Monty Hall revisited, with a twist- survey

[Cyrus]

(Caveat : Monty Hall must announce beforehand that he will open up one wrong door for you. Monty will not be opening up doors randomly. In other words, it's a basic assumption that Monty knows where the car is.)

The easiest way to overcome my intuition about the Monty Hall, which tells me that it's the same whether I switch or stay with my original choice, is to imagine a thousand doors. Not just three dors, but a thousand.

I choose one door at random, and then Monty opens up immediately 998 doors revealing goats! There's one door closed. Plus the one I picked. But I'm not switching.

We do that with Monty three or four times and then I wise up...

[Easy E]

I only needed one coloring book picture, not the whole book!

[img]/forums/images/icons/wink.gif[/img]

I think I'll just stick with my "switch and slap" strategy... much easier to remember than figuring out when a dependent condition isn't an influencing condition....

[DPCondit]

Okay, I know you'd like it phrased to a more reduced mathematical principle, unfortunately, I can't think of how to do that at the moment. I guess all I can do here (at least at this moment) is show examples [img]/forums/images/icons/tongue.gif[/img] .

Maybe someone more knowledgeable than I can reduce it to a more elegant mathematical rule.

Good luck,
Don
(no more coloring book pictures on this one)

[Duke]

Say you choose A, out of obtions A, B, C.

1/3 of the time, the teacher says that A is wrong, 1/3 B, 1/3 C.

Now, you know that 2/3 of the time, the teacher will tell you that one of the other 2 options was wrong. And, since there are 2 of them, half of that time... the one you may want to switch to is the correct one.

This happens 1/3 of the time.

Well, look back at what it started as... you had a 1/3 chance to pick the right one from the beginning.

See now?

I believe that this is as clear as I can possibly make it. I'm sure you understand that it doesn't matter whether or not it's a letter/number/door/monster/hottie.

Just remember in this case that 1/3 of the time you WILL want to switch. If she tells you that your answer is wrong, then you probably want to switch, right?

Feel free to PM me if you still have any questions.

~D

[DPCondit]

Right, that is a good way of putting it. Much more concise than my monsters, hotties, dimes, and pennies.

And because in Monty Hall, you know ahead of time, that a wrong answer (other than yours) will be revealed 100% of the time, instead of 2/3rds of the time, in other words, 1/3rd of the time MORE often that would be dictated by chance alone. THEN, the remaining door must be a goat 1/3rd of the time LESS often than would be dictated by pure chance.

In other words, if x + x = 2x, then, 2/3x + 4/3x = 2x

That's it.

Don

[DPCondit]

Let's reduce the whole thing to a more simple formula:

It is a given in both situations that A, B, and C individually have 1/3 probability of being right (1/3P we'll call it).

You choose A, let's say:

In Monty Hall, MH MUST choose either B or C and expose it as wrong, so instead of 1/3P it becomes 0/3P, or zero probability, (because he must expose a wrong answer between B & C). If one of the two choices becomes 0/3P, then the other choice (the unexposed choice between B & C) must be increased by an equal amount, hence 2/3P, this does not change your original probability of 1/3P on your first choice because he is forced to choose between B & C, and cannot choose A.

Therefore A=1/3P B=1/3P and C=1/3P, whichever one between B & C Monty chooses becomes 0/3P, the other must become 2/3P, because B + C must equal 1/3P + 1/3P, or 2/3P, therefore 1/3P + 1/3P = 2/3P + 0/3P, and A remains unchanged at 1/3P, because A is not a choice for MH, and obviously you must switch to the choice that is 2/3P.

If all are random, then the answer to be exposed remains a 1/3 probability of being correct, if a wrong answer therefore is exposed, it becomes 0/3P (for the wrong answer exposed). Therefore your choice and the other unrevealed choice are affected equally, (because your answer being exposed was just as likely as any other, 1 in 3, in MH, he must choose between 2 answers, here it is randomly chosen among 3 answers with equal weight). Probability of being right for each of 3 choices is 1/3, therefore if C is revealed to be a wrong choice, then A & B must both be increased by 1/6th each.

