Quiz Time

[Marshall]

How are you choosing both? Only one door has been opened.

[flatandy]

Your initial choice looks like this

X X X

Your first choice is now

(X)

leaving

(X X)

Your next choice is basically choosing either to:

open your one door

open both of the other doors. The fact that the host has already opened one of them doesn't change your odds. If you switch, both doors on side 2 are open.

[perrykneeham]

Hmm .... I would add a practical note. It is easier to move goats than cars.

[Marshall]

The third door was opened by the host already, so if you go ahead and open door 1 you will have still effectively opened two doors.

[Marshall]

(I'm sure once I get this I'll feel pretty stupid.)

[perrykneeham]

Marshall - read the link. It does make sense, but it does feel a bit like gamblers semantics.

[Marshall]

The Marilyn vos Savant link? I read that. Still don't get how the odds don't change for your door but do for the other unopened one.
I'll go back and read it again...

[flatandy]

Neither goat nor car move.

The odds as you start are 33% per door. You pick one door - X , which is 33%. The other-two-doors (Y & Z) are 66%.

This much is obvious.

The host eliminates a "definitely not car" (Z) door among the other-two. The odds of the other-two (Y & Z) don't change when he does this. It's still 66%.

-

Some reasoning:

The odds at the start of the unopened door of the other two (Y) was 33%, but because the host knows there's a goat behind door Z he's not randomly eliminating one of the choices. He's knowingly eliminating a bad one, tilting the scales, meaning that all 66% chances of (Y&Z) are now just on door Y

[Marshall]

Okay that managed to pierce my noggin, thanks.

Does seem very counter-intuitive though. Because of the way we are taught to think about these things maybe.

[flatandy]

Yeah, it's fascinating and counter-intuitive.

And of course in the actual game show situation, you still have a fair chance behind the door you first chose, so a lot of people stick with "instinct" and win.

[perrykneeham]

Hang on - I had this earlier, but isn't the switcheroo in the assumption that none of the probability is transferred to the original choice?

I follow the logic, but has anyone actually modelled this?

[perrykneeham]

Ah no. That's it - the host has knowingly eliminated the bad choice. Has he though? What if he knows he's got two goats?

[flatandy]

He's still definitely eliminating a goat, even if he has two goats. If he knew he had a car and a goat and eliminated a goat, then the odds would be 100% for the switch.

[happyhammerhead]

He knows what he's got from the start. The 'game' doesn't work otherwise.

[perrykneeham]

Yeah, I'm back on track now. The host has chosen not to select a door that had a car behind it whether either did or not.

It's a con, so it's hardly surprising that it's counterintuitive. It sort of relies on our assuming that the host isn't making an informed choice.

[voice]

Well yeah, the host can't be the one making random uniformed choices cos the car could get eliminated. He has to know he will always eliminate the goat.

[wetkingcanute]

Here ya go peeps

[ootlg]

The math doesn't work because the host doesn't open Door 3 if it's an Auto door - or any door if it's an Auto door. The basic math works - 33% chance of winning increased to a 50% chance when the goat behind Door 3 is eliminated.

[ootlg]

link to above again

[Marshall]

One thing still bothering me.

The odds as you start are 33% per door. You pick one door - X , which is 33%. The other-two-doors (Y & Z) are 66%.

The host eliminates a "definitely not car" (Z) door among the other-two. The odds of the other-two (Y & Z) don't change when he does this. It's still 66%.

So why can’t you likewise say X and Z together are 66%, then when Z is eliminated that leaves X at 66%?