Monty Hall revisited, with a twist- survey

[Easy E]

Don't want to bring this up again, but in the Sunday Parade magazine, someone wrote in about the Monty Hall/intuition "puzzle".
He THEN said that one of his students posed this variation:
Suppose I am taking a multiple-choice test. On one question, I randomly choose 1 of 3 answers. Then the instructor says that one of the remaining two choices is incorrect. Should I switch my choice?

The "world's smartest woman" (or is that human? I forget what Marilyn vos Savant claims) said that the answer was: it doesn't matter.
She did NOT, however, explain her reasoning.

What am I missing here, because it seems like the exact same thing? Does it have to do with the way tests are scored, or that one wrong/right answer won't make a difference?


Monty Hall Test




Extra credit for detailed answers.

[pudley4]

If he doesn't specifically eliminate one answer, then it doesn't matter what you do.

Notice that whether you originally chose the correct answer or not, one of the remaining two choices will always be incorrect; so the instructor's statement will always be true.

Now, if he says "exactly one of the two remaining answers is incorrect", you'd obviously change your answer.

[Easy E]

My editing needs some work (thanks, pudley). Someone on UPF caught this as well

My editing of the problem IS the problem.

"You pick A. Teacher says C is wrong. Do you switch to B"
is the short version.

Marilyn's take was that, unlike the Monty Hall choice, it doesn't matter if you switch.

Is this correct?

[DPCondit]

With the Monty Hall problem, there are 3 possibilities:

Car, goat, goat.
Goat, car, goat.
or Goat, goat, car.

by eliminating one of the goat choices THAT YOU DID NOT PICK (and knowing the goat choices, and your choice ahead of time), the likelihood is 2/3 of picking a car if you switch, and 1/3 if you don't.

When the teacher tells you C is wrong, he does not know whether you picked C or not, so it does not help you to switch. The difference is he is just eliminating one wrong answer, for all he knows it is your answer. Now walk through it using the goat, car, goat method, and you will see. Monty Hall deliberately picks a wrong answer that you did not pick to give you a choice whether to switch, the teacher just eliminates a wrong answer with no regards to whether or not you picked it.

Don

[Glenn]

Easy-

The difference is that in the problem with the test, the instructor does not know what you picked when he tells you C is wrong. So she is right. In the Monty Hall problem, the possibility that you chose the wrong door and the host tells you that you are wrong does not exist, where it does in the quiz problem.

[Easy E]

So, let me see if I have this straight:

In Monty Hall, your pick has a 2/3 chance to be wrong. When Monty reveals wrong answer (door C) in the remaining choices, KNOWING that you picked door A, the chance that you should switch is 67%, not 50%

In the test, your original pick also has a 2/3 chance of being wrong. The instructor reveals wrong answer (#C) in the remaining choices, WITHOUT knowing that you picked #A. Now, your chance of being wrong with #A is 50%, not 67%.

And the difference is because, not knowing what you picked, the instructor could have told you that #A was the wrong answer, where Monty Hall eliminated revealing your door A as a goat?

Someone please get through my little ole brain how the dependency is gone in the second situation, given that the instructor did NOT pick #A as the answer to reveal, even though they could? I understand that the instructor's relevation gives #A and #B a 50/50 chance... but why did that change YOUR odds of #A being wrong originally, since you did NOT have any more info than in Monty Hall- except for that the instructor's choice was random and Monty's was not?

Am I putting too much weight on the wording that says you picked #A and the instructor did not, even though s/he could have?

I hope there's a mathematical principle behind this, because the verbiage is giving me a headache...

[Easy E]

Why does the dependency in Monty Hall NOT result in the same value, as you two are saying, as the dependency in the testing situation (you pick, teacher reveals a wrong choice out of the two that you didn't pick)

[DPCondit]

Behind 2 doors are horrible, murderous space monsters, behind one door is the most beautiful woman imagineable. To save your life, you must pick the one door out of three that is in front of the beautiful woman. The rooms and doors are completely sound-proof and smell-proof, there is no information that you can observe that will help you make an informed decision.

Without uttering a word, you decide to go with door number two. However, before you can voice your choice, a gigantic space meteor instantly crashes into the room behind door number 3 revealing a hideous space monster. Fortunately, the monster is killed by the meteor. Is this a sign? Should you switch doors? The other two rooms and doors are not exposed in any way.

