A question for the Wizard about the Monty Hall paradox.

I used the Monty Hall paradox to liven up the discussion on another message board, and during the discussion something occurred to me and I was wondering about the official answer.

For those who are not familiar with this puzzle, check it out at the Wizard of Odds website.

I completely understand how a contestant's probabilities will change from 1/3 to 2/3 after a losing door is revealed, but using the Let's Make a Deal game show might not be the best model for this puzzle, and here's why.

On the show, there were always two contestants who were allowed to pick doors. Let's call them Jim and Helen. So Jim picks a door and his chances are 1 out of 3 that he has picked the car. But when it's Helen's chance to pick, she only has two doors left to choose between because the third door has been taken out of play by Jim, which makes Helen's chances 1 out of 2.

If Helen is the one who is then given the option to switch later on, there's no way her chances can have been improved by opening another door because she only had a 50/50 chance from the get-go.

Is my logic wrong? I have a feeling it must be because this puzzle has been explained by the best of the best, but always using a single-contestant scenario, which is not the way the actual game was played.

I thank you in advance for your (hopefully simple) explanation.

You fail to include in the probability that the first contestant already picked the winning door, which would mean Helen's chances are 0/2. I would guess that is why Helen still has a 1/3 chance.

1/3 of the time (Contestant A picks winning door)= 0% chance
2/3 of the time = 50% chance

(1/3*0) + (2/3*.5) = 1/3

Your link doesn't work. Here's link right to the problem:
http://wizardofodds.com/askthewizard/122

Your example of Jim & Helen is flawed.

But that's OK. The text of the original question is flawed too.

On Let's Make A Deal, Doors are only used in the final deal with two contestants, and Monty NEVER offers either contestant the chance to switch.

During the game, Monty works with one contestant at a time, and often shows three curtains and gives the contestant the chance to switch after showing one.


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NOTE: The entire paradox is flawed!

The scenario never occurs except in two specific conditions:

1 - The contestant has picked the top prize or the goat prize. The exposed curtain is always the nice consolation prize.

2 - There is no goat (BOTH of the 'loser' curtains contain nice consolation prizes, giving Monty free will to show either if the contestant picked the top prize, or to show the other, if the contestant picked one of the consolation prizes).


Note: Either Monty knows what's behind each curtain, or a stage hand tells him which to expose, after the contestant makes a selection. I.E. Monty's choice is NOT random.


While these 'specific' conditions may be frequesnt, they are far from guaranteed. If the contestant picks the consolation prize, and the other curtains contain the car and the goat, no curtain is revealed.

If you've watched the show enough, you know that, time permitting, Monty showed the other curtain when the expose/choice thing doesn't happen. Quite often, there was no goat (or other zonk item) behind any curtain.

I also have a problem with the paradox as presented. Consider the last lines from the page I linked:

Quote:

Player picks door 3 --> Host reveals goat behind door 2 --> player switches to door 1 --> player wins.

So by not switching the player has 1/3 chance of winning. By switching the player has a 2/3 chance of winning. So the player should definitely switch.

Doesn't this explaination assume that the player has switched to door number 1?

What's to prevent him from switching to door number 2? Doesn't a 2/3 chance imply two equal choices?

Sorry about that link error, thanks for fixing it.

You're right about the original puzzle being incorrect in some of the details according to the way the real show was played, but mathematically it's still valid.

So since my question was flawed because I was using the wrong scenario, why not make my question a new scenario and work out the probabilities using the facts as they stood? It is not an intermediate deal but the final deal, there are two contestants and three doors, and the second contestant, Helen, is given the option to switch doors after the first contestant's (Jim) choice has already been revealed as not being the car.

Do both contestants have a 1/3 chance of picking the car at the beginning or are Helen's chances different than Jim's? If she were given the option to change her door once she knew that Jim did not win the grand prize, could she improve her chances of getting the car?

Quote: tsmith

Do both contestants have a 1/3 chance of picking the car at the beginning or are Helen's chances different than Jim's? If she were given the option to change her door once she knew that Jim did not win the grand prize, could she improve her chances of getting the car?

Again, then my above math would work. She has a 1/2 chance of picking it only if Jim didn't, so it's 1/3. Her odds would NOT change by switching if Jim's door was revealed to be a failure. Her odds would now be 1/2. Remember, the variable that changes the odds in the original problem is Monty will ALWAYS open a door that does not have the car. That variable is not present in your problem.

I used this problem every semester when I was teaching, even though my class was not really a math class. It got the students minds' going.

Quote: DJTeddyBear

Doesn't this explaination assume that the player has switched to door number 1?

What's to prevent him from switching to door number 2? Doesn't a 2/3 chance imply two equal choices?

Door Number 2 has been opened by Monty and revealed as not a winner.

DJ, they did indeed do the problem as it's written (not by the original poster, but by the Wizard), though in the early days of the show, with only ONE contestant. Player picks door. (1/3 chance). Monty reveals loser (ALWAYS). Two doors left. Player should switch because he still only had a 1/3 chance he was right the first time, but 2/3 that he was WRONG the first time, so he should change. The way my students used to get it was to add more doors. Let's say there are 10 doors. You pick Door 5 (1/10 chance of being right). Then Monty shows you 8 losers, and only Door 2 is left. So what are the odds you were right the first time? 1/10. Should you switch? Obviously!

Quote: cclub79

Quote: DJTeddyBear

Doesn't this explaination assume that the player has switched to door number 1?

What's to prevent him from switching to door number 2? Doesn't a 2/3 chance imply two equal choices?

Door Number 2 has been opened by Monty and revealed as not a winner.

That doesn't answer my questions.

What's to prevent him from picking door number 2?
and
Doesn't a 2/3 chance imply two equal choices?

Quote: DJTeddyBear

Quote: cclub79

Quote: DJTeddyBear

Doesn't this explaination assume that the player has switched to door number 1?

What's to prevent him from switching to door number 2? Doesn't a 2/3 chance imply two equal choices?

Door Number 2 has been opened by Monty and revealed as not a winner.

That doesn't answer my question.

What's to prevent him from picking door number 2?

The rules of the game. It would be like choosing a category in Jeopardy that's already done. The door has been revealed to hide nothing behind it. I guess if you wanted to, you could say "I choose the $0 option and to lose the game after all of this effort!" but I don't find it probable enough to change the odds.

The rules of the game prevent me from selecting the exposed door?

What if Monty exsposes the winning door?


Since the 'Rules of the game' specify that Monty will always reveal a loosing option, then the game itself always starts as a 50/50 chance that you will either pick the car or one of the goats.

After you make your selection, Monty reveals a goat, leaving you with two choices. It's 50/50.

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Or to use your logic, based upon your assertion that I can't select the revealed door:

While it's true that I have a 2/3 chance of selecting a goat, once one door is revealed to contain a goat, one of those chances is removed. So one of my chances to have selected a goat is eliminated. So now I had only a 1/3 chance of selecting a goat. Does that mean I now have a 2/3 chance of having selected the car to begin with?