preselected even before you chose your number. [Obviously, the selection of the 999,998 worthless tickets can't be done by a strictly random mechanical process, but rather one that knows to leave the winning ticket in the hat.] Now there are two possibilities:

1. By huge fluke, you picked the same number as the preselection process, and the other ticket is another loser OR

2. You picked one of the 999,999 losers, and the other ticket in the hat is the winner.

Do you REALLY not see that the latter possibility is more likely? You are being fooled by the belief that Monte's selection is random like a drawing from a hat. It isn't, because (IIRC) the rules of the show were that he couldn't show you the big prize.

James Erlandson, you are completely correct. When the third door is eliminated as a correct answer its "chance" of being the right answer can be conceptualized as either disappearing, leaving a 50/50, or being divided between the remaining doors, leaving, once again, a 50/50. Door one has as much right to credit for the added info about the situation as door two. The "scientific" "mathematical" "look at the probabilty chart" folks who disagree with you would be right if Monty said "I've had the stage hands move the box behind door three to join the box behind door two. NOW do you want to change your choice?" Under those circumstances it makes some statistical sense to change your choice. Not as posed.

Sigh (to Erlandson et al).

List the possibilities, as they develop over time. Then enumerate them, and work back to get the odds at the end of the process. I'll start you off (upper case letter indicates door with prize, and without loss of generality your first choice was a/A):-

- A b c;

- a B c;

- a b C;

After the extra door was opened you have the following possible sets of unopened doors (your work starts here...)

Keep going like that.

Or, from an information theory point of view, remember that you started with 1/3 odds on your first door and that you received no new information from the revelation (since it was always possible to open another door). So, if your first choice is just as likely after the revelation, the remaining door must have the rest of the probability - 2/3.

Look chums, say Monty only asks his damned question when you DO pick the correct door. In that case, gee, switching ain’t so good, three doors, or three million doors. This is not mathematics or economics, it’s show biz. But, but, it’s not FAIR…

James Erlandson

Your example, in which my wife takes over is correct: ignorant about the process, she is indifferent about the doors. But the door that is picked first does not have its probabilities enhanced by the process. Try this James:

A million closed cans, each with a small worhtless stone in them, except for one, which holds a diamond.

Choose one at random, it probably holds a stone.
Put all the others into a machine, that by some technical process sorts out the cans with stones in them. The machine is constructed to leave the can with the diamond. If there is not a cans with a diamond among the cans put into the machine, it will leave one can holding a stone.

Offered to change your randomly picked can for the can the machine leaves - after sorting out a million minus two cans with stones in them - shouldn't you accept?

James Erlandson

Try this. You have one million closed cans, all but one containing a worthless stone, one containing a diamond.

Pick one at random: probably just a worthless stone in it.

Put the other cans in a machine that is constructed to separate the cans with stones from cans with diamonds. If there was a can containing a diamond inserted into the machine, this will be returned and the other cans discareded, If there was only cans with stones inserted, one can with a stone will be returned.

Would you choose the can you picked, or the can the machine returned?

Mats:If you were not given the option to switch, then your odds of winning from the outset are still 3:1, however, the odds of winning from the point of which the wrong door is revealed is 2:1.

The question, I think is that if always given the option to switch with one wrong door eliminated, is it a better play to stay, or switch? In essence, it only is a two door game. The decision to pick the first door, is meaningless. The REAL desicion is to keep the door or switch. Two optionns, both equally qualified with the information give. To stay or to switch comes down to superstition rather than logic.

People of intelligence tend to make things a whole lot more complicated than they really are. This is not that complicated. (If Monty became an active player, and did not always showed a wrong door depending on the circumstances, everything does change, however in this puzzle this is not the case.)

Running through the possibilities is the best way to figure out the answer, IMO. Using P.M.'s notation (capitol letter for prize door), we start with A b c.

There are three possibilities: You pick A, b or c. If you pick A and switch you lose. If you pick b or c and switch you win. Switching wins in 2/3 cases.

