View Full Version : Monty Hall revisited, with a twist- survey

Easy E
05-19-2003, 09:19 AM
Don't want to bring this up again, but in the Sunday Parade magazine, someone wrote in about the Monty Hall/intuition "puzzle".
He THEN said that one of his students posed this variation:
Suppose I am taking a multiple-choice test. On one question, I randomly choose 1 of 3 answers. Then the instructor says that one of the remaining two choices is incorrect. Should I switch my choice?

The "world's smartest woman" (or is that human? I forget what Marilyn vos Savant claims) said that the answer was: it doesn't matter.
She did NOT, however, explain her reasoning.

What am I missing here, because it seems like the exact same thing? Does it have to do with the way tests are scored, or that one wrong/right answer won't make a difference?

Monty Hall Test

Extra credit for detailed answers.

05-19-2003, 11:44 AM
If he doesn't specifically eliminate one answer, then it doesn't matter what you do.

Notice that whether you originally chose the correct answer or not, one of the remaining two choices will always be incorrect; so the instructor's statement will always be true.

Now, if he says "exactly one of the two remaining answers is incorrect", you'd obviously change your answer.

Easy E
05-19-2003, 01:07 PM
My editing needs some work (thanks, pudley). Someone on UPF caught this as well

My editing of the problem IS the problem.

"You pick A. Teacher says C is wrong. Do you switch to B"
is the short version.

Marilyn's take was that, unlike the Monty Hall choice, it doesn't matter if you switch.

Is this correct?

05-19-2003, 02:14 PM
With the Monty Hall problem, there are 3 possibilities:

Car, goat, goat.
Goat, car, goat.
or Goat, goat, car.

by eliminating one of the goat choices THAT YOU DID NOT PICK (and knowing the goat choices, and your choice ahead of time), the likelihood is 2/3 of picking a car if you switch, and 1/3 if you don't.

When the teacher tells you C is wrong, he does not know whether you picked C or not, so it does not help you to switch. The difference is he is just eliminating one wrong answer, for all he knows it is your answer. Now walk through it using the goat, car, goat method, and you will see. Monty Hall deliberately picks a wrong answer that you did not pick to give you a choice whether to switch, the teacher just eliminates a wrong answer with no regards to whether or not you picked it.


05-19-2003, 02:27 PM

The difference is that in the problem with the test, the instructor does not know what you picked when he tells you C is wrong. So she is right. In the Monty Hall problem, the possibility that you chose the wrong door and the host tells you that you are wrong does not exist, where it does in the quiz problem.

Easy E
05-19-2003, 03:16 PM
So, let me see if I have this straight:

In Monty Hall, your pick has a 2/3 chance to be wrong. When Monty reveals wrong answer (door C) in the remaining choices, KNOWING that you picked door A, the chance that you should switch is 67%, not 50%

In the test, your original pick also has a 2/3 chance of being wrong. The instructor reveals wrong answer (#C) in the remaining choices, WITHOUT knowing that you picked #A. Now, your chance of being wrong with #A is 50%, not 67%.

And the difference is because, not knowing what you picked, the instructor could have told you that #A was the wrong answer, where Monty Hall eliminated revealing your door A as a goat?

Someone please get through my little ole brain how the dependency is gone in the second situation, given that the instructor did NOT pick #A as the answer to reveal, even though they could? I understand that the instructor's relevation gives #A and #B a 50/50 chance... but why did that change YOUR odds of #A being wrong originally, since you did NOT have any more info than in Monty Hall- except for that the instructor's choice was random and Monty's was not?

Am I putting too much weight on the wording that says you picked #A and the instructor did not, even though s/he could have?

I hope there's a mathematical principle behind this, because the verbiage is giving me a headache...

Easy E
05-19-2003, 04:06 PM
Why does the dependency in Monty Hall NOT result in the same value, as you two are saying, as the dependency in the testing situation (you pick, teacher reveals a wrong choice out of the two that you didn't pick)

05-19-2003, 09:33 PM
Behind 2 doors are horrible, murderous space monsters, behind one door is the most beautiful woman imagineable. To save your life, you must pick the one door out of three that is in front of the beautiful woman. The rooms and doors are completely sound-proof and smell-proof, there is no information that you can observe that will help you make an informed decision.

Without uttering a word, you decide to go with door number two. However, before you can voice your choice, a gigantic space meteor instantly crashes into the room behind door number 3 revealing a hideous space monster. Fortunately, the monster is killed by the meteor. Is this a sign? Should you switch doors? The other two rooms and doors are not exposed in any way.

It doesn't make any difference, it is even money either way. Ok? Same as if a teacher tells you a choice is wrong in the multiple choice question without any knowledge of whether you picked this choice or not.

Now using the same three doors and choices, your host suddenly takes a liking to you. He will tell you a door, other than the one you picked, that is guarding a monster, after you voice your choice, and give you the option of switching to a different door. He is from the planet Clorox, and is unable to tell a lie. You choose door number 2, He instantly tells you that there is a monster behind door number 3, and asks if you wish to switch.

Do you see the difference? Now you switch.

There is a 1/3rd chance of the doors being arranged as M-W-M, and 2/3rds of being either W-M-M or M-M-W, so therefore when tells you there is a monster behind door 3, there is a 2/3rds chance of the arrangement being W-M-M, because the alien knows the arrangement and your choice already. If the door is instead exposed by a meteor, it gives you less information, because it could have just as easily been your door that got crushed by the meteor.

Put 3 coins on a table, a dime, and two pennies. In whatever order you choose, it is irrelevant. Now, the object is to end up with the dime.

Let's say the order is Penny-dime-penny.

First, "pick" the first coin, then, eliminate a wrong answer (the last penny), then switch, you win the dime.

Next, "pick" the second coin, and eliminate one of the pennies, then switch, wrong answer, you win a penny.

Next, "pick" the last penny, eliminate the other penny, and switch, you win a dime.

See, doing this, you are right 2/3rds of the time, try it any order you want, It also works with nickels and quarters /forums/images/icons/laugh.gif


Easy E
05-19-2003, 10:11 PM
I didn't need
I understood that, having removed a choice, the odds were 50/50.

What I DIDN'T understand is why, with the test variation as written (You pick penny, instructor picks the OTHER penny, though by accident) that the odds were only 50/50.

Yes, it was an uninformed choice by the instructor, but the NET result was the same: You picked in one set (with a 33% chance of being correct) and the instructor eliminated a choice in the other set (which had 67% chance of containing the right answer BEFORE the instructor unknowingly eliminated a choice), but the remaining choice in the second set LOST a 17% chance of being correct.

In Monty Hall, the informed dependent choice, when you KNOW that your door will not be opened, reveals the 2:1 shot.. but when the instructor, by "accident," takes an "independent" action THAT RESULTS IN THE EXACT SAME SCENARIO... it's now 1:1 .....

