The gameshow Lets Make a Deal had a seemingly simple final round..
There were 3 doors.. Behind 1 door was a car, behind the others were dummy prizes.. A contestant was to pick a door that they thought had the car.. When they did this, the host would open one of the other two doors, which was inevitably a gag-gift door.. At this point the contestant was given the option to trade the door that they chose, for the remaining unopened door..
Which is the smarter decision if there is one, and why?

 

Practically, there was nothing one could do with three unknowns and no governing logic.

From a statistics standpoint, from which I think you are eluding, there can be an argument about the cumulutive probabilies involves in a single choice, as opposed to two choices consecutively.

Steve

 

The unknowns are not absolute though.. The host knows where the car is, and will never reveal that door, this is information on an unknown..

 

The gameshow Lets Make a Deal had a seemingly simple final round..
There were 3 doors.. Behind 1 door was a car, behind the others were dummy prizes.. A contestant was to pick a door that they thought had the car.. When they did this, the host would open one of the other two doors, which was inevitably a gag-gift door.. At this point the contestant was given the option to trade the door that they chose, for the remaining unopened door..
Which is the smarter decision if there is one, and why?

Pretty much the same question is presented in the movie "21". In it, Kevin Spacey's character asks a prospective new member of the gambling club/business that Spacey has. The answer, which I didn't follow, involved conditional probabilities.

 

Could be a similar dilemma, I haven't seen that movie (or I don't remember seeing it)..
Who here would trade when given the option, and who would keep their original, and who don't think it matters?

 

Hey Mark, now that you mention it, that particular part in that movie was really bothering me! I cant see how changing the decision at the next option (and accounting for variable change as he put it) gives more chance. After one door is iliminated, its a 50% chance which door is correct. (At least in my mind it is)

 

Sounds like it is pretty much the same problem..
What if you were to look at it as the host opening all but one of the unchosen doors.. With 3 possible doors, the question doesn't change..

 

BTW, this question was posed to a number of ppl, and 12% chose to trade their choice, 88% chose to keep their choice.. Not relevant to the question, just a tidbit..

 

If you are a betting man who fancies his chances you switch

Btw, the Maths proves me right too.

Dave

 

Are you sure that isn't just some rationalization for a lack of committment on your part? Doesn't seem that intuitive..

 

Are you sure that isn't just some rationalization for a lack of committment on your part? Doesn't seem that intuitive..

It is an quirk of probability theory - think about the probabilities of winning associated with your initial guess when you do either stick or switch.

Dave

 

I know, I understand it.. With 3 elements the problem is counter intuitive, but like I said, if you scale it up it makes more sense.. The host was kind enough to reduce the set having a 2/3 probability of containing the car to a single element.. The set still holds the initial 2/3 probability though, even though it now contains only one element, and the initial choice constitues the set having the remaing 1/3 probability throughout the game..
Like I said, it would be more intuitive if there were 100 doors, the conestant chose one, then the host revealed 98 losing doors that weren't chosen and gave the option to switch for the single unopened/unchosen door..

 

Like I said, it would be more intuitive if there were 100 doors, the conestant chose one, then the host revealed 98 losing doors that weren't chosen and gave the option to switch for the single unopened/unchosen door..

But then that defeats the object of this "trick".

Btw, for those that have never seen this before and are interested, have a look for something called the "Monty Hall Problem".

Dave

 

I wasn't actually believing it, and even tho that I saw the maths of it and I understood the logic I was still sure you had 50-50.. So I tried with a couple of friends and switching gave way more wins.

 

A simulation, is almost always required to convince ppl.. Apparently physicists and engineers etc are more likely to get this wrong than young elementary school kids.. (I wonder if the young kids just opt to trade because its more exciting)

 

It's the Monty Hall problem, and you should always switch, because it gives you a 2/3 chance to win.

You can try it for yourself:
http://math.ucsd.edu/~crypto/Monty/monty.html

 

To put it into slightly less mathematical terms. You have a 1 in 3 chance to make the right guess right off the bat. You have a 2 in 3 chance you'll make the wrong guess right off the bat.

If you assume you made the wrong guess initially (for which there is a higher probability that you did) then Monty reveals the second wrong guess leaving you with the correct guess as the second option.

Makes sense, kind of a neat problem.