Please Explain (actually explain) The Monty Hall Problem

[PeroK]

I thought of another angle. Imagine the game is played simultaneously by three players, each with their own instance of the game. In each game the car is behind the same random door for all three players.

The first player always chooses door 1 and sticks; the second player always chooses door 2 and sticks; and, the third player always chooses door 3 and sticks.

If stick wins 50% of the time, then each player must win 50% of the time, and the car must be behind each door 50% of the time. Which is impossible.

 

[Ibix]

If stick wins 50% of the time, then each player must win 50% of the time, and the car must be behind each door 50% of the time. Which is impossible.

I really like this one. I realise it's the "stick only wins if you guessed right first time" explanation turned round a bit, but it explicitly forces you to fit three things evenly into two boxes if you want to believe 50/50. Neat.

 

[FactChecker]

I thought of another angle. Imagine the game is played simultaneously by three players, each with their own instance of the game. In each game the car is behind the same random door for all three players.

The first player always chooses door 1 and sticks; the second player always chooses door 2 and sticks; and, the third player always chooses door 3 and sticks.

If stick wins 50% of the time, then each player must win 50% of the time, and the car must be behind each door 50% of the time. Which is impossible.

But there might be a detail missing here. If Monte opens a door blindly, and it happens to be a goat by luck, then 50% is, indeed, correct. How does that fit into this visualization? This seems to be missing any insight into how Monte's door selection is involved.
ADDED: Legitimate probabilities can be up to 1/2 or 2/3, depending on Monte's decision rules.

 

[Ibix]

But there might be a detail missing here. If Monte opens a door blindly, and it happens to be a goat by luck, then 50% is, indeed, correct. How does that fit into this visualization? This seems to be missing any insight into how Monte's door selection is involved.

If Monty opens a door at random he could open the same random door for all three players. One player sees the car (and what happens next in this game decides what the overall outcome actually is), and of the other two 50% win and 50% lose. You might actually prefer nine games for this example - the car in the same place in all, and all nine combinations of door choice and accidental door open. Again, three contestants see the car, three win and three lose.

Monty Fall doesn't usually define what happens in the case where a car is revealed, although the possibility that it might have happened if he'd tripped differently is what drives the different probabilities.

 

[FactChecker]

Monty Fall doesn't usually define what happens in the case where a car is revealed, although the possibility that it might have happened if he'd tripped differently is what drives the different probabilities.

Suppose Monte makes his decision blindly and the game is declared invalid and restarted if he opens the prize door. Then, for valid games that continue, the probability of both remaining doors is 1/2.
Depending on what rules Monte follows, there are legitimate probabilities of 1/2 and 2/3. I don't see how that can be clearly addressed in your approach.

 

[Dale]

But there might be a detail missing here. If Monte opens a door blindly, and it happens to be a goat by luck, then 50% is, indeed, correct. How does that fit into this visualization?

It fits fine, but is not as compelling.

For the player that has the car behind their door, Monty cannot spoil the game and that player wins (since they are all sticking).

For the other 2 players there is a 1/2 probability each that Monty spoils their game. So, on average 2*1/2=1 is spoiled per round, and the remaining unspoiled player loses.

So of the unspoiled games there is indeed a 1/2 probability of winning and losing.

 

[DaveC426913]

This thread could have been two posts long. All we had to do was ask Facebook.

 

[Dale]

The probability of a Monty Hall thread being only 2 posts long is substantially less than 1/3.

 

[DaveC426913]

This thread could have been two posts long. All we had to do was ask Facebook.
View attachment 370118

I just realized this didn't land the way I thought it would. (especially since it's too low-rez to actually read the comments)
FBusers are usually dumb at best, outright ign'ant at worst.

But the comments in this Monty Hall posts were bang-on. Every one was "Always switch. This is old news." and "Switching doubles your odds from 1/3 to 2/3rds", etc.

 

[FactChecker]

FBusers are usually dumb at best, outright ign'ant at worst.

Hay!!! I take that personally. ..... well, ok.

But the comments in this Monty Hall posts were bang-on. Every one was "Always switch. This is old news." and "Switching doubles your odds from 1/3 to 2/3rds", etc.

If that was the FB consensus, I am surprised and impressed.

 

[DaveC426913]

If that was the FB consensus, I am surprised and impressed.

Yes. By a landslide. And not just in an intuitive sense; most seemed to know it probabilistically.

 

[DaveC426913]