Glenn

My "diamond in the can" is also a "two door game", keep or swich. I remove the superstition though by changing Monty for a machine.

But I don't get your "odds of winning from the point of which the wrong door is revealed" reasoning. You give me no option to switch, then my odds of winning the game is that "from the outset", 3:1. I cannot improve it later on by responding to the new information.

I don't know what you refer to - in your change-prohibited game - with "the odds of winning from the point of which the wrong door is revealed", or how this, however it is defined, happens to coincide with the odds for the changer's game with the non-playing Monty.

James Erlandson says:

" I will be happy to play this game for money as long as we play a statistically significant number of times. I'll be Monty."

You're on, James. I'll be happy to be the contestant, in which case--if we play a thousand times with a prize of $1--you'll owe me about $667.

But if your version of the odds were correct, you'd owe me $500. When do we start?

Mats:My reading of this problem, is that Monty will always show an incorrect door.

In a nutshell, the first decision is unimportant. The REAL decision is this. You are given a door. One contains a prize, the other not. It is completly random. If you switch or not is a matter of superstition or luck, as it is 50/50.

The 3-door game is a red herring. In terms of the show, it is intended to give a feeling of suspense to the proceedings.

Now, if I am reading the question wrong, and Monty:

1. Does not always reveal a door.

2. Is actvitly working against you.

Things get into game theory, and do not work out so nicely. In this case, we would need much more information to make a rational decision.

Smith: 'The "scientific" "mathematical" "look at the probabilty chart" folks who disagree with you would be right'

I'm a little bewildered why you've put all these words in quotes. The people explaining why you should switch (assuming the game rule Monty always reveals a bogus prize) ARE scientists and mathematicians, and we're doing our best to explain this rather tricky concept to lay readers. Do you also talk about the "mathematicians", "scientists", and "look at the horizon preople" who refute the naive observation that the Earth is flat?

Your model of dividing the "chance" is WRONG, for pretty much the same reason as Erlandson's. If you look, you will see that the rule Monte never reveals the real prize isn't encoded in your model. IF (big IF) Monte picks from the remaining two doors RANDOMLY (and therefore sometimes shows you the big prize, making your "switch" opportunity useless and spoiling the suspense), THEN IF he shows you a bogus prize, it doesn't matter whether you switch or not. Your analysis is correct for this game, but it isn't the actual game. In the actual game, IF you have picked a bogus door (which will happen 2/3 of the time), THEN Monte can only show you ONE of the other doors (and the other one has the prize). If you have picked a bogus prize, the "chance" (in your terminology) that the door Monte revealed had the prize is ZERO. Monte knows where the prize is AND NEVER REVEALS IT. He is not a random actor, and models based on his being a random actor ARE WRONG.

Switch who is picking the doors and it makes more intuitive sense (at least to me it does).

Monty picks two doors, looks at both and throws away the bogus prize (by showing it to the audience). Now we know Monty's seen two prizes and there was a 2/3 probability he picked a winner and a loser, discarding the loser, and a 1/3 probability that he picked two losers. If he has to get rid of a loser that means that 2/3 of the time he throws away his only loser and 1/3 of the time he throws away his choice of the two losers.

Now you have a choice do you want the door Monty never chose with its 1/3 odds, or the door Monty has left with its 2/3 odds. I'll take what Monty's holding every time.

Switch who is picking the doors and it makes more intuitive sense (at least to me it does).

Monty picks two doors, looks at both and throws away the bogus prize (by showing it to the audience). Now we know Monty's seen two prizes and there was a 2/3 probability he picked a winner and a loser, discarding the loser, and a 1/3 probability that he picked two losers. If he has to get rid of a loser that means that 2/3 of the time he throws away his only loser and 1/3 of the time he throws away his choice of the two losers.

Now you have a choice do you want the door Monty never chose with its 1/3 odds, or the door Monty has left with its 2/3 odds. I'll take what Monty's holding every time.

