The Monty Hall Problem Revisited

phyti said:
There are 2 doors containing a goat. thus there are 2 goats existing simultaneously. They could have ear tags 1 and 2 to avoid any confusion.
Click to expand...
Dude, c'mon! Maybe just forget about the goats for the sake of analysis--it's just win or lose, with the goats being a loss. You're making this way more convoluted than is necessary.
 
I'll also note that there was a somewhat similar scenario to this one involving a deck a cards, in the season 3 finale of Yellowjackets. Spoiler alert not needed, as I'm not revealing anything critical, but Shauna goes, "It doesn't change the odds, how did you get into AP Stats?"

I'm on team Young Natalie and Adult Misty, for the record.
 
phyti:

You couldn't find a flaw in my analysis in post #99, I take it.

I think we're done here, then.
 
Sarkus;

Your post 67.
"Simply being a possible outcome does not mean that it happens with the same frequency as all other outcomes. It's known as the equiprobability fallacy (or "the fallacy of uniform probability assumption")"

The coin toss has 2 outcomes H or T. By definition of faces, the edge is not a face.
On the left, a grid represents all possible paths to points on the diagonal.
A path consists of horizontal units H and vertical units T. Values on the diagonal are number of possible paths to that point. There are more paths near the diagonal indicating more short sequences than long ones. The total number of paths =2^6=64.
If all diagonals were drawn back to the origin, the graph would be an inverted Pascal triangle.

On the right, a random sequence in red. There are variations from the ideal case but as the number of tosses increases, the probability will approach the ideal value of 1/2.
Statistical probability applies to large numbers of events, not individual events. If a decision is made based on a single coin toss, the one outcome decides, and there is no need of statistics.
The abstract concept of probability is a convention that is applied to events in the real world when the answer is unknown.

As probability applies to the MH game, in the real world, there is no 1/3 of a car behind each door.
Probability states the location of the car is uniformly distributed as 1/3 per door, predicated on the fact that the whole car is contained in the (set of 3 doors), with a value of 1.
After the host opens 1 door revealing a goat, and excluding the door from play, the player is offered a 2nd guess.
Applying the same method,
probability states the location of the car is uniformly distributed as 1/2 per door, predicated on the fact that the whole car is contained in the (set of 2 doors), with a value of 1.
The probability of an event does not become a property of the event or any object in the event. In the MH game, it depends on number of possible games.

coin toss.gif
 
JamesR;

Your 3 scenarios post 76.

player choice red
host opens blue

s1
A B C
A B C
C

s2
A B C
A B C
C

s3
A B C
A B C
B

s1 A or C
s2 B or C
s3 C or B

Where is C or A?

It's here,
s4
A B C
A B C
A

win car 2 of 4, stay or switch.

The host is restricted to open 1 door per game. Opening 2 doors would end the game. Each game is played the same way each time and with the same frequency, meaning the same player choice and same host choice. So you need s4.
The logic of a Venn diagram.

mh Venn.gif
 
forum;

The problem is when the player chooses the door containing the car 'c'.
Fig.1 shows 2 different games (e for event), before the host chooses door 2 or door3.
Fig.2 shows 4 possible results from those 2 games.
Fig.3 shows Savant's interpretation of the 2 distinct prizes 'a' and 'b' as identical, 'g' for goats, even though there are 2 in different locations.
Fig.4 shows her list of 3 games, based on location of the car.
By excluding 2 games resulting in goats, switching is an advantage.

monty hall 22.gif
 
phyti said:
View attachment 6670
Click to expand...
Do you accept that the there is a 1/3 chance that the player could pick goat1? So why does this chart put it at 1/4 (follow the red line).
Similarly with goat 2 (follow the green line).

There is 1/3 chance the player initially picks the car, yet the blue and black lines sum to 1/2.

See, as explained previously, you are ignoring that initial probability and working fallaciously solely on the possible outcomes. But, as explained previously, the outcomes do not occur with the same frequency.

Apply 1/3 to each of the 3 initial choices that the player can make. In the case of the car, the two lines (blue and black) must also sum to 1/3. Therefore, since each are equally likely, they each have 1/6 chance.
Correct your analysis in this manner and you will get the correct result.

I'm not sure if your inability to comprehend is trolling or stupidity on your part. But either way, you remain wrong, and your analysis, as explained by others and myself, remains fallacious.
 
