I wrote this thing on another blog and I thought it would be nice to write my findings here. For those who don't know it has to do with a twisted variation of the Monty Fall problem that bugged me for days. I first wrote about it in my previous blog entry. I am not the guy who spends a lot of time in endless discussions on twisted quizzes or counter-intuitive problems cause usually I am more practical, like who solves sudoku or IQ tests when I actually manage to write a new algorithm that is both clever and usefull? But this one for some reasons hit me and I became a little obsessed with it. I still don't know if it's 1/2 (I'd say, inspired by PHd comics that it's pi/2 :) but the simulation with my own rules does it even if the mathematicians prooved the problem differently (they didn't even defined if something happens when the host accidentally reveal the car,.. oh well just read at the end ps or just get familiarized with the monty hall/fall problems, except if you don't want to waste your time :)

For the Monty Fall problem (the second variation) there is too much controversy and that is I think because the description is kinda weird and I can't think of a way I could possibly simulate that in reality nor can I have a good insight (yet) of what is the difference or what hard math or a very deep perception of statistics or probability that I could possibly lack. Till now I thought this was also 2/3 and tried to find out why they say (Marilyn, the PDF with the probability math I still haven't read, etc) it's 1/2.

One think that is not explained is what happens if the host accidentally opened the car. This is not defined. It says that he slips in a banana and randomly happens to get the goat. So, does he always randomly slips and gets the goat (so that the game show makes sense) or is revealing the goat just a specific run (and what happens in this case) and in the other runs the host would possibly reveal the car too. Yet we are asked if in the specific run that a goat is revealed (but a car could be revealed in other runs too) what would happen?

Although to be able to check it in a simulation a rule must be set for what happens if the host chooses the car. So I invented my own two different rules for my simulation:

1) The unfair rule: The game goes on and the player switches from a goat to a goat and looses anyways. So, the host can accidentally reveal the car and make the player instantly loose. In a C programm the simulation really gets us the proposed 1/2. That's easy.

2) The rerun rule: If the host randomly reveals the car then that run is simply discarded (and not counted in the total runs) and we setup a new one and try again. Now in this one I initially made a 2/3 but when I explained the monty hall/fall problem to my brother he surprisingly came with nice ideas to use in my simulation. My mistake was that when I had a new run with the 3 doors, getting the car I started the simulation again but with the same contents in 3 doors. That means, if I had GOAT CAR GOAT and the host opened the 2nd door with the car, I reran with the same GOAT CAR GOAT (didn't randomized again a new set of doors) and randomly the host would select the 3rd door and let the run continue. But this was like canceling the mistake of the host and selecting always the goat, bringing it back to the old monty fall (not hall) problem, thus 2/3.

He actually told me:

CAR | GOAT GOAT

GOAT | CAR GOAT

GOAT | GOAT CAR

"In the occasion of the second or third row the host has a 50% probability to fuck up and select CAR. In the 1st row the mistake will never happen. So if you change your simulation programm so that a wrong choice actually reruns the game by shuffling the car/goats order in the doors, you will get twice times the first row than the others. In the first row if you switch you certainly loose. So you have two cases that switching makes you loose (two times the 1st row) and two other rows (2nd and 3rd) that switch makes you win. This is 2/4 aka 50%."

And so it actually worked in simulation for both rules getting 50%.

p.s. Although I don't think that the people who originally invented the Monty Fall version of the problem thought of these rules for the solution. If they had they would be more precise. Somehow I feel that the Monty Fall is a slight variant of the Monty Hall deliberately made to get a 1/2 effect maybe to justify the several PHDs who did it wrong on the first problem and say that, they understood the Monty Fall variation from the description, not the Monty Hall description. They say that Marilyn had an unclear description of the initial problem. But Marilyn's description is just plain right. It's the Monty Fall variation that has an unclear or misleading or not complete (needs more data) description for me. It's just a counter-intuitive problem that most of us reply 1/2 at first sight. Did all of us thought of the Monty Fall problem? No. We just didn't see the whole image, just hide the revealed car and worked independently of the initial choice. A common pitfall. I don't believe that all those PHDs and especially other non-mathematical people just got it different. They got it wrong with a first guess as I initially and my brother and my friends did.

## Thursday, July 09, 2009

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## 2 comments:

I agree that it's hard to believe all those PhDs were visualizing a "random Monty". More likely, they assumed that Monty didn't always open another door in the first place, in which case the game becomes more psychological, like rock-paper-scissors. (Does this guy only let me switch when he knows that would make me lose? I dunno.)

Even more likely, though, is that the letter-writers just plain didn't think through the question. Going by what they said, they simply assumed that the second choice is a "separate" problem. The only way that really makes sense is if the doors are "re-randomized" so they have an equal chance of winnitude.

All that said, the people telling Marilyn she was wrong do have a statistical excuse of sorts: namely, someone who read that original column and agreed with Marilyn would be far less likely to write to her than someone who thought she was wrong, especially if that second someone thinks the answer is "obviously" 1/2 and therefore the answerer is dumb. Even before the internet there was SIWOTI syndrome, but not (as commonly) SIROTI.

Oh, I feel like I should add that I too got the original question wrong, same way everyone does. As for the random Monty thing, that the odds would be different didn't make any sense to me at first either, but I feel like I get it know.

In the original version, the door Monty picked told you something important (2/3 of the time, he's constrained to showing you the only other goat), but the fact that he picked a goat didn't (he was always going to do that).

In the banana version, the door he picks tells you nothing (it can be either with equal probability), but the fact that he picked a goat is signifigant (it happens 100% of the time you pick a car but only 50% of the time you pick a goat).

As it happens, it doesn't actually matter what the procedure is after he picks a car; all that matters is that Monty picking a car can happen with 1/3 probability, but didn't happen this time. If he always lets you keep a revealed car, then seeing a goat means fifty-fifty, and if he never lets you keep a revealed car, seeing a goat means fifty-fifty.

One weird exception is if he is able to "reset" the game by erasing your memory of his showing you the car. Then he just "keeps guessing" until he gets it right (only taking two guesses), and the effect is identical to intelligently picking the car in the first place, making it the vanilla game. However, if the fuck-up Monty instead resets your memory to before you even picked a door, then it becomes the banana game. (This is equivalent to a non-memory-modifying Monty saying, "Oops, I showed you the car. Let's call that one a draw, not write it down, reset the doors, and try again".)

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