1/3 + 1/3 + 1/3 = (1/3 + 1/6) + (1/3 + 1/6) + (1/3 - 1/3) = 1/2 + 1/2, therefore each remaining choice has 50% chance of being correct.

I apologize for any redundancy of anything anyone else may have posted, but this seems like a good formula for reducing the essence of the problem.

Thank you to the Duke for getting it back on the right track after my earlier storytelling atrocities.

I am not claiming my random choice formula is any better than Duke's, just putting it here with the MH formula for completeness.

Don

[Nottom]

I think it comes down to whether the professot would say one of the answers is incorrect if both remaining answers were incorrect. If so then it doesn't matter, if not then you should switch.

Bah just read, the addendum. Since he just eliminates an answer you didn't pick, it doesn't change. It bumps you chances from 1-in-3 to 1-in-2 but changing you choice at this point doesn't affect that

[DPCondit]

Given that 1 in 3 chances is always correct in either scenario, the only difference between Monty Hall and the Professor is:

Monty Hall may only choose a wrong answer from among two choices

&

The professor chooses a wrong answer from among 3 choices.

If you choose from among only 2 choices, then you must look at the one "unpickable" choice separately from the other two, such as (1/3) + (1/3 + 1/3), now if one choice among the latter two is found to be wrong (whether by random chance or preordained knowledge, but specifically your choice is excluded from the picking) then it becomes zero and the other becomes 2/3. (1/3) + (1/3 + 1/3) = (1/3) + ([1/3 - 1/3] + [1/3 + 1/3]) = (1/3) + (2/3) = 1. Of course your choice is no different from the other remaining one if he is choosing from all 3 with no regards to what you picked.

So that is the real difference, choosing from only 2, or choosing from all 3 without specifically excluding your answer.

Don

[Al Mirpuri]

You are asked to pick one out of three doors. Behind one of the doors is a prize. Can you pick the correct door (the one with the prize behind it)? The doors are labelled A, B, C. You pick B (for the sake of the argument). Monty Hall informs you that C (for the sake of the argument, he could just have easily said A) does not have the prize behind it. He now tells you that you may forgo your original choice B and choose A. Should you choose A?

It has been answered that yes you should. The reasoning goes like this: whatever you picked had a 1/3 chance of being correct, what remained unpicked had a 2/3 chance of being correct. Once one of the unpicked doors has been shown to not have the prize behind it you should switch to the other door as the whole of the 2/3 chance now falls on that door. Remember it was 2/3 that the unpicked doors had the prize behind them.

It has also been answered that you should not switch your allegiance as after one door has been shown to not have the prize behind it you are left with two doors and only one of them has the prize behind it; a fifty-fifty proposition.

The reason it is called the Monty Hall Paradox is because two equally persuasive pieces of reasoning produce differing answers. It is not a mathematical puzzle but a philosophical conundrum.

[Cyrus]

"The reason it is called the Monty Hall Paradox is because two equally persuasive pieces of reasoning produce differing answers. It is not a mathematical puzzle but a philosophical conundrum."

I'm sorry but I respectfuly disagree.

This is indeed a mathematical puzzle (problem). It has nothing to do with philosophy. For a better and more immediate understanding of the purely mathematical nature of the dilemma and its solution, please check my hint at an earlier post.

It is called a paradox because the answer runs counter ot our intuition. And not because there are 2 equally plausible solutions. No, one solution (switching) is correct and the other (staying put) is incorrect.

When a problem presents us with an intuitively obvious but actually wrong solution, it is called a paradox. Or when it offers no solution at all, on the basis of the stated conditions of the problem, although at first examination it looked like it can be solved.

<ul type="square">paradox, n.

1. A seemingly contradictory statement that may nonetheless be true: the paradox that standing is more tiring than walking.

2. One exhibiting inexplicable or contradictory aspects: "The silence of midnight, to speak truly, though apparently a paradox, rung in my ears"; (Mary Shelley).

3. An assertion that is essentially self-contradictory, though based on a valid deduction from acceptable premises.

4. A statement contrary to received opinion.

--From the American Heritage Dictionary, www.dictionary.com[/list]