The BEST and FINAL answer on the MONTY HALL PARADOX!

[KidCharlemagne]

Ok, So I’m a little behind the times, but I just finished reading through the Cecil Archive on the Monty Hall Paradox (Search to refresh your memory). There was a long debate whether or not Cecil was right, and at the end, alot of readers weren’t satisfied with the explanation. I love Cecil (in that way) but his explanation for this was pretty lame. I got the feeling even he didn’t “get it” even though he was right. Anyway, I got to thinking about it and drew a diagram below that makes it absolutely clear to anyone (I think). Tell me if you agree .
Ok, In the following example, I start with the prize behind door number three (marked by a “!” after the X). It doesn’t matter which door it starts behind so I just picked any. Then, on the y axis, I listed the three possible situations after you have made your inital choice (since you have the option of three doors). Now, knowing that Monty will not open the door with the prize, and will not open the door you have chosen, watch how this new info changes the odds: (Unfortunately after having previewed this post, I saw that it didn’t format the table correctly, just try and see it as a 3X3 grid with Door# on X axis and Possible Situations on the Y.)
Door1 Door2 Door3
Poss1 X* O X!

Poss2 O X* X!

Poss3 O O X!*

In Poss1 you choose door number 1 (X*). Monty only has the option of opening door 2 (O) (because you have chosen door1 and the prize is behind door3 (X!). In this situation, switching will DEFINATELY win you the prize.
In Poss2, you choose door2 (X*), Monty MUST open door1 (O), and once again switching DEFINATELY wins you the prize.
In Poss3, Monty can open either door1 or door2 and switching will Always make you lose.
If, after the new information, you switch, you will win 2 in the 3 situations. Hence the odds of winning = 2/3.
Tada! Ok someone give me kudos for this cause I crave validation. Now for the icing on the cake (granted this isn’t quite as accessible as the proof, but you’ll get it).
Why does this intuitively FEEL wrong? The reason you think the odds on the second choice are 50/50 is because it appears that you have 2 choices (stay or switch) and Monty has 4 possible doors he can open (Shown by the 4 O’s) in any given situation. The difference is that 2 of monty’s available door choices are available in ONE of the possible situations, that being Poss3 on the table. Since in each possibilty you can only be wrong once, Monty’s two options in Poss3 doesn’t make for 2 potential losses for you, only one. Most people look at it from the perspective of how many doors can Monty choose from versus how those choices affect YOUR odds.

.Having said that, I haven’t figured out something about the two children question (refer to same article). While the odds statistically do say there is a 2/3 chance of the next child being male, clearly, (as far as I know), betting on this in real life will be a losing bet. In the Monty Hall bet you will actually win money (Yes play the game with friends and play the contestant). I can’t think anymore after doing that stuff above so will someone explain how the children paradox works statistically but doesn’t seem to work in real real life?

[Chronos]

On “Let’s Make a Deal,” you pick Door #1. Monty opens Door #2–no prize. Do you stay with Door #1 or switch to #3?

Well, KidCharlemagne, the good news is that you’re right. The bad news is that it doesn’t matter. Studies have conclusively shown that it is scientifically impossible to convince anyone of the answer to the Monty Hall problem.

The reason you’re still puzzled about the “other child” problem is that it’s not well-posed. The answer can be either 1/3 or 1/2, depending on how you got your information (which isn’t usually specified in the problem). For instance: A major newspaper columnist surveys her readers. She tells them that any person who has two children, at least one of whom is a boy, should write to her and tell her the gender of the other child. Any random respondant will have two children, one of whom is a boy. As predicted, two thirds of her respondants said that the other child was a girl, and one third said that the other child was a boy.

On the other hand: You meet a woman on the subway, and she’s got a boy with her who’s obviously her son. She mentions that she has two kids, but doesn’t say anything about the other one. You now know that she has two children, at least one of whom is a boy. If you want to take a guess at the gender of the second child, there’s a fifty-fifty chance, either way.

By the way, let me just point out that it’s very poor etiquette to expect everyone who reads the thread to search for the column on their own, rather than just providing the link yourself, especially considering that you were just reading the column, anyway, and didn’t even need to search.

[KidCharlemagne]

Can I request a username chg to Fauxpas? First I posted in 3 forums then didnt give people a link (though i don’t know how cause im the guy those “For Dummies” guides are written for. As you might be able to tell by the grammar of that previous sentence.

