Descret statistics and experiment design.

Monty Hall problem keeps popping back up recently, most of them with a catchy headline “even geniuses/ Ph.D. got it wrong”.

Well since it was originated from one of the Berkeley profs, where I study now. I might jump in to support the intellectual exercise.

Monty Hall problem keeps popping back up recently, most of them with a catchy headline “even geniuses/ Ph.D. got it wrong”.

Well since it was originated from one of the Berkeley profs, where I study now. I might jump in to support the intellectual exercise.

Some people made their names by sounding as if she is smarter than the smartest Mathematicians by claiming "they were all wrong" (i.e. here: The Time Everyone “Corrected” the World's Smartest Woman and here http://marilynvossavant.com/game-show-problem/)

As much as a catchy title, the real math is much more interesting... The truth is to claim to be "the" smartest, it needs a lot of assumptions.

Here is why: (according to Wikipedia)

Here is why: (according to Wikipedia)

"The behavior of the host is key to the 2/3 solution. Ambiguities in the "Parade" version do

**not explicitly**define the protocol of the host. However, Marilyn vos Savant's**solution**(vos Savant 1990a) printed alongside Whitaker's question**implies**"**So she implied something in her solution that was not explicitly mentioned in the text. While those implied assumptions are given support, her answer would be correct.**However, it is also fair for people to argue, if those assumptions does not hold, other answers might be correct.

Ok, according to Wikipedia, Marilyn later corrected (clarified) her question in a followup post. However, that also means "most people" who disagreed with her answer, weren't wrong, they were merely given a vague question that only got clarified after they had provided their solutions. Her "implied" original posted question which is

**still**available to see on her website:
Suppose you’re on a game show, and you’re given the choice of three doors. Behind one door is a car, behind the others, goats. You pick a door, say #1, and the host,

**who knows what’s behind the doors,****opens another door, say #3, which has a goat.**He says to you, "Do you want to pick door #2?" Is it to your advantage to switch your choice of doors?

*Craig F. Whitaker*

Columbia, MarylandColumbia, Maryland

For this particular question text, take it literally without assuming information out of the text, the answer should be "I don't know" like most Mathematicians PhD argued. (see host behavior section of the Wikipedia page)

But, Marilyn's solution would be correct if she spells out "The HOST will NEVER open a door with the car"!!!

(in other words, there WAS NO chance that the door the host opens first would be the car. This is the key, you did NOT lose a 1/3 chance because the HOST will never give it to you, which means this had not been a 1 chose 3 game ever.

But, Marilyn's solution would be correct if she spells out "The HOST will NEVER open a door with the car"!!!

(in other words, there WAS NO chance that the door the host opens first would be the car. This is the key, you did NOT lose a 1/3 chance because the HOST will never give it to you, which means this had not been a 1 chose 3 game ever.

What does that mean? Why most Mathematicians got it "wrong"?

Well, there is a thing called Mathematical model ... basically, that's the process of expressing a question in mathematical equations. Here is what went wrong. Marilyn built a mathematical model that "implied" the player KNOWS the host will always open a door with Goat. The other PhDs, did not build this into their model.

That's why Paul Erdös kept asking a "Common Sense" solution -- he was able to figure it out, but unfortunately for whatever reason Andrew Vazsonyi refused to understand it, instead he decided to stick with his decision tree approach which means he stuck with his mathematical model refusing to understand if his model correctly reflects the actual question. (https://web.archive.org/web/20140413131827/http://www.decisionsciences.org/DecisionLine/Vol30/30_1/vazs30_1.pdf)

So those "smarter than smartest" people have been implying information that was not explicitly provided in the original text, and they even built computer simulations with this assumption built in!

Since Paul Erdös' commonsense solution was lost by Andrew Vazsonyi, here is my common sense explanation and why I call Marilyn's posts intellectual scam.

Here is the game:

There are 3 doors, one of them has a car behind it. you didn't know which one (BUT the host KNOWS), so you randomly chose one, you THINK you have 1/3 chance of winning the car (ACTUALLY YOU DON'T, keep reading!).