It seems to me that a required dependency, the fact that the teacher did NOT reveal your answer #A as the wrong answer, is being ignored here for an unfathomable reason, just because it COULD have been random but ended up NOT being random, as the problem was presented.

And they wonder why Monty Hall confuses everyone so much.... sigh.

Well, I follow it, but I don't "get" it. Thanks, Don, but MY solution to the Monty Hall problem and its variations, from here on in, will be to:

a) ALWAYS switch, since it may be better and may be breakeven and then

b) SLAP THE BEJEZUZH out of the fool who dared to bring it up, then walk off with my monsterous goat penny, muttering to myself....

05-19-2003, 11:28 PM
I'm sorry if I'm pounding too hard, I was just trying to paint a picture.

Let's try the pennies again, this time (order doesn't matter, but let's just use penny-dime-penny again).

Now pick the first coin, then reveal one coin at a time in order (to simulate choosing randomly):

1st coin, yours, you must choose one of the other two doors, its a 50/50 wash

2nd coin, the dime is revealed, no guesswork at all now.

3rd coin, if you switch you win

Now, same scenario, pick the second coin and reveal them one at a time:

1st coin, you switch, you lose

2nd coin, the winner is revealed, no guesswork

3rd coin, you switch you lose

Same scenario, third coin is chosen, reveal one coin at a time:

1st coin, switch and win

2nd coin, dime revealed, no guesswork

3rd coin, you must choose among the remaining two, no advantage, 50/50 wash.

So, lets count up the wins and losses for switching, win twice, lose twice, two 50/50 washes, and no guessing the 3 times that the winner is revealed. So, you see when picked randomly (using every possible choice) there is no advantage to switching or not switching (except when the winner is revealed and there is absolutely no guesswork involved).

There, I think that illustrates the random choice switching scenario better. Seriously, I'm not trying to be a pain-in-the-***, just to paint a clear picture.


05-20-2003, 07:25 AM
(Caveat : Monty Hall must announce beforehand that he will open up one wrong door for you. Monty will not be opening up doors randomly. In other words, it's a basic assumption that Monty knows where the car is.)

The easiest way to overcome my intuition about the Monty Hall, which tells me that it's the same whether I switch or stay with my original choice, is to imagine a thousand doors. Not just three dors, but a thousand.

I choose one door at random, and then Monty opens up immediately 998 doors revealing goats! There's one door closed. Plus the one I picked. But I'm not switching.

We do that with Monty three or four times and then I wise up...

Easy E
05-20-2003, 12:18 PM
I only needed one coloring book picture, not the whole book!


I think I'll just stick with my "switch and slap" strategy... much easier to remember than figuring out when a dependent condition isn't an influencing condition....

05-20-2003, 01:54 PM
Okay, I know you'd like it phrased to a more reduced mathematical principle, unfortunately, I can't think of how to do that at the moment. I guess all I can do here (at least at this moment) is show examples /forums/images/icons/tongue.gif .

Maybe someone more knowledgeable than I can reduce it to a more elegant mathematical rule.

Good luck,
(no more coloring book pictures on this one)

05-20-2003, 05:02 PM
Say you choose A, out of obtions A, B, C.

1/3 of the time, the teacher says that A is wrong, 1/3 B, 1/3 C.

Now, you know that 2/3 of the time, the teacher will tell you that one of the other 2 options was wrong. And, since there are 2 of them, half of that time... the one you may want to switch to is the correct one.

This happens 1/3 of the time.

Well, look back at what it started as... you had a 1/3 chance to pick the right one from the beginning.

See now?

I believe that this is as clear as I can possibly make it. I'm sure you understand that it doesn't matter whether or not it's a letter/number/door/monster/hottie.

Just remember in this case that 1/3 of the time you WILL want to switch. If she tells you that your answer is wrong, then you probably want to switch, right?

Feel free to PM me if you still have any questions.


05-20-2003, 07:03 PM
Right, that is a good way of putting it. Much more concise than my monsters, hotties, dimes, and pennies.

And because in Monty Hall, you know ahead of time, that a wrong answer (other than yours) will be revealed 100% of the time, instead of 2/3rds of the time, in other words, 1/3rd of the time MORE often that would be dictated by chance alone. THEN, the remaining door must be a goat 1/3rd of the time LESS often than would be dictated by pure chance.

In other words, if x + x = 2x, then, 2/3x + 4/3x = 2x

That's it.


05-21-2003, 07:37 PM
Let's reduce the whole thing to a more simple formula:

It is a given in both situations that A, B, and C individually have 1/3 probability of being right (1/3P we'll call it).

You choose A, let's say:

In Monty Hall, MH MUST choose either B or C and expose it as wrong, so instead of 1/3P it becomes 0/3P, or zero probability, (because he must expose a wrong answer between B & C). If one of the two choices becomes 0/3P, then the other choice (the unexposed choice between B & C) must be increased by an equal amount, hence 2/3P, this does not change your original probability of 1/3P on your first choice because he is forced to choose between B & C, and cannot choose A.

Therefore A=1/3P B=1/3P and C=1/3P, whichever one between B & C Monty chooses becomes 0/3P, the other must become 2/3P, because B + C must equal 1/3P + 1/3P, or 2/3P, therefore 1/3P + 1/3P = 2/3P + 0/3P, and A remains unchanged at 1/3P, because A is not a choice for MH, and obviously you must switch to the choice that is 2/3P.

If all are random, then the answer to be exposed remains a 1/3 probability of being correct, if a wrong answer therefore is exposed, it becomes 0/3P (for the wrong answer exposed). Therefore your choice and the other unrevealed choice are affected equally, (because your answer being exposed was just as likely as any other, 1 in 3, in MH, he must choose between 2 answers, here it is randomly chosen among 3 answers with equal weight). Probability of being right for each of 3 choices is 1/3, therefore if C is revealed to be a wrong choice, then A & B must both be increased by 1/6th each.

1/3 + 1/3 + 1/3 = (1/3 + 1/6) + (1/3 + 1/6) + (1/3 - 1/3) = 1/2 + 1/2, therefore each remaining choice has 50% chance of being correct.

I apologize for any redundancy of anything anyone else may have posted, but this seems like a good formula for reducing the essence of the problem.

Thank you to the Duke for getting it back on the right track after my earlier storytelling atrocities.

I am not claiming my random choice formula is any better than Duke's, just putting it here with the MH formula for completeness.


05-25-2003, 01:50 AM
I think it comes down to whether the professot would say one of the answers is incorrect if both remaining answers were incorrect. If so then it doesn't matter, if not then you should switch.

Bah just read, the addendum. Since he just eliminates an answer you didn't pick, it doesn't change. It bumps you chances from 1-in-3 to 1-in-2 but changing you choice at this point doesn't affect that

05-25-2003, 07:10 PM
Given that 1 in 3 chances is always correct in either scenario, the only difference between Monty Hall and the Professor is:

Monty Hall may only choose a wrong answer from among two choices


The professor chooses a wrong answer from among 3 choices.