[zut]

*Originally posted by Chronos *
The reason you’re still puzzled about the “other child” problem is that it’s not well-posed. The answer can be either 1/3 or 1/2, depending on how you got your information (which isn’t usually specified in the problem). [pertinent examples deleted for brevity]

Hum. Well, I suppose it depends on what your definition of “well-posed” is. The important difference between the examples that you give is whether or not there is any way to differentiate between the two children. In the first example (newspaper columnist), there is no differentiation, and thus the probability split is 1/3-2/3. In the second example (subway), the childeren can be differentiated as “the one on the subway” and “the other one”, and thus the probability split is 1/2-1/2.

Now, in Cecil’s original column, the problem is posed thusly:

There is a family with two children. You have been told this family has a daughter. What are the odds they also have a son, assuming the biological odds of having a male or female child are equal? (Answer: 2/3.)

I would argue that in this expression of the problem, there is specifically no differentiation, and thus the problem is well-posed. Of course, this fact could be emphasized in the formulation (“You have been told this family has a daughter, and that is all you know about her,” or some such), but nonetheless it is there already. I think it’s more the inherent non-intuitiveness of the problem, and the sometimes subtle nature of the differentiation, (and, admittedly, the sometimes sloppy wording that actually changes the intended nature of the problem) that makes it confusing.

[foolsguinea]

Don’t change your name, KidCharlemagne, your name is much too cool. Although we may call you “KidFauxPas” for short.

Hmmm…
Any boy has a 1/2 probability for each of his siblings of that sibling being male.
Any mother has a 1/2 probability for each of her children that that child be male.
Any woman with 2 children has a 1/4 probability of both being male, and a 1/2 probability of them being different sexes.
So far, all this is consistent.

If Mrs. Smith has a son Billy, and she has another child, that child has a 1/2 (or 1:1) probability of being male.

Of Mrs. Smith’s two children (hereafter “A” & “B”), each has a 1/2 (1:1) probability of being male.
Where A is male, 1/2 probab. of B being male.
Where B is male, 1/2 probab. of A being male.
Given total actual incidence of all cases where only A is male (a distribution we will call possA), B is male (possB), or both are male (possM), we find that possA: possB: possM = 1:1:1 .
But the above is a deliberately incomplete sample. The actual set of probabilities includes “possF” where both children are female. possA: possB: possM: possF = 1:1:1:1 .

Given a probabilistically ideal set of families deliberately chosen to exclude possF, only 1/3 of those families will be possM. That is, possM: possA+possB = 1:2 .

But if I meet a woman accompanied by one child, & disregard any case where the accompanying child is female or the woman does not have exactly one other child, then I find she has a 1:1 (1/2) probability of having another son.
(I have excluded, by my ignorance, the cases where I see the girl but there is a boy.)
In any case where I only know the sex of one of her two children, and then only randomly, this is the probability.

But say I am headmaster of a boy’s school. For all the boys in my school that have exactly one sibling, each boy has a 1/2 (1:1) probability of having a brother.
However, I have no possF families among my students; and if I count the number of student’s mothers to determine what proportion of these women have two sons, this number will be modified, as a woman may have both of her sons at my school.
The 1/2 ratio can only be maintained if I count each woman once for each son she has at my school.
Counting each woman exactly once, I find that between 1/2 and 1/3 have two boys. Given the nature of my sample, the actual probability is between these two ratios. Some women with two sons will not have both in my school.
So in this real-world scenario, the number is between 1/2 and 1/3, not exactly either.

If I have no particular reason to know more mothers of boys or more mothers of girls, then of women I know, the 1:2:1 ratio would (ideally) apply (or 1/4 + 1/2 + 1/4).

But, still, in this case, of boys I know with one sibling, 1/2 would (ideally) have a brother.

So it depends on your sampling process.
A sibling’s probable sex is always a 1:1 probability.
But when you come from the parent, or family unit as a whole, the real underlying probability ratio is 1:2:1. This may be modified to 1:1, 1:2, or an indeterminate number in between, depending on what the actual question/situation is.

[foolsguinea]

All of which is different from the controlled situation in the Monty Hall problem.

Thank you, and why do I bother? It’s still clear as mud.

Hmmm… OK, the answer is NO!

[Chronos]

Quoth zut:

Now, in Cecil’s original column, the problem is posed thusly:

There is a family with two children. You have been told this family has a daughter. What are the odds they also have a son, assuming the biological odds of having a male or female child are equal? (Answer: 2/3.)

That’s still not well-posed, though, unless we know something about why someone told you that. Did you ask your mutual friend “Does that family have a daughter?”, and he (knowing all about the family) said “yes”? In that case, it’s the one third/two thirds situation. On the other hand, maybe you asked your friend to tell you everything he knew about that family, and he (having knowledge of only one family member) said “They have a daughter”. In this case, it’s fifty-fifty.

[sailor]

I’ve said it before and I’ll say it again: This has been beaten to death. Last time I said it I was wrong and Chronos was right. Here is the definitive answer.