Patrick R. Sullivan's "two doors for one" explanation just made me think that this is about regret theory:

Glenn,

Monty is automatic, so we can just change the show a little. Instead of being offered to change after Monty shows a goat behind one of his doors, you are offered to overtake Monty's two doors, his task of "goat showing", and then to take what's behind the other door you did not show.

Since Monty's role is automatic, its unimportant wheather you or Monty performs it. So you do it.

Then, as the goat is assumed to be virtually worthless compared to the Cadillac, you can actually skip the "goat showing", just taking what is behind your two doors, without going through the irrelevant goat showing procedure.

Here we arive at Patrick's argument: Keep your door or change it for Monty's two doors. It's all random, so keeping wins 1/3, changing 2/3.

Now, in the original setting, changing means you have been manipulated to take part in Monty's game. You dismiss your own original choice for a door that is basically pointed to by him. Losing in this situation means a large loss of presitge, you have to regret your actions. Not changing is OK, if you lose it is just bad luck.

Overtaking Monty's role, you might feel that you have more control over the game, in the business of opening doors and showing goats, you might even feel that changing your first door is something that goes with your role. You might be able to rationalize rather than regret if you lose.

In Patrick's setting, choosing one door or two doors, is just a choice between two good things, regret does not get into it and it is simple to chose two doors for one.

Read the actual problem statement. Monty has no choice but to show you a door in this problem, *regardless of what you saw on TV*. This was the subject of a Marilyn vos Savant column years ago. Upshot: she was right (though admitted that she wasn't as clear as she could have in the original column that in this problem, Monty must always show you the donkey) a bunch of loudmouthed statisticians were wrong, just like several people on this board. If you don't believe it, you and a friend can try it for 25 rounds with 3 cards. If Monty *must* show you another door, it does not matter whether he is evil or not.

Or you can simulate it:
http://www.stat.sc.edu/~west/javahtml/LetsMakeaDeal.html

Read the actual problem statement. Monty has no choice but to show you a door in this problem, *regardless of what you saw on TV*. This was the subject of a Marilyn vos Savant column years ago. Upshot: she was right (though admitted that she wasn't as clear as she could have in the original column that in this problem, Monty must always show you the donkey) a bunch of loudmouthed statisticians were wrong, just like several people on this board. If you don't believe it, you and a friend can try it for 25 rounds with 3 cards. If Monty *must* show you another door, it does not matter whether he is evil or not.

Or you can simulate it:
http://www.stat.sc.edu/~west/javahtml/LetsMakeaDeal.html

P. M. Lawrence's listing of possibilities is
incomplete. His argument might be enough to
convince some that the answer is 1/3.

Those who see that might find the following
extension amusing.

Suppose Monty must follow this procedure:

After the contestant picks, Monty must show
a door with a goat. If the contestant picks
the door with the prize behind it, Monty has
a choice. In this case, Monty picks the
lowered numbered door with probability p.

Does whether or not switching improves your
chances depend on p?

Anonymous is starting from a different set of assumptions from most other analysts. Usually in these kinds of puzzles one assumes that unspecified outcomes are equally likely, since one has no knowledge to base any other assumption on. For example, what's the probability that the coin I just flipped is heads? For me, it's 1 or 0, and I'm not telling you which. For you, you might as well model it as a probability 1/2. So when Monty is faced with the case where you've picked the correct door, his behavior is unspecified and he might be following a rule such as picking the smallest numbered door. But we don't know that. He might also be picking the largest numbered door. We best model the situation, with our lack of knowledge, as saying Monty picks the door randomly.

In these kinds of problems, this kind of assumption sometimes gets stated explicitly, and sometimes it's left as an unstated obvious assumption. Once we've agreed the assumption is there, we should agree on the predictions (switch!). A discussion about what language is sufficient for justifying the assumption is, to me, uninteresting.

Final point: we can do better than anonymous's example for revealing hidden assumptions. What if the game were that the prize is always behind door 3. Then the best strategy is simple: choose door 3, or if you've somehow muddled that, switch to door 3. Presumably in the model we've assumed the door with the prize is equally likely to be 1, 2, or 3.