If you choose from among only 2 choices, then you must look at the one "unpickable" choice separately from the other two, such as (1/3) + (1/3 + 1/3), now if one choice among the latter two is found to be wrong (whether by random chance or preordained knowledge, but specifically your choice is excluded from the picking) then it becomes zero and the other becomes 2/3. (1/3) + (1/3 + 1/3) = (1/3) + ([1/3 - 1/3] + [1/3 + 1/3]) = (1/3) + (2/3) = 1. Of course your choice is no different from the other remaining one if he is choosing from all 3 with no regards to what you picked.

So that is the real difference, choosing from only 2, or choosing from all 3 without specifically excluding your answer.


Al Mirpuri
05-31-2003, 06:12 AM
You are asked to pick one out of three doors. Behind one of the doors is a prize. Can you pick the correct door (the one with the prize behind it)? The doors are labelled A, B, C. You pick B (for the sake of the argument). Monty Hall informs you that C (for the sake of the argument, he could just have easily said A) does not have the prize behind it. He now tells you that you may forgo your original choice B and choose A. Should you choose A?

It has been answered that yes you should. The reasoning goes like this: whatever you picked had a 1/3 chance of being correct, what remained unpicked had a 2/3 chance of being correct. Once one of the unpicked doors has been shown to not have the prize behind it you should switch to the other door as the whole of the 2/3 chance now falls on that door. Remember it was 2/3 that the unpicked doors had the prize behind them.

It has also been answered that you should not switch your allegiance as after one door has been shown to not have the prize behind it you are left with two doors and only one of them has the prize behind it; a fifty-fifty proposition.

The reason it is called the Monty Hall Paradox is because two equally persuasive pieces of reasoning produce differing answers. It is not a mathematical puzzle but a philosophical conundrum.

05-31-2003, 08:21 AM
"The reason it is called the Monty Hall Paradox is because two equally persuasive pieces of reasoning produce differing answers. It is not a mathematical puzzle but a philosophical conundrum."

I'm sorry but I respectfuly disagree.

This is indeed a mathematical puzzle (problem). It has nothing to do with philosophy. For a better and more immediate understanding of the purely mathematical nature of the dilemma and its solution, please check my hint at an earlier post (http://www.twoplustwo.com/forums/showthreaded.php?Cat=&Board=exchange&Number=267357 &page=5&view=expanded&sb=6&o=14&fpart=).

It is called a paradox because the answer runs counter ot our intuition. And not because there are 2 equally plausible solutions. No, one solution (switching) is correct and the other (staying put) is incorrect.

When a problem presents us with an intuitively obvious but actually wrong solution, it is called a paradox. Or when it offers no solution at all, on the basis of the stated conditions of the problem, although at first examination it looked like it can be solved.

<ul type="square">paradox, n.

1. A seemingly contradictory statement that may nonetheless be true: the paradox that standing is more tiring than walking.

2. One exhibiting inexplicable or contradictory aspects: "The silence of midnight, to speak truly, though apparently a paradox, rung in my ears"; (Mary Shelley).

3. An assertion that is essentially self-contradictory, though based on a valid deduction from acceptable premises.

4. A statement contrary to received opinion.

--From the American Heritage Dictionary, www.dictionary.com (http://www.dictionary.com)

05-31-2003, 07:19 PM
I fail to see the paradox.

I am not going to post formulas again, but just a simple explanation.

The difference between Monty Hall, and the Professor, is that Monty Hall can only choose from the two doors that you did not pick, this is a given. The Professor shows an answer that could have been any one of the three.

Therefore, and this should be clear enough by this time, if Monty Hall can only choose from answers that you did not pick (either by flipping a coin, or deliberately, this point is irrelevant, the only relevant point, is that he is not allowed to choose your pick), therefore, your answer still has exactly a one third chance of being correct. If he had chosen to expose this wrong choice without knowing which one that you had picked, then that changes everything, now he is exposing simply a wrong choice (either randomly or deliberately, once again, this point is irrelevant, the only relevance is that he has not excluded your answer from being exposed, and doesn't know what you picked), and therefore both remaining choices are affected equally, and are 50/50 shots.

So perhaps you could not understand the earlier explanations (hopefully you understand this one), but there simply is no paradox here. Just a problem that is somewhat counterintuitive upon first examination.


06-01-2003, 12:36 AM
In mathematics, the term "paradox" takes on a more specific meaning than it does in everyday language. A mathematical paradox is a fundamental contradiction which cannot be resolved by the theory in which it arises, and which thus shows that the theory is incomplete, and that the contradiction may only be resolved if a change were made to the theory, or if the theory were replaced by a new theory. A true paradox presents two plausible statements which cannot simultaneously be true, yet cannot be individually proven to be either true or false by the existing theory.

When a problem presents us with an intuitively obvious but actually wrong solution, it is NOT generally called a mathematical paradox. The wrong solution is simply termed a "fallacy". We also do not have a paradox simply because a problem offers no solution at all, on the basis of the stated conditions of the problem. If the problem cannot be resolved due to insufficient information, even though it is not immediately clear that extra information is required, the problem is termed "ill-posed". The term "conundrum" can also be used to describe this type of problem.

Sometimes the term "paradox" becomes attached to a problem which was once thought to be a paradox, but which was later found to be merely a fallacy, or perhaps a conundrum. For example, the so-called "twin paradox" offered to disprove Einstein's theory of relativity is actually just based on a fallacy, so it is not a true paradox. On the other hand, the "EPR paradox" offered by Einstein to disprove quantum theory could not be fully addressed without taking a particular interpretation of quantum theory which required additional experimentation to verify, hence it is a true paradox. Recent experiments offer strong validation for the proposed interpretation of quantum theory, but this specific interpretation was required to resolve the paradox nonetheless. This could be considered a resolved paradox; however, it is still not fully resolved to everyone's satisfaction. Schrödinger’s cat is also a true paradox which experimental evidence may soon resolve. Russell's paradox was a true paradox which imposed radical changes to logic and set theory. There are other paradoxes. True paradoxes are usually truly mind blowing, almost by definition.

The Monty Hall problem is far from a paradox. If we are not told how Monty chooses the door to show us, then the problem is ill-posed and cannot be answered without imposing our own assumptions. If we are told that a) Monty is constrained to show us a door that we did not choose, b) Monty must show us an incorrect door, and c) when we choose correctly, Monty chooses between the two incorrect doors to show us with equal probability, then the problem is not a paradox, and it is not ill-posed. It is completely and quite easily solvable. If someone were to argue after being given all this information that it shouldn't matter if we switch, then that person would be committing a fallacy, plain and simple. If we are told something different regarding conditions a, b, or c, then the problem can still be solved, but the answer may be different. Condition c is generally ignored by every discussion of this problem I have ever seen, yet it is absolutely essential information.

