If you change to the door Monty left you, 66.666...% of the time, you will not get a goat.

If you stick with the door you had originally, 66.666...% of the time, you will get a goat.

0.999... of a goat.

steve wrote:If you change to the door Monty left you, 66.666...% of the time, you will not get a goat.

If you stick with the door you had originally, 66.666...% of the time, you will get a goat.

0.999... of a goat.

Show Your Work (math rocker)

Three symmetrical cases. Assume (not-goat) behind door A:

Choose A, Monty will leave you with a goat.
Choose B, Monty will leave you with (not-goat).
Choose C, Monty will leave you with (not-goat).

I don't understand the above posts at all.

I say switch to door 2.

If door 1 was a car then the one Monty showed was immaterial. So you have 50/50, so this scenario is irrelevant.

But if door 1 was a goat then Monty obviously showed you the other goat to keep you guessing/playing.

Unless Monty doesn't play like that. I have to confess I don't know Monty too well.

But I switch to door 2.

i'll go with trevor. isn't it 50/50 between the other two doors?

edit: sorry, tanx. that picture... avatar, whatever.

This is a good brain-teaser.

I had to look it up.

do tell...

6079smith wrote:do tell...

OK.

Spoiler alert:

I had to look it up. I remembered enough probability to know it wasn't 1/2 or 1/3 chance if you switched, but I couldn't articulate it.

All I have done here is boil down what I thought were the most easily understandable solutions.

Do not read below this line if you want to figure it out for yourself.
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If you do not switch, you still only have your original 1/3 shot at winning. You are not capitalizing on the new information you have.

If you switch, the only way you DON'T pick the car is if you hit your 1/3 chance the first time. Which means you have a 2/3 chance of picking the goat.

Or:

You pick Door 1. 1/3 chance you picked the car. 2/3 chance the car is behind Door 2 or Door 3.

Monte shows you Door 3 doesn't have the car. It's out of the running. He knew it would be out of the running.

Now there is a 2/3 chance the car is behind Door 2 ALONE.

You only had a 1/3 chance to begin with. That doesn't change if you stay put.

It's essential that Monte ALWAYS reveals a goat. That is: he will never show you a car. Otherwise, your initial 1/3 chance holds, and switching is pointless.

The monty hall problem: smacks of freshman year discrete mathematics. MIT genius tried to explain it with decision tree: he proved that it was always better to pick again but mostly because we didn't know what the hell he was doing. The conclusion: do what you're told or flip a coin. QED.


If you want the probability/math explanation:

Here's the pure math assuming the premise of the game is known:

If E, F are two events, Pr[E|F] = Pr[E&F] / Pr[F]

E = the prize in in the door picked
F1 = Monty opens the first other door
F2 = Monty opens the 2nd other door

Pr[E] = 1/3
Pr[F1] = 1/3 * 1/2 + 1/3 * 0 + 1/3 *1 = 1/2
Pr[E & F] = 1/6
Pr[E|F1] = 1/3

E = prize isn't in door

Pr[E | F1] = Pr[E & F1] / Pr[F1] = (1/3) / (1/2) = 2/3

Pr[E | F2] = 2/3

So always pick the other door. You can make a convaluted tree to trace the results of each decision, but with math this obtuse, who needs it?

Without commenting on the answers so for...

for bonus points...

What if it turns out that Monty doesn't always knowingly open a door with a goat? What if he just randomly chooses one of the other two doors, and this time it thappens to reveal a goat? Should you switch or keep your original choice?