Now to anonymous's proposed model: if, when he has a choice (i.e. when you've guessed correctly first try) Monty is always going to choose the lower-numbered door, and if the prize is equally likely to be behind any of the 3 doors, then here's your strategy:

if you choose 2 or 3, then if Monty opens 1 you've got 50-50 odds. Switch or not at your whim. But if Monty opens 2 or 3 (whichever you ddin't choose) then switch and you'll win.

if you choose 1, then if Monty opens 2 you've got 50-50 odds, but if Monty opens 3 then switch and win.

Interestingly, regardless of which door you pick, if you play your best strategy you have a 2/3 chance of winning, which is the same odds you had in the "Monty picks randomly between doors" case.

This did the rounds when I was at graduate school. Almost everyone assumed there was no reason to change at first. On prompting mathematicians and engineers worked it out more or less correctly. Philosophers and economists however stuck with it until they played the game (using three playing) cards for money. Of course the sample was too small for this to mean anything.

"P. M. Lawrence's listing of possibilities is incomplete. His argument might be enough to convince some that the answer is 1/3..."

Well, I DID say "I'll start you off... your work starts here... Keep going like that". It's SUPPOSED to be incomplete, because making people work it through will show them better than just telling them. If I had wanted to put more of a hint, I'd have said work it through while annotating a decision tree (you first)/(Monty)/(you last).

Richard's approach is quicker and clearer but less general, and so less instructive for similar problems. But it IS right, and my suggesting a decision tree could have made people less likely to spot that better special approach.

By the way, one can apply the methods of the "telephone Indian"/"inverted pyramid" swindle, as tried in Damon Runyon's "Lemon Drop Kid", to Newcomb's paradox about choosing one or two of two boxes that will have either $10 or $1000 DEPENDING on whether you chose only one box. You can torment graduate students by setting them up with fake evidence that you really can do the trick. Then, study the victims to see their likely reactions.

A bit like what Brad deLong is doing right here, I suppose.

A meta-comment.

I used to think the reason it was possible to make money playing cards was that other people were too lazy to do the work involved. An economist would perhaps put it that people were prepared to pay for the pleasure of playing cards without having to think very much about it.

But looking at these comments, it seems to me that such an explanation is inadequate. There are commenters who are putting in a great deal of effort justifying a clearly incorrect (and at the card table, a money-losing) view of the world.

"There are commenters who are putting in a great deal of effort justifying a clearly incorrect (and at the card table, a money-losing) view of the world."

P.T.Barnum had an intuitive understanding of transitional but sustained versus equilibrium issues. Many modern economists could use this. He had an exhibit with a lion lying down with a lamb (the secret? put another lamb in every so often), but more famously - "there's one born every minute", explaining why you can't provide perfect enough information to clear the stock of suckers.

"There are commenters who are putting in a great deal of effort justifying a clearly incorrect view of the world"

It is particularly puzzling given that Brad Delong begins this by calling the problem infamous, claiming that most people have defective intuition for it, and then giving an exhaustive proof in the case that you choose A to begin with (repeat 2 times if not convinced it's the same for B and C). To list out the cases is only 6 lines, so I guess anyone who still disagrees is still trying to go about it by intuition rather than by proof despite being warned.

I would be curious to know if anyone who still insists that choosing (and sticking with) door A every time leads to winning 1/2 of the time has taken out pencil and paper.

Hint to the still doubtful: this is a contradiction with the given fact that the prize has a 1/3 chance of being behind door A.

Once more for reference: the problem as stated omits the crucial premise that the host opens a low-value prize door and offers a switch to the remaining one pursuant to a rule that he always do so, and that the contestant knows of that rule. The problem people are being beaten up on for not getting is, ironically, not the one stated but the one that should have been stated.

In response to Ken Doran, the wording of the problem really the main topic. Everyone knows what the interesting set of assumptions are:

* the prize is equally likely to be behind each door

* Monty will never open a door that reveals the prize

* In the case where the contestant guessed right the first time, so Monty has a choice which door to open, he will choose randomly between the two choices.