[december]

*Originally posted by KidCharlemagne *
**
Why does this intuitively FEEL wrong? The reason you think the odds on the second choice are 50/50 is because it appears that you have 2 choices (stay or switch) and Monty has 4 possible doors he can open… **

Good analysis, Kid.

Another reason that th solution feels intutively wrong is that it requires you to make a questionable assumption. In order for the “official” solution to be correct, you must know for sure that Monty Hall alwaysopens a door without a prize behind it and offers a choice. (BTW this assumption was not even stated in the original problem, as written by Marilyn.) However, if you believe that Monty might or might not open a door, then the problem becomes more difficult, because your answer depends on what you think Monty’s strategy is.

E.g., if you think Monty is trying to save prize money, then maybe he would open a door and offer you a chance to switch only when you were right. Then, you’d be 100% sure of winning be NOT switching.

[zut]

*Originally posted by Chronos *
[Cecil’s formulation is] still not well-posed, though, unless we know something about why someone told you that. Did you ask your mutual friend “Does that family have a daughter?”, and he (knowing all about the family) said “yes”? In that case, it’s the one third/two thirds situation. On the other hand, maybe you asked your friend to tell you everything he knew about that family, and he (having knowledge of only one family member) said “They have a daughter”. In this case, it’s fifty-fifty.

To be honest, I hadn’t thought of it in quite that way. However, I still argue that the question is well-posed as written, strictly because it is written as a probability question.

Allow me, by way of explanation, to tangentialize for a moment. Suppose I pose this probability question: “If I flip a fair coin exactly twice, what is the chance that I’ll get two heads?”
“Easy,” you say, “one in four.”
“Well, not quite,” I say. “It just so happens that I already flipped the coin the first time, and it came up heads, so actually the probability is one in two.”
“Ridiculous!” you scoff. “You can’t leave out information from the original question.”

And of course you’d be right. My point, of course, is that it’s understood that all pertinent information is included in the original statement. The additional information (that I’ve already completed the first flip) changes the nature of the problem. In the same way, the additional information in your daughter problem example (that my friend knows only of one family member) changes the nature of the problem.

In the coin flip example, I shouldn’t have to explicitly state that I’ve not flipped any coins yet (although that would further clarify the problem). In the daughter problem, I shouldn’t have to explicitly state that I have no further information about the family (although that would further clarify the problem).

I have a feeling, though, that in some sense I’m arguing semantics. I agree that the daughter problem, as usually stated, is unclear, and that the inexactitude of language, coupled with the non-intuitiveness of the problem, can be confusing. However, the daughter problem, as stated, leads to an answer of 1/3-2/3, and the answer can be 50-50 only if you introduce additional information, which makes it a different problem.

[Wumpus]

december – The original Monty Hall problem predates Maralyn’s column on it. It was originally formulated by statisticians to demonstrate Bayesian probabilities in terms that the average college student could grasp. (Martin Gardner evidentally published a version of it in Scientific American way back in 1959.)

Anyway, in the original problem, it was indeed stated that Monty must open another door, and he never opened the door you picked or the winning door.

(This, of course, was the way the actual game show worked. There was a time when Everyone Knew That, just like Everyone Knew who JR and Bobby were, or how to do the Hustle.)

BTW, Maralyn got this one right back in 1991. See:

http://www.dartmouth.edu/~chance/course/topics/Monty_Hall.html

[jharding]

I’ve had five semesters of graduate level statistics courses and both the examples (Monty Hall & children) continue to cause a thick fuzz to grow around my brain. Probably the most valuable nugget of knowledge I’ve taken away from all those classes is that statistics are at least as much about precise language as they are about mathematics.

Forget about all your nifty equations, if the problem isn’t stated explicitly different answers may follow.

[KidCharlemagne]

Foolsguinea, I was just curious if you liked the name because it’s generally melodious or if you knew it was the name of a Steely Dan song. (with the greatest guitar solo possibly ever recorded)

[december]

*Originally posted by Wumpus *
**Anyway, in the original problem, it was indeed stated that Monty must open another door, and he never opened the door you picked or the winning door.

(This, of course, was the way the actual game show worked. There was a time when Everyone Knew That, just like Everyone Knew who JR and Bobby were, or how to do the Hustle.) **

I agree with the 1st paragraph above. Howver, the 2nd paragraph is contradicted by a quote from a site called “Marilyn is Right”

There has been some discussion about whether or not Monty always offered the switch and if his doing so depended on whether the contestant chose the right door to begin with. Monty said that he did not always offer the switch. Sometimes he would offer it to lure the contestant away from the right door, and to keep the audience off balance sometimes he would bluff and offer the switch when the contestant had chosen wrongly to see if he could make him be stubborn.

see: http://www.icdc.com/~samba/marlright/monty.htm I never watched the show myself, so I cannot confirm the accuracy of this quote.

