MJ: Professor Mickey Rosa's Game Show Problem: Thoughts please

[21forme]

> Here's where you're wrong:

> At the start, the odds were 99 to 1 he was wrong.

> The host can't open the door with the car without
> giving away the game -- he can only open doors with
> goats. Thus the "equal likelihood" concept
> goes out the window.

> Now with only two doors left, the original proposition
> (i.e., 99 to 1 the door he picked was the wrong one)
> HAS NOT CHANGED! It is STILL 99:1 he is wrong.

> If we switches doors now, the odds will switch to 99:1
> in his favor.

This is where I don't get it - you now are left with 2 doors - one goat and one car. You know the car is behind one of the two. How can his odds still be 99:1? Doesn't effect of removal change those odds, as each door is removed?

[Don Schlesinger]

> It would seem that each birth is an independent trial.
> Each birth has a 50% of being either a girl or a boy.
> It doesn't matter what sex the first sibling is, the
> second sibling will still have a 50% chance of being
> either a girl or a boy.

Are you answering question #1 or #2, or both, with the same answer?

Don

[OldCootFromVA]

> Doesn't effect of removal change those odds, as each
> door is removed?

No, because the host cannot randomly open doors; he can only open doors he knows to hide goats.

[OldCootFromVA]

Let's go back to the 3-door example for the sake of simplicity.

There are a limited number of unique permutations because all the others are merely rotations or reflections.

Case 1: contestant has picked #1; car is behind #1
Case 2: contestant has picked #1; car is behind #2
Case 3: contestant has picked #1; car is behind #3

In case 1; the host cannot open door #1 revealing the car; he can only open #2 or #3. Obviously, in this situation, changing is bad.

In case 2: the host cannot open #1 because that's the picked door and cannot open #2 because that's where the car is; so HE CAN ONLY OPEN DOOR #3.

In case 3: same deal, except the host can only open door #2.

As you can see switching is bad in only one case, but good two cases -- i.e., on average, you're twice as well off if you switch.

To repeat a point in the previous message, the fact that the host is limited in which doors he can open prevents the original proposition (i.e., only 1 chance in 3 of being right) from changing.

[Doug]

Don,
I was answering Dog Hand's question. I have to re-read yours and give it some thought.

[Doug]

Okay, after careful consideration (and fully ready to get slapped down) my answer stands for both questions.

[young gun]

Don, you post the correct answers to this anywhere?

[Don Schlesinger]

> Don, you post the correct answers to this anywhere?

No, I'm sorry. The thread got down on the page, and, frankly, I forgot!

I'm pleased to say that your "revised answer," above, is correct for both!

The answers are, indeed, different. Hope everyone who gave this some thought had fun.

Don