For examples of mathematical fallacies and paradoxes, and the definitions I have given here, see this entertaining and very readable text:

Mathematica Fallacies and Paradoxes (http://www.amazon.com/exec/obidos/tg/detail/-/0486296644/qid=1054440172/sr=8-2/ref=sr_8_2/102-5944473-0437740?v=glance&amp;s=books&amp;n=507846)

06-01-2003, 06:01 AM
I should clarify something about the EPR paradox. I made it sound as though quantum theory needed to be modified in order to resolve this paradox. This was not the case. In fact, some clever math by a guy named Bell showed that the apparently impossible effect predicted by the paradox was real and could be derived directly from quantum theory. This paradox did not expose a contradiction within quantum theory, but an apparent contradiction between quantum theory and special relativity, both of which are intended to be parts of a larger consistent theory of the physical universe. Experiments have shown that quantum theory comes through with flying colors, and that the results predicted by quantum theory really do occur, even though these results are extremely bizarre, so much so that Einstein proposed the paradox because he felt that they cannot possibly occur. In fact, these results do not explicitly contradict the specific statements made by relativity. It is not understood, however, how such results are able to co-exist with relativity. No established theory exists which explains how such results can occur without violating relativity, and the ones that have been proposed are even more bizarre than the paradox itself, i.e. parallel universes, backwards time travel, reversal of cause and effect, etc.

I consider EPR to be the single most interesting topic in the universe, and I've given a lot of thought to bestowing that title upon it. It is highly relevant to probability, and I intend to post more on it when I get time to do it justice. The whole issue is not even known to most people because the true nature of quantum theory is usually obscured by sophisticated mathematics. Even those who study quantum theory as part of undergraduate physics or engineering courses never even come close to appreciating its bizarre implications, and this is partly because the implications are so bizarre as to strongly resist a correct interpretation. It is not widely known, for example, that this field has already produced what can accurately be described as teleportation on a small scale in a laboratory, and plans are underway to teleport atoms and even molecules the size of viruses. EPR also attracts a lunatic fringe interested in extrasensory perception and so forth, but what I am telling you is very real hard science.

In a nutshell, EPR stands for Einstein-Podolsky-Rosen, and the paradox predicts that certain types of particles must under certain conditions exhibit an instantaneous effect on each other, even if those particles are separated from each other by the vast expanses of the universe. Experimental evidence confirms that this happens. Relativity tells us that no energy or information can travel faster than the speed of light, therefore such instantaneous action at a distance would seem impossible. It turns out that no information has ever been observed traveling faster than the speed of light in these experiments because the result of these interactions can only be verified after the fact. Therefore, the particles seem to have found a loophole in the laws of relativity which allow them to perform this feat without being caught violating the theory. How this is possible is the big mystery, and is truly a paradox.

06-01-2003, 01:17 PM
Elsewhere in this thread, a noted contributor to these pages makes the important distinction between a paradox in everyday, layman's terms and in mathematical terminology. Though not as qualified a mathematician as that noted contributor, I should probably clarify that the Monty Hall problem is indeed, as the contributor mentions, not a paradox in the mathematical sense : all the information necessary to solve the problem is clearly given to the solver and the problem has but one solution, i.e. always switching.

The Monty Hall problem, nonetheless, has entered everyday language also, if not mostly, as a paradox. A quick web search (http://www.google.com/search?as_q=Monty+Hall+paradox&amp;num=100&amp;hl=en&amp;ie=IS O-8859-1&amp;btnG=Google+Search&amp;as_epq=&amp;as_oq=&amp;as_eq=&amp;lr=lang _en&amp;as_ft=i&amp;as_filetype=&amp;as_qdr=all&amp;as_occt=body&amp;a s_dt=i&amp;as_sitesearch=&amp;safe=images) brings forth a plethora of sites that denote the problem as a paradox, and quite a lot of them are mathematical websites. Some of the texts point out that the problem is seemingly a paradox, others don't. (For instance, a paper out of MIT begins with the words : "The Monty Hall problem is based on an apparent paradox that is commonly misunderstood, even by mathematicians." Even professional mathematicians will often say that there is no difference between switching and non-switching, as the paper tellingly reports. In other words, mathematicians can be betrayed by their intuition --and not by their calculations-- which leads them to the wrong answer.)

The mathematical definition of a paradox is this : An argument is a set of statements (aka the premises) that leads to another statement (aka the conclusion) which follows from those statements. A paradox is an argument that involves a set of apparently true premises P1, ... , Pn and a further premise Q such as that we can derive a contradiction from both [P1 &amp; ... &amp; Pn &amp; Q] and [P1 &amp; ... &amp; Pn &amp; notQ]. The Monty Hall does not fit this definition.

Important clarification :

Mathematicians distinguish between logical paradoxes, such as the paradox by Russell, which was also mentioned by the distinguished contributor, and semantic paradoxes, such as the liar's, aka the Cretian's, well-known paradox. It is important to note that, although of course mathematics is a most robust science, if I may say so, semantic concepts creep into it unavoidably, despite the purists taking offense at such an infamy, as this is clearly demonstrated in those semantic paradoxes. (Other examples of semantic paradoxes are Berry's paradox, Grelling's paradox, Richard's paradox, etcetera). Russell's paradox which belongs in the former category and befuddled set theory, led to that theory coming forth with a better definition of sets, i.e. sets to be defined by their members rather than by general conditions. In other words, the premises were immproved in order to accommodate, and thus nullify, a mathematical paradox as such.


PS : In my above post I also argued that the Monty Hall problem has nothing to do with philosophical conundrums. One could stretch our analysis of human perception to accomodate philosophical queries about "reality" and "denomination", the stuff paradoxes are made of, but that would be, as I said, stretching it : there is no philosophical aspect to the Monty Hall.

PPS : The poster who dissents to calling the Monty Hall a paradox (and, in mathematical terms, he is of course correct, if I may say so once more) also showed in his post examples of true mathematical paradoxes. In every single one of them, the scientists inform us that we will most probably, if not certainly, resolve those paradoxes when we are able to construct &amp; carry out experimental verification and/or when our knowledge expands. (And this should include the EPR paradox, which in our contributor own's words "is the single most interesting topic in the universe".) In this sense, would we be correct to argue, only half in jest, that everything that, on the basis of the current totality of out knowledge, i.e. our current premises, we cannot understand or resolve, is a scientific paradox ? In other words, is it correct to say that everything we cannot yet understand is a temporary paradox? (But this would impertinently presume that eventually we will be able to understand everything! I am sorry, Father, I'll take 20 Hail Marys and fast. /forums/images/icons/frown.gif )

PPPS : Although this may come as a surprise to some /forums/images/icons/tongue.gif , it is not my mission in life to forever debate semantics, semiology or nomenclature on the Probability page of the twoplustwo.com website. Hence, and since my lowly contribution here is already taking too much bandwidth, this is my last post in this thread.