In this case, the answer is that you should switch, and that by switching you have a probability 2/3 of winning, and this is interesting because it's counter-intuitive at first glance.

Ken Doran's point is well-taken that the problem at the top isn't that clearly stated. The statement is

"You choose a door Monty then counters by showing you what is behind one of the other doors (which is a bogus prize), and..."

Now, if this is read as being a description of what happens in all cases, then you can infer that Monty is never revealing the real prize. You still have to assume (by virtue of ignorance) the other two points listed above, but then you're led to the conclusion that you should switch.

If instead this is being read as a description of a single person's experience -- lo and behold, Monty opened a bogus door, how about that! -- then the person cannot operate on the assumption that Monty will always open a bogus door. In this case, the odds become 50-50 and it's irrelevant whether they switch or not (anyone bothered by how odds can change from 1/3 to 1/2 by Monty's action, skip to the bottom for an explanation).

I think the most natural reading of the problem statement is the first one, that it's a general situation, therefore a rule that Monty must open the door. But I'm willing to defer to any other opinions on the matter, because the issue is not what is interesting.

Now: if I only have a 1/3 chance of picking the right door, how can my odds go to 1/2 by Monty revealing an empty door? (Remember, this is the case where I don't get to assume it's a general rule.) Break it into two cases:

1/3 of the time you've chosen correctly, and so Monty will reveal a bogus prize. You should stay put to win.

2/3 of the time you've chosen incorrectly, so the prize is behind one of the other two doors. In this case, 1/2 the time Monty will open the door with the prize, game over. 1/2 the time Monty will open the door without the prize and you should switch to win.

So the net chances are:

1/3 you have the correct door

(2/3)(1/2) = 1/3 Monty opens the correct door, game over

(2/3)(1/2) = 1/3 game is not over, and you do not have the correct door

Hence a 50-50 chance, for this case.

Apologies for the botched editing. Post above should start "the wording of the problem isn't really the main topic"

Does anybody know what Monty actually did in practice? Some have suggested he did not always give a switch. By following Ben's method above I would think he would have made the show more exciting and unpredictable. Otherwise those who watched the show regularly might have known to always switch. But then again if in practice there were a large number of contestants who played it improperly then that adds some tension to the game as well.

I see it's on at 9AM daily on the Game Show Channel.

Or Monty might be "malevolent", as some have put it, and being trying to entice you off a winner, when he could and would have simply ended the game if you had chosen a loser. The problem as stated does not exclude that possibility.

Well, for those interested in questions that depend on the subtle details of what may be assumed, here's another example:

In a discussion, a stranger states he has two children. You notice a blonde girl nearby and ask if she is his daughter, and the stranger says "yes". What are the odds that the other child is a boy? Answer: 2/3.

In contrast, suppose the stranger states that he has two children, and you ask if either are blonde. The stranger says "yes, one of them is blonde", and you ask if the blonde child is a girl, and the stranger says "yes." What are the odds that the other child is a boy? Answer: 1/2.

In one case you know he has a blonde daughter, and the odds are 2/3 he has a son, and in the other case you know he has a blonde daughter (and that the other child isn't blonde) but the odds are 1/2 he has a son. That one's kinda fun, you can mix up some geneticists if you pose it well. Now we'll see if I got the semantics right...

The following link appears to be a pretty good source page. It contains a letter from Monty Hall that, while somewhat cryptic at least as excerpted, fairly clearly denies that the now famous problem represents what happened on the show.
http://www.letsmakeadeal.com/problem.htm

Ben, I don't believe you are allowed to state the problem like that because the chance of someone's daughter being nearby is not independent of how many daughters you have. You need to ask the question of both children at once "is one of your children a girl?" for that to work out as 2/3.

Lots of parents with two kids take one for a stroll while the other is in school. Those with 2 girls are more likely to have a little girl with them. I don't see how this differs from the way you state the problem.... do you allow for the possibiliy that the daughter is not standing nearby?

