A question for the Wizard about the Monty Hall paradox.

Monty will NEVER reveal the winning door. It's an essential part of the puzzle. It says that either he knows where the car is, or a stagehand tells him which door to reveal.

It's still 1 in 3 when you first choose.

Here are the scenarios:

Let's say the car is behind Door number 3 and you are going to switch no matter what...

You choose Door 1. Monty reveals Door 2 is a loser. You change to Door 3 and win!
You choose Door 2. Monty reveals Door 1 is a loser. You change to Door 3 and win!
You choose Door 3. Monty reveals Door 1 is a loser. You change to Door 2 and lose.

Let's say the car is behind Door number 3 and you are going to stick no matter what...

You choose Door 1. Monty reveals Door 2 is a loser. You stick with Door 1 and lose.
You choose Door 2. Monty reveals Door 1 is a loser. You stick with Door 2 and lose.
You choose Door 3. Monty reveals Door 1 is a loser. You stick with Door 3 and WIN!

You can replace the numbers of the doors, but you'll get the same result every time. You win 2/3 of the time you switch. You'd only win 1/3 if you don't. The only variable not included is whether Monty reveals 1 or 2 when you actually did choose correctly, but it does not change the numbers.

Ah, but there are actually SIX scenarios, assuming the car is behind door #3:

You choose Door 1. Monty reveals Door 2 is a loser.
You choose Door 1. Monty reveals Door 3 is a winner.
You choose Door 2. Monty reveals Door 1 is a loser.
You choose Door 2. Monty reveals Door 3 is a winner.
You choose Door 3. Monty reveals Door 1 is a loser.
You choose Door 3. Monty reveals Door 2 is a loser.

Quote: cclub79

The only variable not included is whether Monty reveals 1 or 2 when you actually did choose correctly, but it does not change the numbers.

yes it does


If the host only gives you the option of changing when you are right, it is not correct to switch

Its correct to switch with 3 assuptions
1. The host knows which door is correct
2. the host will give you the option to switch every time
3. the host will never open the door with the car

Quote: Bigsooner

Quote: cclub79

The only variable not included is whether Monty reveals 1 or 2 when you actually did choose correctly, but it does not change the numbers.

yes it does


If the host only gives you the option of changing when you are right, it is not correct to switch

Its correct to switch with 3 assuptions
1. The host knows which door is correct
2. the host will give you the option to switch every time
3. the host will never open the door with the car

Correct, but those 3 assumptions are vital parts of the puzzle, which means it's always correct to switch.

Quote: DJTeddyBear

Ah, but there are actually SIX scenarios, assuming the car is behind door #3:

You choose Door 1. Monty reveals Door 2 is a loser.
You choose Door 1. Monty reveals Door 3 is a winner.
You choose Door 2. Monty reveals Door 1 is a loser.
You choose Door 2. Monty reveals Door 3 is a winner.
You choose Door 3. Monty reveals Door 1 is a loser.
You choose Door 3. Monty reveals Door 2 is a loser.

2 & 4 never occur. It's just part of the puzzle that he doesn't reveal the winner.

If it helps, think of it like this: After choosing a door, you are offered the option to trade the contents of the door you chose for the contents of BOTH other doors. This is, in essence, what your choice boils down to.

If you do not switch, you will win the car 1/3 of the time, and a goat 2/3 of the time.

If you do switch, you will win two goats 1/3 of the time, and a goat+car 2/3 of the time.

Assuming the value of a goat is $0 (sorry goats and goat-lovers), switching has the greater expected value.

I recommend the Wikipedia article for more on this.

And if you like things like this, I recommend one of my all-time favorites: the Two envelopes problem.

Thank you for the cover, JB. I understand the difficulty to grasp this problem. It took me awhile and it took many of my students a long time as well.

Quote: cclub79

Quote: DJTeddyBear

Ah, but there are actually SIX scenarios, assuming the car is behind door #3:

You choose Door 1. Monty reveals Door 2 is a loser.
You choose Door 1. Monty reveals Door 3 is a winner.
You choose Door 2. Monty reveals Door 1 is a loser.
You choose Door 2. Monty reveals Door 3 is a winner.
You choose Door 3. Monty reveals Door 1 is a loser.
You choose Door 3. Monty reveals Door 2 is a loser.

2 & 4 never occur. It's just part of the puzzle that he doesn't reveal the winner.

EXACTLY!

So there are FOUR possible scenarios. 2/4 are winners, 2/4 are losers. It's a 50/50 proposition!

You should check out the wikipedia article, there is a section called "Why the probability is not 1/2" But again, it fools a lot of people:

>>>>>>>
When first presented with the Monty Hall problem an overwhelming majority of people assume that each door has an equal probability and conclude that switching does not matter (Mueser and Granberg, 1999). Out of 228 subjects in one study, only 13% chose to switch (Granberg and Brown, 1995:713). In her book The Power of Logical Thinking, vos Savant (1996:15) quotes cognitive psychologist Massimo Piattelli-Palmarini as saying "... no other statistical puzzle comes so close to fooling all the people all the time" and "that even Nobel physicists systematically give the wrong answer, and that they insist on it, and they are ready to berate in print those who propose the right answer."
<<<<<<<<<<<

Quote: DJTeddyBear

Quote: cclub79

Quote: DJTeddyBear

Ah, but there are actually SIX scenarios, assuming the car is behind door #3:

You choose Door 1. Monty reveals Door 2 is a loser.
You choose Door 1. Monty reveals Door 3 is a winner.
You choose Door 2. Monty reveals Door 1 is a loser.
You choose Door 2. Monty reveals Door 3 is a winner.
You choose Door 3. Monty reveals Door 1 is a loser.
You choose Door 3. Monty reveals Door 2 is a loser.

2 & 4 never occur. It's just part of the puzzle that he doesn't reveal the winner.

EXACTLY!

So there are FOUR possible scenarios. 2/4 are winners, 2/4 are losers. It's a 50/50 proposition!

You are using the fact that there are 4 possible scenarios to mean all 4 are equally likely to occur. The first 2 occur 2/3 of the time (when you didn't pick the winner originally), and the second 2 occur 1/3 of the time (when you did pick the winner originally)