06-01-2003, 04:05 PM
Some time ago I posted about what is known as "Simpson’s paradox". It is named as such in statistics books since this has become popular nomenclature, but they will also tell you that it is not a paradox at all, but simply a counterintuitive arithmetic fact.

Simpson's Paradox (http://www.twoplustwo.com/forums/showflat.php?Cat=&amp;Board=exchange&amp;Number=28197&amp;Foru m=All_Forums&amp;Words=simpson&amp;Match=Entire%20Phrase&amp;S earchpage=5&amp;Limit=25&amp;Old=allposts&amp;Main=28197&amp;Searc h=true#Post28197)

There are many examples of misnomers in science and mathematics. Centrifugal force is not a real force, most random numbers are not really random, and there is no such thing as rms power no matter what your stereo specs say.

When mathematicians refer to something as a paradox which is not really a paradox, they are either doing so because of popular terminology, or because they do not really understand the problem under consideration. Mathematicians are not experts in all areas of mathematics, and not all mathematicians are competent statisticians.

As to the issue of whether all paradoxes will eventually be resolved when we finally know everything, mathematics has proven that this can never happen if we limit our knowledge to that which can be proven by mathematical logic. There will always be statements that can neither be proved nor disproved by logic, otherwise the theories in which those statements arise would become inconsistent. On the other hand, theories about the universe need not depend on the completeness of mathematical logic. For example, physics is not mathematics. Physics uses mathematics, it even creates its own mathematics, but theories of physical reality do not constitute a formal logical system in the mathematical sense. It is not of particular concern to a physicist if the mathematical theory which he uses is incomplete in the sense that there are statements which can neither be proved nor disproved, unless those statements come up in a physical context. For this reason, one might suppose that physicists may at least have a shot at eventually believing they understand everything about the known physical universe, but they will never convince mathematicians that they know everything.

06-02-2003, 04:38 AM
I don't even consider the Monty Hall problem, with all information given, to be an example of a paradox in the everyday sense of the word. At least it is not for me, and not for anyone who thinks about the problem the right way. If you never switch, you will be wrong 2/3 of the time, so if you always switch you must be right 2/3 of the time. That is indisputable. This is the simplest way to think about the problem, and there is nothing counterintuitive about it. To say it doesn't matter if you switch would be to suggest that Monty opening a door suddenly makes you win at a higher rate when you do nothing differently, which would be truly counterintuitive, and for good reason since it would be clearly wrong. This notion can be immediately dismissed. People who are not used to thinking about conditional probability can confuse themselves by thinking about this problem other ways, but you don't even need conditional probability to analyze this. There are other problems where you do need conditional probability which are far more counterintuitive.

Al Mirpuri
06-02-2003, 08:04 AM
The only factual inaccuracy, as opposed to a disagreement on the value of any specific point, I could find with your post was that you deny that the Monty Hall Paradox has anything to do with philosophy. This is not so. I first came across the Monty Hall Paradox in a philosophical dictionary whilst studying for a philosophy degree.

My post has generated a response that is admirable for its intellectual rigour. There is obviously much to discuss. As humans, I think the truth is beyond us; so let the debate rumble on.

For those of you who think that mathematics and science are rigorous disciplines that allow no room for disagreement or subjectivity (the 'I'm right, You're wrong' school) can I provide two examples. First, any equation concerning Pi attains an arbitrary certainty for Pi cannot be fully calculated and this spurious exactitude is given to Pi as there is a very human need for certainty. Secondly, positive co-ordinates are plotted up and right on graphs whilst negative co-ordinates are plotted down and left on graphs and this is because we are a right-handed species that believes heaven is above us whilst fearing left-handers and thinking hell is beneath us. In short, mathematics is value-laden (just as science is).

06-02-2003, 02:28 PM
For those of you who think that mathematics and science are rigorous disciplines that allow no room for disagreement or subjectivity

I certainly think that the science of mathematics is rigorous, the physical sciences somewhat less so, and other sciences less so still. I certainly do NOT think that there isn't room for disagreement and subjectivity in any of these sciences. Subjectivity in mathematics arises from the creation of axioms which assign arbitrary truth or falsehood to statements that cannot be proven, and also from the making of arbitrary definitions. Different mathematicians may make different arbitrary choices of definitions and axioms based on their opinions of what should make mathematics most elegant, or most powerful, or they may do so simply to explore where a different set of axioms will lead. Once a particular choice of axioms and definitions are agreed upon however, there can be no valid dispute as to what has been proven and what has not been proven.

The truth of the Monty Hall problem is absolutely NOT "beyond us". It is completely known, quite trivial, and agreed upon by all competent mathematicians. The two different answers to the Monty Hall problem you gave are each only valid for a specific set of conditions which must be stated in the problem, and they are not simultaneously valid for any completely specified set of conditions. Failure to state a complete set of conditions is the only valid reason why there ever would be any debate among competent mathematicians. If there is not enough information specified to answer the question (and I believe this is almost always the case because my condition c is virtually always ignored) then different mathematicians could make their own separate assumptions to fill in the missing information. Then and only then would each of these answers have the same validity. Even then there would be no cause for a debate because once each mathematician states his assumptions, then all other competent mathematicians must agree that his answer follows from those assumptions. If there is not this agreement, then it means necessarily that some of the mathematicians are not competent, either because they analyzed the problem wrong, or because they failed to state their assumptions. If a philosophy book states otherwise, then the author of that book may be a competent philosopher, but he would be an incompetent mathematician. Incompetence does not justify declaring mathematical conclusions to be subjective.

See my posts below for the proper way this problem should be stated, and the one and only one correct solution given this problem statement. If after reading my problem statement and explanation you still believe that not switching is a valid solution, then there will be nothing more I can do to impart to you an understanding of this matter; however, I do have a game I want to play with you in which I will give you 3-2 odds on what you will believe to be an even money shot. You should be eager to play this game with me for as long as I like. I will also be eager to play this game with you until one of us is broke, because I will know with absolute certainty that you are a 2-1 underdog. Gambling is one arena where I welcome people to disagree with me.

Now for your two examples of subjectiveness in mathematics. The fact that pi is irrational does not mean that we can not state formulas involving pi which are completely exact and certain under the axioms of mathematics. Pi is a very precisely defined quantity. The fact that we graph things a certain way is an arbitrary convention, and does not produce subjective truths.

In the physical sciences, different scientists may hold different opinions as to what they believe to be true. It is never really possible to prove that a theory is true in the physical sciences. The best we can do is mount experimental evidence for a particular theory so that the theory becomes verified to a greater degree. It IS possible to DISprove a theory with a single experimental result however, so long as the experiment is reproducible by other scientists, and so long as there is no dispute about the accuracy of the experiment.

06-03-2003, 01:28 AM
Check this out.
monty hall problem (http://cartalk.cars.com/About/Monty/)

Al Mirpuri
06-03-2003, 06:30 AM
Since I replied to your post, I have gone back to the source of my original knowledge of what I termed the 'Monty Hall Paradox'. Having done so, I now hold that 'Monty Hall' is a problem and not a paradox and that correct reasoning leads to a switching of choices everytime.

