Universal Translator

Thursday, December 15, 2011

The Game Show Example

That last “solution” to the problem I posed in Reading Marx – Part IV (the one about labor value shifting from the destroyed paper warehouse to the remaining one) reminded me of something I thought I’d share just because I find it interesting.  You may already know this – it was mentioned early in that Kevin Spacey movie 21.  It’s the game show example designed to illustrate how changing information can change probability, and how non-intuitive this is for most of us to understand.

The Set-Up:  You’re participating in a game show like Let’s Make a Deal.  You are presented with three doors and told that behind one of those doors is a brand new car.  The other two doors have goats.  You have to pick a door blindly, and you get to keep as your prize whatever you find behind that door. 

The Pitch:  You choose, say, Door A.  But before showing you what’s behind Door A, the host has them open up Door B and behind door B there is a goat.  So the car is behind either Door A or Door C, and you’ve chosen Door A.  The host turns to you and says:  “Are you sure you want to stick with Door A?  If you like, you can switch and pick Door C.  What do you want to do?”

The Answer:  You should always switch to Door C, because by doing so you will double your odds of winning the car.

Wait -- what?  Here’s how it works.  After you pick Door A, imagine the three doors as divided into only 2 different sets:  YourSet and NotYourSet. 

YourSet only has one door, Door A, and behind it is either a car or a goat.  We know that the probability YourSet contains a car is only 1 out of three.  Which means the probability YourSet contains a goat 2 out of three.

NotYourSet has two doors, Doors B and C.  We know the probability NotYourSet contains a care is 2 out of three.  But also . . . we know that the probability that there is at least one goat is in NotYourSet is 100%.  By the rules of the game NotYourSet has either 1 goat or 2 goats but it’s got at least one goat.

So . . . . before the host opens Door B, this is what we know for sure:

                                    YourSet                                   NotYourSet

Odds a Car is in
this Set:                               1/3                                               2/3

Odds Only Goats
Are in this Set:                   2/3                                               1/3

Odds at least One
Goat is in this
Set:                                      2/3                                           100%                                                 

Suppose before opening the door in NotYourSet and showing you the goat the host had asked you if you’d like to switch not doors, but sets.  That is, you can stick with your choice of Door A, or you can pick Doors B and C collectively and you will win the car if it is behind either one of those two doors.  Obviously, you would switch immediately because the odds are twice as great the car is behind either Door B or Door C than they are that the car is only behind Door A.

Stated another way, the odds are twice as great the car is in NotYourSet than they are that the car is in YourSet.  And showing you a goat behind one of the doors in NotYourSet doesn’t change those odds, because you always knew there was at least one goat in NotYourSet.

So there is still a 2 out of three chance the car is in NotYourSet.  It’s just that now you know that – between the two doors that comprise NotYourSet – Door B definitely does not have the car.  That means that all of the “2 out of 3” probability collectively owned by NotYourSet that previously had been split between Doors B and C now is contained behind Door C only,

So you should always switch doors, because by doing so you double your chance of winning the car.

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