It doesn't make any difference, it is even money either way. Ok? Same as if a teacher tells you a choice is wrong in the multiple choice question without any knowledge of whether you picked this choice or not.

Now using the same three doors and choices, your host suddenly takes a liking to you. He will tell you a door, other than the one you picked, that is guarding a monster, after you voice your choice, and give you the option of switching to a different door. He is from the planet Clorox, and is unable to tell a lie. You choose door number 2, He instantly tells you that there is a monster behind door number 3, and asks if you wish to switch.

Do you see the difference? Now you switch.

There is a 1/3rd chance of the doors being arranged as M-W-M, and 2/3rds of being either W-M-M or M-M-W, so therefore when tells you there is a monster behind door 3, there is a 2/3rds chance of the arrangement being W-M-M, because the alien knows the arrangement and your choice already. If the door is instead exposed by a meteor, it gives you less information, because it could have just as easily been your door that got crushed by the meteor.

Put 3 coins on a table, a dime, and two pennies. In whatever order you choose, it is irrelevant. Now, the object is to end up with the dime.

Let's say the order is Penny-dime-penny.

First, "pick" the first coin, then, eliminate a wrong answer (the last penny), then switch, you win the dime.

Next, "pick" the second coin, and eliminate one of the pennies, then switch, wrong answer, you win a penny.

Next, "pick" the last penny, eliminate the other penny, and switch, you win a dime.

See, doing this, you are right 2/3rds of the time, try it any order you want, It also works with nickels and quarters [img]/forums/images/icons/laugh.gif[/img]

Don

[Easy E]

I didn't need
I understood that, having removed a choice, the odds were 50/50.

What I DIDN'T understand is why, with the test variation as written (You pick penny, instructor picks the OTHER penny, though by accident) that the odds were only 50/50.

Yes, it was an uninformed choice by the instructor, but the NET result was the same: You picked in one set (with a 33% chance of being correct) and the instructor eliminated a choice in the other set (which had 67% chance of containing the right answer BEFORE the instructor unknowingly eliminated a choice), but the remaining choice in the second set LOST a 17% chance of being correct.

In Monty Hall, the informed dependent choice, when you KNOW that your door will not be opened, reveals the 2:1 shot.. but when the instructor, by "accident," takes an "independent" action THAT RESULTS IN THE EXACT SAME SCENARIO... it's now 1:1 .....

It seems to me that a required dependency, the fact that the teacher did NOT reveal your answer #A as the wrong answer, is being ignored here for an unfathomable reason, just because it COULD have been random but ended up NOT being random, as the problem was presented.

And they wonder why Monty Hall confuses everyone so much.... sigh.

Well, I follow it, but I don't "get" it. Thanks, Don, but MY solution to the Monty Hall problem and its variations, from here on in, will be to:

a) ALWAYS switch, since it may be better and may be breakeven and then

b) SLAP THE BEJEZUZH out of the fool who dared to bring it up, then walk off with my monsterous goat penny, muttering to myself....

[DPCondit]

I'm sorry if I'm pounding too hard, I was just trying to paint a picture.

Let's try the pennies again, this time (order doesn't matter, but let's just use penny-dime-penny again).

Now pick the first coin, then reveal one coin at a time in order (to simulate choosing randomly):

1st coin, yours, you must choose one of the other two doors, its a 50/50 wash

2nd coin, the dime is revealed, no guesswork at all now.

3rd coin, if you switch you win

Now, same scenario, pick the second coin and reveal them one at a time:

1st coin, you switch, you lose

2nd coin, the winner is revealed, no guesswork

3rd coin, you switch you lose

Same scenario, third coin is chosen, reveal one coin at a time:

1st coin, switch and win

2nd coin, dime revealed, no guesswork

3rd coin, you must choose among the remaining two, no advantage, 50/50 wash.

So, lets count up the wins and losses for switching, win twice, lose twice, two 50/50 washes, and no guessing the 3 times that the winner is revealed. So, you see when picked randomly (using every possible choice) there is no advantage to switching or not switching (except when the winner is revealed and there is absolutely no guesswork involved).

There, I think that illustrates the random choice switching scenario better. Seriously, I'm not trying to be a pain-in-the-***, just to paint a clear picture.

Don