Al Mirpuri
06-03-2003, 06:37 AM
Since I made my original post, I have gone back to the source of my original knowledge of what I termed the 'Monty Hall Paradox'. Having done so, I now hold that 'Monty Hall' is a problem and not a paradox and that correct reasoning leads to a switching of choices everytime.

I do, however, stand by the remarks I made in Philosophy &amp; Mathematics.

Louie Landale
06-04-2003, 05:56 PM
If he just says that one of the two remaining are incorrect you gain no information because you already knew that at least one of the remaining two were incorrect. But lets assume that he tells you which one is incorrect.

Yes, this is the classic Monte Hall problem.

I didn't read the other threads, but the "solution" to the Monte Hall problem resides in Monte's motivation. If he ALWAYS makes the offer then yes you should switch for a 67% chance. If he ALWAYS wants you to win then you should switch for 100% chance. If he ALWAYS wants you to lose you should never switch for 100% chance. If you have no idea about Monte's motivations then it doesn't matter whether you switch or not for a 50% chance.

Actually, you should assume Monte wants you to win if the previous contestants lost, and should assume Monte wants you to lose if previous contestants won, but that's another issue... Well, actually you should mumble "switch" as you point to YOUR door, then try to win the subsequent law suit... /forums/images/icons/smile.gif

So you both are correct. Yes, its the classic Monte Hall problem; AND the worlds smartest woman is also correct since you have no idea about the motives of the instructor. AND you are correct, she is probably not as smart as she thinks ... but who is.

- Louie

PS. Someone who claims to be the worlds best whatever is probably really good at whatever.

06-06-2003, 02:15 PM
With regard to the EPR paradox and action at a distance, I think (as a layman, but having read a lot) that you are misstating the "speed of light" limitation. What general relativity says is that nothing can ACCELERATE beyond the speed of light (that is start from less than SOL and wind up traveling faster than SOL). In fact there is nothing in General Relativity that prevents something from traveling beyond the SOL if it has always been traveling that fast. Thus action at a distance is not a paradox, but a consequence of the fact that the "information" passing from one particle to the next has always traveled faster than the speed of light, which, of course is not observable.

The Monty Hall problem and the effect of knowledge might be more easily understood in a simpler situation, known to bridge players as "the law of restricted choice" when deciding whether or not to finesse for a card.

You are missing the Q and J of a suit, and have no knowledge of the location of either of them. After the J is played, should you finesse for the Q?

With no knowledge about who played the J (you turned to look at Gus Hansen play 32o on WPT when it was played), the answer is that its a pure guess...either one could have it. But what if you know the player on your left played the J, and if he had both QJ he is equally likely to play either one? Then the answer is to assume the player on your right has the Q and not finesse. Why? The original distributions were QJ/nothing, Q/J, J/Q, or nothing/QJ, with equal probability and not knowing who played the J, all of those are possible and teh Q could be on either side.

But what if you know that left played the J? The original distribution was clearly not nothing/QJ or Q/J..left couldnt have played the J. So the original distribution was QJ/nothing or J/Q. If it was J/Q, left was "restricted" to playing the J every time. If it was QJ/nothing, then left had an equal choice to play Q or J, and only half the time would he have played the J. Thus its twice as likely that right has the Q (the half of the time it was J/Q versus the 1/4 of the time that it was QJ/nothing AND he chose to play J).

06-06-2003, 05:19 PM
With regard to the EPR paradox and action at a distance, I think (as a layman, but having read a lot) that you are misstating the "speed of light" limitation. What general relativity says is that nothing can ACCELERATE beyond the speed of light (that is start from less than SOL and wind up traveling faster than SOL). In fact there is nothing in General Relativity that prevents something from traveling beyond the SOL if it has always been traveling that fast. Thus action at a distance is not a paradox, but a consequence of the fact that the "information" passing from one particle to the next has always traveled faster than the speed of light, which, of course is not observable.

With all due respect, if that was all that was required for an explanation, do you think that Einstein or I would have put this forth as a paradox? /forums/images/icons/smile.gif

It is true that relativity does not prohibit a particle from traveling faster than the speed of light if it has always been traveling faster than the speed of light (tachyons). This does not resolve the paradox of EPR however. In EPR the information which is apparently being instantaneously communicated is created by us, and the two communicating particles are also created by us. This information would necessarily need to accelerate not only to the speed of light, but to infinite speed in order to be conveyed instantaneously from one particle to the other. This is precluded by relativity, which is why Einstein put it forth as a paradox.

In fact, no information has ever been conveyed faster than the speed of light using EPR. It is possible to transmit messages by EPR, and there are some advantages to doing so, but this is always done at sub-light speeds, because the information which was apparently coupled instantaneously can only be verified after the fact via a conventional communication channel which operates at sub-light speeds. If anyone ever succeeds in sending a message faster than the speed of light by EPR or any other method, then the theory of relativity will indeed have been violated. We cannot do this, but somehow the particles can.

While we cannot observe that EPR depends on a violation of relativity, no verified mechanism exists which explains EPR consistent with relativity, and that is why it is a paradox. There have been theories involving a channel in which information flows backwards through time or where cause and effect are reversed. There is also an interpretation of quantum mechanics called the "many worlds view" in which all possibilities occur simultaneously in parallel universes, and this theory shows promise. None of these are established theories. They all require verification, and some theory is needed to resolve the paradox of how relativity is not violated in EPR.

Al Mirpuri
06-27-2003, 07:26 AM
The thought has occurred to me that using Bayes' Theorem you get the answer that both of the remaining doors (out of the initial three) have an equal 1/3 chance (even tho' there are only two of them at this point, the point being that it is an equal chance) of being the door behind which the prize lies. The reasoning is thus: the picked door was originally 1/3, whilst the unpicked door is 2/3 x 1/2 = 2/6 = 1/3. This equation being the likelihood of the prize being behind the two doors but not behind any particular one door - the one that is revealed not to have the prize behind it. Yes, I am aware that the door revealed not to have the prize behind it is not chosen at random.

What does anyone think? I am especially interested in what BruceZ has to say about this.

06-28-2003, 12:04 AM
"The thought has occurred to me that using Bayes' Theorem you get the answer that both of the remaining doors (out of the initial three) have an equal 1/3 chance (even tho' there are only two of them at this point, the point being that it is an equal chance)"

Well, 1/3 + 1/3 = 2/3. What event is supposed to occur at the remaining 1/3 chance? /forums/images/icons/smile.gif Either you can claim that each of the 2 remaining doors has a 50% chance -or- that each door has a 1/3rd of a chance, in which case we have to specify what happens at the ummm third 1/3rd. The probabilities must sum to 1.0 or 100%.