[jshore]

KidCharlemagne:

Well, everyone has there own “best” way to see the result of the Monty Hall problem. I like yours. Another one I like even better is to say that your original chance of picking the right door is 1/3; since MH always can and will open one of the other two doors to reveal no prize, this action cannot give you any new info about what it behind your door. Ergo, the probability of the prize behind your door is still 1/3 and hence it must be 2/3 behind the other door. [Another way to see this is to realize that MH has really just given you a choice between sticking with the 1 door you originally chose (hence 1/3 prob of winning) or winning the prize if it is behind EITHER of the other two doors (hence, 2/3 prob or winning).]

This analysis also makes clear the distinction that occurs if the problem is stated differently: E.g., that MH opens one of the other two doors completely at random (w/o regard to whether or not the prize is behind it). In such a case (and assuming that you are lucky enough to be the contestant when the door he opens does not reveal the prize), then the odds really are 50-50. How can the odds of the prize being behind your original door switch in this case? The answer is that you have gained new information in this case because in some (1/3) of the cases when MH does this, he will open a door that reveals the prize. The odds work thusly:

1/3 = (2/3)*(1/2) + (1/3)*0

The 1/3 on the lefthand side is the odds of you picking the right door initially. The 2/3 represents the fraction of the time MH opens a door and does not reveal the prize, the 1/2 represents the odds that your door is the correct one in these cases, the following 1/3 represents the fraction of time MH opens a door and does reveal the prize, and the 0 represents the odds that your door is the correct one in these cases.

[december]

*Originally posted by Wumpus *
**Anyway, in the original problem, it was indeed stated that Monty must open another door, and he never opened the door you picked or the winning door.

(This, of course, was the way the actual game show worked. There was a time when Everyone Knew That, just like Everyone Knew who JR and Bobby were, or how to do the Hustle.) **

Wumpus I’ve been told that the show did indeed work just the way you said, and my disagreeing source was wrong.

There’s an analogy to the two children question in the game of bridge, where it’s referred to as “restricted choice.” In this game, your tow opponents were each dealt 13 cards. At a certain point in the play, you [declarer] discover that one of the opponents was dealt the Jack of a Spades. You know that the Queen of Spades is held by one of the opponents/ What’t the probability that the player with the Jack of Spades also holds the Queen?

(You should also be aware that if a player holds both the Jack and Queen of a suit, there’s no advantage to playing one rather than the other.)

I’ll withhold the answer to give you smarties a chance to analyze the problem.

[Chronos]

Near as I can figure, december, it’s 48%, provided that all you know about your opponents’ hands is that one of them has the JS and one has the QS, and both are still holding 13 cards each. Am I right?

Oh, yeah… This also assumes that your opponents didn’t choose somehow to let you know they had those cards. I don’t know enough about how to play bridge to know if this would likely be a factor.

[december]

My restricted choice example was incomplete. Let me expand it.

You have the Ace and King of spades. At a certain point, an opponent plays the Jack of Spades in a situation where he’d have played a low Spade if he had one. So, you know he must have held either the Jack of Spades and no other Spades or the Jack and Queen of Spades and no other Spades. If he held the Jack alone, he had no choice but to play it. If he held the Jack and Queen of Spades, he might have played either of them with 50-50 odds. Given that he played the Jack of Spades, what are the odds that he also has the Queen of Spades?

[Chronos]

You stated earlier that we knew that one opponent held the QS… Do we, in fact, know that? And how many cards are left in everyone’s hand at the time this situation comes up? The other players’ hands are significant here, because there’s only one QS in the deck, and someone has to be holding it if it hasn’t already been played (which I think we can assume, else we wouldn’t be asking the question at all), and if one player has more cards than another, that player will have a higher chance of holding any given card.

Yes, I know that the whole point of a probability problem is that we have incomplete information, but you need some amount of information, at least. Otherwise, you get problems like “I hand you some cards. What’s the chance that one of them is the two of diamonds?”.

[december]

*Originally posted by Chronos *
You stated earlier that we knew that one opponent held the QS… Do we, in fact, know that?

Yes, there are two opponents and one of them holds the QS.
Yes, I know that the whole point of a probability problem is that we have incomplete information, but you need some amount of information, at least.

This complaint is somewhat valid. However, to deal with the essence of the problem, change it to a multiple chioce question. The three choices are:**
[]approximtely 1/3 []approximtely 1/2 approximtely 2/3 **