Now, the host will open a door, BUT he would *NEVER* open the door you chose!!! (That's why you never had a 1/3 chance of winning, he will NOT doors randomly, instead, he is using his knowledge to play tricks on you, hence you now have a chance to play it back at him, but this was not explicitly spelled out to begin with).

There SHOULD be the following possibilities:

Now, the host will open a door, BUT he would *NEVER* open the door you chose!!! (That's why you never had a 1/3 chance of winning, he will NOT doors randomly, instead, he is using his knowledge to play tricks on you, hence you now have a chance to play it back at him, but this was not explicitly spelled out to begin with).

There SHOULD be the following possibilities:

1) you selected the door with the car: 1/3 of the chance (but since the HOST won't open it, even though you have 1/3 chance of choosing the door with the car, you DON"T have 1/3 chance winning)

Here, knowing you are not going to win right away (he will always pick a goat door), you LIMITED

the host's choice to 2 doors, (because he will NOT open door 1 whatsoever.)

So remember, if you were correct, the HOST has 2 options to play tricks on you.

Here, knowing you are not going to win right away (he will always pick a goat door), you LIMITED

the host's choice to 2 doors, (because he will NOT open door 1 whatsoever.)

So remember, if you were correct, the HOST has 2 options to play tricks on you.

2) you selected 1 of the 2 doors with goats, you have 1/3 of the chance to be here as well, however, because your choice was a goat, AND the host will NEVER open the door with the car, you LIMITED the host's choice to ONLY ONE.

Yes, remember, here, the host HAS to open the goat that you did not select!!! He has NO other options, because he would NEVER open the car in the first round.

Yes, remember, here, the host HAS to open the goat that you did not select!!! He has NO other options, because he would NEVER open the car in the first round.

3) you selected the other door with goats, this is the same as 2), which is also 1/3 chance.

OK, you have 1/3 chance choosing a door with the car, and allow the host play tricks on you with 2 doors.

HOWEVER, you have 2/3 of the chance choosing a door with a goat, which FORCE the host to open the other goat door for you (remember, in this situation, the host actually has no options to play tricks other than open the OTHER door with goat).

Because in the cases (#2 & #3 you selected a goat door) you forced the host to open the other goat door, you know in this 2/3 of the chances, switching is a sure-fire of the car.

It's a little bit complicated for #1, because if you already chose the door with the car if you switch, you would have lost the car, so in this 1/3 of the cases, you should stay.

BUT since you don't know which you chose the first round, you really could not make the adjustment like we did here, BUT you know in 2/3 of the cases (where your chose was a goat), switch is a sure-fire of the car, but in the 1/3 of the cases (your selection was the car), staying would be better... Since you don't know, you might just switch, because it has higher chance 2/3 .vs. 1/3

OK, you have 1/3 chance choosing a door with the car, and allow the host play tricks on you with 2 doors.

HOWEVER, you have 2/3 of the chance choosing a door with a goat, which FORCE the host to open the other goat door for you (remember, in this situation, the host actually has no options to play tricks other than open the OTHER door with goat).

Because in the cases (#2 & #3 you selected a goat door) you forced the host to open the other goat door, you know in this 2/3 of the chances, switching is a sure-fire of the car.

It's a little bit complicated for #1, because if you already chose the door with the car if you switch, you would have lost the car, so in this 1/3 of the cases, you should stay.

BUT since you don't know which you chose the first round, you really could not make the adjustment like we did here, BUT you know in 2/3 of the cases (where your chose was a goat), switch is a sure-fire of the car, but in the 1/3 of the cases (your selection was the car), staying would be better... Since you don't know, you might just switch, because it has higher chance 2/3 .vs. 1/3

OK, the interesting thing comes here:

According to this text, the host opens another door shows you a goat -- was that by chance? The text didn't say! It did mention the host is aware of where the car was, however, it did not specify that the host WILL NEVER open the door with the car!!!