Since it cannot be (1/3)+(1/3) , it must be either (1/2)+(1/2) or something else, such as (1/3)+(2/3).

Using Bayes' Theorem:

The Theorem uses a posteriori knowledge to adjust a priori odds.

Out of 3 doors, A, B and C, you pick A. Monty opens door B and reveals a goat.
Do you stick with A or switch to C?

The a priori probability that the prize is behind any one of the 3 doors, A, B, or C, is
P(X) = 1/3

The probability that Monty Hall would open door B if the prize were behind A, is
P(Monty opens B|A) = 1/2

The probability that Monty Hall would open door B if the prize were behind B, is
P(Monty opens B|B) = 0
and that's because Monty Hall will ALWAYS open a door with a goat (that's a basic premise of the Monty Hall parad--- umm problem).

The probability that Monty Hall would open door B if the prize were behind C, is
P(Monty opens B|C) = 1
for the same reason as above.

The probability that Month Hall would open door B is then

P(Monty opens B) =
[P(A)*P(M.o. B|A)] + [P(B)*P(M.o. B|B)] + [P(C)*P(M.o. B|C)] =
1/6 + 0 + 1/3 =

Then, by Bayes' Theorem,

P(probability of prize being behind A |if Monty opens B)=
P(A|Monty opens B) =
P(A) * P(M.o. B|A) / P(M.o. B) =
(1/6)/(1/2) =


P(probability of prize being behind C |if Monty opens B)=
P(C|Monty opens B) =
P(C) * P(M.o. B|C) / P(M.o. B) =
(1/3)/(1/2) =

I think this has been posted on this forum before. In any case, it has been posted countless times on the web.

06-28-2003, 05:46 PM
The probabilities of the winner being behind each of the unopened doors cannot both be 1/3 since these two probabilities must sum to 1. This is because the winner must be behind one of these doors since it cannot be behind the door Monty picked. Here is the correct solution by Bayes' theorem:

Bayes' theorem states that for any events A and B with P(B) non-zero:

P(A | B) = P(AB) / P(B) = P(B | A)*P(A) / P(B)

Where P(A | B) means the probability that A occurs given that B occurs, P(B | A) means the probability that B occurs given that A occurs, and P(AB) means the probability that both A and B occur = P(B | A)*P(A). This can be understood by noting that P(A | B) is the fraction of the time that both A and B occur out of those times that B occurs. Bayes' rule or "theorem" is really just a restatement of the definition of conditional probability, and I have simplified the denominator here from the full-blown statement of Bayes' theorem for convenience. The present form is always correct.

Assume we pick door A and Monty opens door B. Then by Bayes' theorem:

P(A is winner | Monty opened B) = P(Monty opened B | A is winner)*P(A is winner) / P(Monty opened B)

All of these probabilities are computed under the condition that we have picked door A. When door A is the winner, we assume Monty will choose betwen the remaining two incorrect doors equally (which is why it is important that this assumption be stated in the problem), so he will open door B half of the time. So

P(Monty opens B | A is winner) = 1/2.

Since door A is the winner 1/3 of the time, we have

P(A) = 1/3.

Monty will open door B half of the times that A is the winner or (1/2)*(1/3) of the time, but in addition Monty will also open door B half of the time that A is not the winner, since half the time that A is not the winner, B will not be the winner, and Monty must open B, while the other half of the time B will be the winner and Monty must open C. So

P(Monty opens door B) = (1/3)*(1/2) + (2/3)*(1/2) = 1/2.

Putting it all together we have:

P(A is winner | Monty opens B) = (1/2)*(1/3)/(1/2) = 1/3.

So when Monty opens door B, the probability that the winner is behind our choice A is 1/3. It follows then that the probability must be 2/3 that the winner is behind C. We can also compute this directly from Bayes' theorem the same way, again remembering that we picked A. Note that when C is the winner, Monty MUST open B, so

P(C is winner | Monte opens B) = P(Monty opens B | C is winner)*P(C is winner) / P(Monty opens B)

= 1*(1/3) / (1/2) = 2/3.

07-01-2003, 02:14 PM
Paradox has been strewed to mean "something that isn't obvious". Whether that is the true meaning, we have debated.

Simple explanation of the "paradox":

If you were correct in choosing which door or penny, and you switch, you lose.
If you were wrong and you switch you win.

Basically, you switch the odds. It may seem illogical, but the math is there.

07-01-2003, 04:10 PM
Much Philosophy is built upon the foundation of discrete math (ie...proving stuff is either true or false).

The rest of philosophy is worthless musing.

A philosophy buff who insults mathematics insults the only thing philosophy has going for it.

07-01-2003, 07:12 PM
"Much Philosophy is built upon the foundation of discrete math, i.e. proving stuff is either true or false."

I am truly sorry that I have to say this but... Philosophy is not an exercise in binary language. It doesn't work that way.

"The rest of philosophy is worthless musing."

You are kindly requested to elaborate a little on what would make Philosophy worthwhile, for you.

07-03-2003, 03:58 PM
Im not sure i understand the whole 'switching' thing. So the way i explained the whole process was like this: because there is an unknown on all of the doors. and monty take away one, leaving still two unknowns. However, one of the other 2 doors is wrong as well, meaning it doesnt matter which of the 2 wrong doors he picked, you are being left with one right and one wrong.
With that said, let us for a moment pretend that a 3 sided solid object exists. On each side is a letter, A, B, C. Respectively, if we toss this figure it will land face down on A B or C 1/3 of the time each, in terms of probability. So now let magic Monty remove on of these sides from our 3 sided solid object leaving us a 2 sided solid object. (Let us also pretend we are betting on which letter will land face down and so monty tosses the 3-sided figure into the air and its supposed to land on the side we want) However, the object, as it is in the air, becomes a 2 sided solid object leaving now only 2 choices. My odds went from 1/3 to 1/2 because he picks to remove a side that I did not bet on and for some reason he knows (although i dont think this matters in the case of the door choosing) it wouldnt land on. Maybe a better example is roulette table with 3 numbers. I place my bet on 1, while the thing is spinning a block of wood falls into the third number slot so the ball cannot land in it leaving 2 places for the ball to land in, 1 and 2. Why should I switch to 2? I would have the same odds choosing either. I think the same principal can be applied to a test with a "right answer" or a door with a monster or woman or the choice of transportation between a car and a donkey.
Im open to being totally wrong here though if the probability between a coin toss does not equate to the probablity where there is a "right answer" pre-determined. Although in a sense, once the coin falls there will always have been a right choice and a wrong one, am i correct?
The other possibility here is that this topic is exhausted and nobody wants to talk about it anymore /forums/images/icons/smile.gif. Hope this helps someone out.

07-03-2003, 09:14 PM
djptolemy wrote: Im not sure i understand the whole 'switching' thing. So the way i explained the whole process was like this: because there is an unknown on all of the doors. and monty take away one, leaving still two unknowns. However, one of the other 2 doors is wrong as well, meaning it doesnt matter which of the 2 wrong doors he picked, you are being left with one right and one wrong.

