Probability Question

Mr_Roger_Bongos

Just reading this article from the 'interesting articles' thread;

This section-
Gameshow with three doors. Behind one door is a brand-new Camaro. Behind each of the other doors, there’s a goat. If you pick the right door, the Camaro is yours.

Host asks;“Do you pick door number one, door number two or door number three?”

Let’s say, door number one. Host directs assistant to open one of the other doors — door number two, say. Behind that door is … a goat.

First, host is going to give you another choice. He’s going to give you both an option, and a dilemma:

“Would you like to stay with door number one,” he asks, “or would you like to switch to door number three?”

You ponder. But it’s a no-brainer, right? With two doors left, there’s a 50-50 chance the Camaro is behind either one. So you might as well stick with door number one.

Now I’m going to quote a man named Jeff Yass, from a book called The New Market Wizards. These words have made him very rich:

“The correct answer is that you should always switch to door number three. The probability that the prize is behind one of the two doors you did not pick was originally two-thirds. The fact that the host opens one of those two doors and there is nothing behind it doesn’t change this original probability, because he will always open the wrong door.”

That’s what you didn’t consider: the host knows exactly where that Camaro is parked, and he wasn’t about to direct the assistant to open a door that revealed it — no, he’s going to stretch out the game, to have fun with you, to see if, in your finger-crossed begging of the fates, you’ll switch doors.

And that’s the thing — your intuition is telling you, with two doors left, that the odds have got to be 50-50. But that’s wrong. The door you picked originally had a one-third chance of yielding that Camaro; the host picking a door he knows doesn’t hide the car won’t make it more likely that your door does.

Back to Jeff Yass:

“Therefore, if the probability of the prize being behind one of those [other] two doors was two-thirds originally, the probability of it being behind the unopened of those two doors must still be two-thirds.”

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Im baffled by the above. I'll never be Jeff Yass, but didn't realise my probability was that bad!

If i KNOW that 1 of the doors is not the car, then surely the odds of door one having the car is 50/50. There are only 2 doors left, it has to be behind one right?

I kinda see what he's angling at, it was 2/3rds before. The host knew it wasn't behind one, but with the removal of one choice, surely the odds of it being behind one of the remaining two doors is 50/50.

So why would you change doors? Because the initial odds were 2/3rds in favour of one of the other doors? - If this isn't correct, can someone put me out of my misery!

Jesus Wept

Yea that sounds correct. It all hinges on whether or not the host knows what door it is behind.

Discussion here.

stevedublin

Its probably easier to understand if you assume the host does NOT know which box contains the star prize and so the box they revealed COULD have been the star prize.

stevedublin

If the host KNOWS the box with the star prize and wants you to lose (not get star prize) the host would not give you the option of switching unless you picked the correct box originally.
It all depends on whether the host must show the contents of one of the remaining boxes or not.

pocketdooz

stevedublin wrote: »

If the host KNOWS the box with the star prize and wants you to lose (not get star prize) the host would not give you the option of switching unless you picked the correct box originally.
It all depends on whether the host must show the contents of one of the remaining boxes or not.

I think this is wrong. Has nothing to do with the Host knowing.

It's all down to math and probabilty. The host/prize etc are just to illustrate the point.

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stevedublin

pocketdooz wrote: »

I think this is wrong. Has nothing to do with the Host knowing.

It's all down to math and probabilty. The host/prize etc are just to illustrate the point.

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well it does confuse things. Whether the host is on "your side" or not influences the outcome, so its not necessarily all down to math and probability.
The host/prize illustration is not a good one imho.

willietherock

just google "Monty Hall Math" effect. Its counterintuitive but correct JUST like saying that if there is 23 people in a room their is a greater than 50% chance that some pair of them will have the same birthday.

stevedublin

looked up wikipedia:

http://en.wikipedia.org/wiki/Monty_Hall_problem

version of the problem is technically ambiguous since it leaves certain aspects of the host's behavior unstated, for example, whether the host must open a door and must make the offer to switch

wikipedia also gives an unambiguous version of the problem, which is more understandable.

pocketdooz

stevedublin wrote: »

well it does confuse things. Whether the host is on "your side" or not influences the outcome, so its not necessarily all down to math and probability.
The host/prize illustration is not a good one imho.

I agree, the host/prize illustration is not a good one. It does confuse things.

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stevedublin

I understand the problem now.
How do I use it to get as rich as Jeffy Ass?

dunkamania

The correct use of probalities can identify mispriced situations, especially when the everyone else has got their maths wrong