IF he opened the door to a goat by chance, then you just missed your 1/3 chance of winning, and were given a second chance.

IF he opened the door to a goat by chance, then you just missed your 1/3 chance of winning, and were given a second chance.

Then why Marilyn could prove she was correct? Well, she assumed the host will never open the car first round. Under that assumption, her solution (switch) is the best solution for the math model she built (here in the explanation of her solution she added "...and the host always opens a loser. ")

This is the key to her "scam", this "always" was not given in the original question (mathematically this means she is adding additional conditions and restrictions and changed the actual model).

This is the key to her "scam", this "always" was not given in the original question (mathematically this means she is adding additional conditions and restrictions and changed the actual model).

OK, now here is the commonsense description of a properly construct the question that Marilyn solved:

a) a car is randomly placed behind 1 of 3 doors, the other two hide a goat each;

b) you have zero information to begin with other than a);

c) you can choose one of the 3 doors as

**round 1**;
d) the host will then

__(he is forced by rule to do so, and he is not allowed to open a door with a car behind it, he cannot choose to skip opening a door either, and__**have**to open a door with a**goat****you KNOW**this), this is**round 2**;
e) you are then asked "switch or stay", this is

**round 3**.And here is the Commonsense Solution:

1) for round 1 above, do you agree you have 1/3 chance choosing right, 2/3 wrong? (of course, no one denies that, right?)

2) now it comes the interesting part, because of "host

**always**open a**loser**", IF your first selection was a loser, then the host is FORCED to open the other loser (so by now both you and the host are holding on to a loser door, but there are only 2 losers, so the leftover must be the winner) And it does not matter which loser you first selected, because by the rule, the host will have to eliminate the other loser anyways, so either one of the 2 losers you pick, the host had to remove the other loser and leave you with the winner.
3) so now if I ask you "how likely was your first round choice wrong?" the answer, of course, is 2/3, but if you were wrong, then the only door left has not been chosen by you nor the host would be the car, so in this 2/3 of the cases, you should switch.

4) because of the host has the option to eliminate one of two losers if you chose correctly during round 1, but this is only 1/3 of the chances which you should stay.

You ask why? I still don't get it? like mentioned before, the host was forced by rule to eliminate one of the 3 possibilities (the 1/3 chance where he opens the door showing the car), and if you know this rule, then here are the "possibilities" (let's say you select door #1)

1) car is behind door #1, the host can freely open either #2 or #3, he is not helping you - you have 1/3 chance

2) car is behind door #2, the host is FORCED by rule to open #3, so if you switch, you get the #2, winner - you have 1/3 chance;

3) car is behind door #3, the host is FORCED by rule to open #2, so if you switch, you get #3, winner - you have 1/3 chance;

See, if you stay, your only hope is it turns out to be 1), 1/3 chance; however, if it was either 2) or 3) above, the host is forced to point you to the winner by open the other loser. So you have two times of these 1/3 chance winning if you switch.

This ONLY works like Marilyn solved IF the host is forced by rule to always open a loser, and you KNOW this rule. Her mathematical model assumed this very specifically, so was her computer simulations -- because the computer model was programmed to "always open a loser", the simulation result supported her claim.

Why it's a scam? Because the original question did NOT specify this "host has to open a door with goat" which in essence change the round 3 to a different question, and most Mathematicians and PhD actually used common sense to think about it like this (without assuming the player knows a rule forces host to always open a goat): If you do NOT KNOW the host opened goat door was the rule of the game (instead, you think you were lucky that he did not open a door with car prove you were wrong at round 2), then there is no reason to switch in round 3"

There have been comprehensive discussion of

**host behavior**and why he opens a door with goat i.e. some variation says host only open a door to goat if he knew you were correct, otherwise he would skip round 2, and ask you if you want to switch without opening a door in which case switch is a sure losing strategy. Of course Marilyn's math model was not built this way.