Ken,

Oh, that's priceless. The model diverges from reality and we end up discussing ad nauseam precisely the aspect of the model that doesn't correspond to reality.

Has anyone who wrote here actually seen the show?

J.

Snsterling, indeed you are correct. I wrote that a little too quickly - when done well it can be quite misleading, making it sound like there is a probability penalty for being non-blonde.

The problem with consulting Monty Hall is that his version isn't really interesting. Few people would be talking about "Let's Make a Deal" if it didn't serve as useful background for posing what IS an interesting puzzle (in the sense that an optical illusion is interesting). Why not enjoy the puzzle for what it is, rather than nitpick a particular phrasing of it, or research the practices of an old game show.

"Has anyone who wrote here actually seen the show?"

Sure. Monty didn't offer to switch doors but he did offer to buy the contestant's door for an increasing amount of cash (or another good but lesser prize). The contestant then had to weigh the payoff of the sure "bird in hand" cash against the lure of keeping the door in hope of getting the grand prize while risking getting stuck with the booby prize.

Monty's offering more and more cash for the door is what got the audience all excited. And when a contestant was deciding whether or not to take the cash for the door, knowing if there was a 1/2 or 1/3 chance that the door was a winner was highly relevant. I'd wager that 95% of the contestants thought they had a 50-50 chance -- they looked like they spent far more time getting costumed up as bannanas and fire hydrants to get called on than they did figuring probabilities --which would've reduced the prize budget for the show.

If you want to see it yourself the original shows are still running on cable, and the show's web site says a new version just started running on NBC. http://www.letsmakeadeal.com/

Ben Vollmayr-Lee has a god point with his example above. It clearly illustrates that sampled data without knowledge of the sampling procedure is of little value. In his first example, randomly sampled data from a known distribution gives you a good update of the mean-value for what is left in the population after the sampled items are taken away. In his second example, sampling from a population which is independent from the one we are curios about, gives us no useful information.

IMHO, this entire discussion is a good illustrattion of the importance in statistic application in knowing how you sample (and how Monty does it!), and from what population you are sampling.

Hello,

Another way to look at this is to assume that there are 1000 doors. You pick a door and then Monty reveals 998 empty doors.
Also it's weird that people didn't switch based on the wrong argument that you when you picked a door, the probability of you getting right is 1/3. Now when an empty door, is revealed, the probability of picking a door correct is 1/2. Thus the other door has a greater chance of holding the prize.
Also reminds me of the problem that poses: someone places a marble which could be either black or white (with equal probability) in a matchbox. Then places a another white marble. Then a white marble is pulled out. What's the probability that the remaining marble is white? (It's amazing how many people will say 3/4 or 1/4).

Hello,

Another way to look at this is to assume that there are 1000 doors. You pick a door and then Monty reveals 998 empty doors.
Also it's weird that people didn't switch based on the wrong argument that you when you picked a door, the probability of you getting right is 1/3. Now when an empty door, is revealed, the probability of picking a door correct is 1/2. Thus the other door has a greater chance of holding the prize.
Also reminds me of the problem that poses: someone places a marble which could be either black or white (with equal probability) in a matchbox. Then places a another white marble. Then a white marble is pulled out. What's the probability that the remaining marble is white? (It's amazing how many people will say 3/4 or 1/4).

OK lets all agree that IF we know ahead of time that Monty HAS TO reveal a goat then it is indeed better (2/3 vs 1/3) to switch. He's essentially offering two doors vs one.

The interesting issue is in the strict interpretation where there is no guarantee. Intuition (and game theory) assumes at least some degree of self-interest (working for your benefit) on the part of all parties. And self-interest implies to at least some degree that all are working against the benefit of others in the sense that there is always some amount of zero-sumness in all situations.

It is in this self-interest context combined with the lack of explicit guarantee that Monty will reveal a goat that leads peoples intuition to what I believe is the correct answer. Again, this is given a strict reading of the problem.