Sarkus said:
I'm not sure if your inability to comprehend is trolling or stupidity on your part. But either way, you remain wrong, and your analysis, as explained by others and myself, remains fallacious.
Click to expand...
Yeah, and sure, Savant could have been wrong, but in this instance she is entirely correct. IN this day and age, you can just google the Monty Hall Problem and there's like a gazillion hits and dozens of solid pieces on this.

I'm partial to this one in part because it gives a breakdown of all the ways in which people go wrong when approaching this problem, but also because the author echoes my sentiment that "it's kinda like a magic trick"--lots of red herrings and things to distract you and throw you off:

www.scientificamerican.com

This Infamous Probability Puzzle Fools Nearly Everyone. Here’s How to Beat It

How to finally wrap your mind around the uniquely counterintuitive Monty Hall dilemma
www.scientificamerican.com www.scientificamerican.com

For what it's worth, Savant is pretty good when it comes to puzzles and logical problems and suchlike. Where she's prone to be a bit weird is when her column ventures into Dear Abby territory, and all that stuff is entirely subjective so it's gonna appear weird to anyone who doesn't share her opinion on matters. But we're not talking opinions here.
 
phyti said:
JamesR;

Your 3 scenarios post 76.
Click to expand...
Try post #99. That's the one I asked you about, back in post #100. Remember?

It seems like you're trying to obfuscate.

Go through my post #99 line by line. Find the error. Quote me and tell me which step in my reasoning is wrong.

If you can.
 
A summary.

Savant did not do a simulation, since it isn't necessary if you have a complete list of all possible games, 'complete' being the critical word.
There are 6 distributions of prizes, but 8 possible games.
Why, because the car being behind door1, allows the host to choose either door2 or door3. One choice per game equals 2 additional games.
Simple logic states (door2 or door3) is different from (door2 and door3).

The player always chooses door1, so the host can only choose door2 or door3.
The game rules restrict the host from opening the player's 1st choice, and from opening a door containing the car. The host's choice is red.

The general game list is the complete list for 3 distinct prizes.
Substitute car for c and goat for a and b.
For the player to stay vs switch is column 1 vs column op.
The win ratio is 4/8 for both.
There is no advantage.

Savant solves the problem in terms of locations instead of choices.
Savant also lacks understanding of probability.
For 3 doors, there is 1 successful guess to win the car and 2 to not win, 1 of 3.
For 2 doors, there is 1 successful guess to win the car and 1 to not win, 1 of2.
There are not 3 choices for choosing 1 of 2 things.
If you disagree show us what it is!
monty hall 30.gif
 
phyti said:
A summary.

Savant did not do a simulation, since it isn't necessary if you have a complete list of all possible games, 'complete' being the critical word.
There are 6 distributions of prizes, but 8 possible games.
Why, because the car being behind door1, allows the host to choose either door2 or door3. One choice per game equals 2 additional games.
Simple logic states (door2 or door3) is different from (door2 and door3).

The player always chooses door1, so the host can only choose door2 or door3.
The game rules restrict the host from opening the player's 1st choice, and from opening a door containing the car. The host's choice is red.

The general game list is the complete list for 3 distinct prizes.
Substitute car for c and goat for a and b.
For the player to stay vs switch is column 1 vs column op.
The win ratio is 4/8 for both.
There is no advantage.

Savant solves the problem in terms of locations instead of choices.
Savant also lacks understanding of probability.
For 3 doors, there is 1 successful guess to win the car and 2 to not win, 1 of 3.
For 2 doors, there is 1 successful guess to win the car and 1 to not win, 1 of2.
There are not 3 choices for choosing 1 of 2 things.
If you disagree show us what it is!
View attachment 6676
Click to expand...
Easy. You show 8 games. Half of which the contestant picked the car. The car only has a 1 in 3 chance of being picked.

You're making the mistake of thinking that all outcomes are equally probable. Someone had a word for that....
 
phyti said:
A summary.

Savant did not do a simulation, since it isn't necessary
Click to expand...
Unless you like getting the right answer...

You can paste your logic till the cows come home. The data shows it is wrong. And the data is much much easier to analyze than any tortured logic. See post 86.

Theories must adapt to fit the facts; facts do not adapt to fit pet theories, no matter how strongly opined.
 
You must log in or register to reply here.
Back
Top