The point is, each choice initially has 1/3 probability, right?

1/3 + 1/3 + 1/3 = 1

If one door is randomly removed each remaining door now has 1/2 probability.

(1/3 - 1/3) + (1/3 + 1/2(1/3)) + (1/3 + 1/2(1/3)) = 0 + 1/2 + 1/2 = 1/2 + 1/2 = 1 , right?

But, if he is only allowed to choose a wrong answer THAT YOU DID NOT PICK, that changes things quite a bit, you must look at it as two different sets, A= what you chose, always worth 1/3, Monty cannot expose your choice, and B = the 2 remaining choices, whereas

A + B = 1, and A = (1/3) and B = (1/3 + 1/3) = (2/3)

Monty must choose only from B, B could be goat-goat, goat-car, or car-goat, therefore assuming no extra information, B must always be worth 2/3, and A must always be worth 1/3. Because HE MUST ALWAYS CHOOSE FROM B, NOT A, A will always be worth 1/3 when an incorrect answer is shown in set B, and therefore the incorrect answer exposed in set B becomes 1/3 - 1/3, and the remaining unexposed answer in set B becomes 1/3 + 1/3. Removing a choice from set B cannot change the probability of your choice (set A), because he could not choose the one you pick.

The first set in parentheses is A, the second is B

(1/3) + (1/3 + 1/3) = 1
(1/3) + ((1/3 - 1/3) + (1/3 + 1/3)) = 1
(1/3) + (0 + 2/3) = 1
(1/3) + (2/3) = 1
A = 1/3, B = 2/3, you must switch.

One choice did not mysteriously disappear, it was taken only from the two choices that you did not pick, and could not have been taken from your pick, your pick is therefore always worth 1/3.

Given the rules of the problem, there is nothing that takes place here to increase the probability of your original guess from 1/3. A and B MUST ALWAYS BE LOOKED AT AS SEPARATE SETS if the wrong answer can only be shown from set B. If a wrong answer can be exposed from all 3 answers then the previous sentence is no longer true.


07-03-2003, 11:50 PM
Look at it this way:

There is a 1 in 3 chance of it being behind any of the doors (A, B, C). So there are 3 possible ways the doors could be organized (Y=yes the prize is here, N=no it is not)

<pre><font class="small">code:</font><hr>
Door A Door B Door C
N N Y</pre><hr>

Each of these three scenarios is equally likely.

Let's say you always choose door A. So 1/3 of the time you are right, but the other 2/3 you are wrong. If you always switch, you'll be wrong 1/3 of the time, but right 2/3 of the time.

If you randomly pick a door, you will be right 1/3 of the time, and wrong 2/3. Again, switch doors and you'll be right 2/3 of the time.

Last way to look at it. We're going to play the game 6 times. We'll keep the setup the same each time - Door A is the winner (but you don't know that). You'll pick each door 2 times.

<pre><font class="small">code:</font><hr>

Try # You pick Monty eliminates You stay (W/L)
1 A B W
2 A C W
3 B C L
4 B C L
5 C B L
6 C B L

Again, you win only 1/3 of the time if you stay, but 2/3 if you switch.

07-04-2003, 11:17 AM
A dismissive comment elsewhere about the worth of Philosophy itself, got me thinking about your enquiry, ie the relation between Philosophy and Mathematics. Others may disagree but personally, I find it impossible to overrate the effect that Goedel's Theorem had on Philosophy (and not just Philosophy of Mathematics but the pure thing proper.) The realization (which amusingly enough came by way of math'matical proof) that a full understanding of the human condition and getting the answers that philosohphers have posed might be impossible, is a monumental event. So, forget trivialities like the Monty Hall, here's the real thing.

I'm not suggesting that this was a novel thought in early 20th century or that this had not been posited before. But Goedel proved it "on the blackboard" and not just by logic. A rarity for Philosophy.


PS : BruceZ had a couple of posts up about Goedel somewhere. The latest one chronologically is the one to read. But I believe that a web search would provide anyone so inclinded with a wealth of information about Goedel's work. Have your anti-depressant handy.

07-04-2003, 11:29 AM
I agree with Marilyn's answer. The reason is you are writing a test, and you should have studied for the exam, and your choice of selecting the answer is based on the information which you studied before taking the test. Therefore, when the professor tells you which one is the wrong answer, it won't matter because, if you know what is the right answer, based on your hard studying before taking the test, you won't switch to the wrong answer. The fact the professor tells you the wrong answer, it only increases your chances that your first pick was the right answer. Of course, if you are taking a test on a subject which you have absolute no knowledge, than your choice is random. Taking an exam for a grade /forums/images/icons/tongue.gif is not random /forums/images/icons/smirk.gif .

07-04-2003, 12:12 PM
You are in front of a thousand closed doors. When you'll pick one, I have promised to open 998 doors showing a goat. You choose door #74. True to my promise, I switch open 998 doors and they all show a goat behind 'em.

One door is left closed, #569. So now there are only 2 doors closed, #74 and #569. I ask you to choose again. You stick to your original choice.

The result is_____________

..OK, we do this again. The prize is moved around a different door -- maybe. Now you choose door #6. I open 998 doors and they all show a goat but door #991 I leave closed.

Again you stick to your original choice.

The result is______________

...We do this again. And again.

The question is no longer where is the prize. It's How many times do you have to go through the above to realize that you gotta be switching ev'ry time??

07-04-2003, 05:02 PM
The solution to this problem does not depend on the fact that you are taking an exam, or whether or not you have studied for it.

The difference between this problem and the Monty Hall problem is that the teacher will tell you one of the wrong answers without knowing which answer you selected. So it's possible he will tell you your answer is wrong.

Let's look at 6 examples again.

<pre><font class="small">code:</font><hr>
Problem # A B C You pick Teacher eliminates You switch
1 Y N N A B L
2 Y N N A C L
3 N Y N A A 1/2 W
4 N Y N A C W
5 N N Y A A 1/2 W
6 N N Y A B W

So you'll end up with 2 guaranteed correct, 2 guaranteed wrong, and a 1 in 2 chance at the other 2. On average you'll end up with 3 correct answers if you always switch, and 3 correct answer if you never switch (except when the teacher tells you you are wrong)

07-06-2003, 03:00 PM

07-09-2003, 03:41 PM
Wow, I totally forgot about this post.

Your scenarios, the 3 sided solid figure changing into 2, or 3 numbers on a roulette table with a block blocking one, are not analogous to Monty's situation. In your scenarios, when one of the choices is blocked, you introduce an element of chance. If Monty didn't know which door had the car, then yes, if he chose a door with a sack of potatoes, then it doesn't matter whether you switch or not, your chances went up to 50% from 33%.

However, Monty knows where the car is. That seemingly minor fact changes the scenario. Because he will always choose a door with nothing to show you is why you should switch.