Many people have stated that the problem is indeterminate because of the lack of a guarantee. One can still pursue an "optimal" solution. The contestant can eliminate any "self-interest" (i.e. nastyness) on Monty's part by simply flipping a coin and choosing to stay on heads and switching on tails. The best probablilty that can be guaranteed is 1/2. In the event of even the smallest amount of self-interest on Monty's part your probability will be less than 1/2.

OK lets all agree that IF we know ahead of time that Monty HAS TO reveal a goat then it is indeed better (2/3 vs 1/3) to switch. He's essentially offering two doors vs one.

The interesting issue is in the strict interpretation where there is no guarantee. Intuition (and game theory) assumes at least some degree of self-interest (working for your benefit) on the part of all parties. And self-interest implies to at least some degree that all are working against the benefit of others in the sense that there is always some amount of zero-sumness in all situations.

It is in this self-interest context combined with the lack of explicit guarantee that Monty will reveal a goat that leads peoples intuition to what I believe is the correct answer. Again, this is given a strict reading of the problem.

Many people have stated that the problem is indeterminate because of the lack of a guarantee. One can still pursue an "optimal" solution. The contestant can eliminate any "self-interest" (i.e. nastyness) on Monty's part by simply flipping a coin and choosing to stay on heads and switching on tails. The best probablilty that can be guaranteed is 1/2.

If you don't eliminate Monty's behavior from the problem, nn the event of even the smallest amount of self-interest on his part your probability will be less than 1/2.

test

Initially 1/3 is the correct probability across the board. When the level of uncertainty becoms 2 doors, then 1/2 is the corrected probability for the two doors rmaining. The likelihood of swift demise has not changed, and so it does not matter which door he choses.

Initially 1/3 is the correct probability across the board. When the level of uncertainty becoms 2 doors, then 1/2 is the corrected probability for the two doors rmaining. The likelihood of swift demise has not changed, and so it does not matter which door he choses.

Initially 1/3 is the correct probability across the board. When the level of uncertainty becoms 2 doors, then 1/2 is the corrected probability for the two doors rmaining. The likelihood of swift demise has not changed, and so it does not matter which door he choses.

Initially 1/3 is the correct probability across the board. When the level of uncertainty becoms 2 doors, then 1/2 is the corrected probability for the two doors rmaining. The likelihood of swift demise has not changed, and so it does not matter which door he choses.

Initially 1/3 is the correct probability across the board. When the level of uncertainty becoms 2 doors, then 1/2 is the corrected probability for the two doors rmaining. The likelihood of swift demise has not changed, and so it does not matter which door he choses.

Initially 1/3 is the correct probability across the board. When the level of uncertainty becoms 2 doors, then 1/2 is the corrected probability for the two doors rmaining. The likelihood of swift demise has not changed, and so it does not matter which door he choses.

Disreguard my last post. In the cas of a malicious host he canoly offer after the conestand picks the winner. In this case the ony way to win is to stay. SO the best strategy when the host IS NOT REQUIRED TO ALWAYS reveal a goat is to STAY

George,

Are you addressing the situation where the host ALWAYS reveals a goat? If so, then it definitely not 50/50. If the host always reveals a goat, he is essentially offering you teh choice of one door (if you stay) vs two doors (if you switch). SO in this case the odds are 33/67.

i faithfully played the game and won money and prizes andnever got anything
i called them so many times and they said we are working on this right its almost 3 years i know of others who never got theirs

to understand this problem, you just have to look at the three situations.

1: you choose the right door...if you stick, you're HAPPY. if you switch, you're SAD.

2:you choose the wrong door...if you stick, you're SAD. if you switch, you're HAPPY.

3:you choose the other wrong door...if yo stick, you're SAD. if you switch, you're HAPPY

stick = 1 HAPPY and 2 SAD
switch = 2 HAPPY and 1 SAD

because there are 2 bogus doors and 1 good door at the start, the probability of choosing a bogus door is 2/3. therefore, you are more likely to have the bogus door than the good door, so a switch would be the better option.