Can someone explain the "Let's Make a Deal" thing again?

[Boredom]

If possible, using actual math AND laymen's terms?

[TomCollins]

What are the odds you were right when you didn't know anything?

Since you didn't know anything, and your chances were 1/3, then the odds you were wrong are 2/3. Every time you are wrong, they show you a goat, and it makes sense to switch. Every time you are right, they show you a goat, and it makes sense to stay.

So if you switch, you will be correct 2/3 of the time, and wrong 1/3 of the time.

For a simpler example, imagine there are 100 doors. You pick one. Monty opens every door but yours and another. The prize must be behind yours or the other door. A little more obvious?

[BruceZ]

[ QUOTE ]
If possible, using actual math AND laymen's terms?

[/ QUOTE ]

Laymen's terms 1: You pick door A, and Monty must show you one of the other doors that has a goat, so he shows you door C. Now he might have had a choice of door B or door C and chose C randomly, or he might have been forced to show you door C because door B has the prize. In the first case, where he chooses randomly, he would only pick door C 1/2 of the time, but in the second case, where door B has the prize, he will show you C ALL of the time. Therefore, the second case is twice as likely as the first case, so door B is twice as likely to have the prize as your door A. So you should switch your pick to door B to have a 2/3 chance of winning, rather than staying with door A which has a 1/3 chance of winning.

Laymen's terms 2: Imagine we play this game a million times. For convenience, we will always pick door A, so Monty will show us either door B or door C with a goat. Now let's say we decide that we will never switch our initial choice. How often will we win the prize? Obviously 1/3 of the time. Therefore the prize will be behind the other door 2/3 of the time. Not much to that one, eh?


Math: Again assume we chose door A, and Monty shows us that door C has a goat. P(A | B) means the probability of A GIVEN B.

By the definition of conditional probability:

P(B = prize | Monty shows C = goat) =
P(B = prize AND Monty shows C = goat) / P(Monty shows C = goat)

By Bayes' theorem, this is:

P(B = prize | Monty shows C = goat) =
P(B = Prize)*1 / [P(Monty shows C = goat | B = Prize)*P(B = Prize) + P(Monty shows C = goat | A = Prize)*P(A = Prize)]

= (1/3)*1 / [1*(1/3) + 1/2*(1/3)] = 2/3.

So the probability that door B has the prize is 2/3.

Note that the common error people make is solving for P(B = prize | C = goat) rather than P(B = prize | Monty shows C = goat). The first probability is 1/2, and this leads to the erroneous conclusion that it doesn't matter if we switch or not.

[jschaud]

if i remember correctly. you pick a door. then monte opens one of the other two showing a goat. then you get to decide to switch or not. if this is the scenario, then it doesnt matter if you switch or not. its still 50/50.

basically he opens a losing door and you now get to pick your door again with 50/50 odds. by not switching you are in effect re-choosing door A vs. door B which is no more likely now to have the prize than it was before. monte only removed a bad choice which equally improved your odds that you picked correctly.


Math: someone figure out the probability of the prize being in A, given C does not hold the prize and i think it will come up 50%, the fact that he is opening B sometimes and C sometimes shouldnt affect the logic.

[BruceZ]

[ QUOTE ]
basically he opens a losing door and you now get to pick your door again with 50/50 odds. by not switching you are in effect re-choosing door A vs. door B which is no more likely now to have the prize than it was before.

[/ QUOTE ]

It is not 50/50, and I just showed you 2 different ways why Monte's choice makes door B more likely, one of which spelled out the proof using Bayes' theorem. You have done nothing to argue against these proofs other than to make false statements contradicting them.

The answer to this question is well-known, the solutions I posted are correct, and this problem has been analyzed to death on this forum. If you don't understand the proofs presented, then ask qustions, or go elsewhere on the net where you can find an explanation that you do understand. Please do not interpret this thread as an invitation to post any more erroneous replies.

[Alan3]

Instead of three doors, let's pretend there are 100 doors.

You choose one.
The host then opens 98 doors knowing that none of them hold the prize.
You saw him do the same thing with all of the previous contestants.
Do you switch from the door you chose to the door that the host didn't open?

What is the chance that you chose the correct door? 1/100
What is the chance that one of the other doors is the correct door? 99/100
If all of the other doors are eliminated but one then there is a 99/100 chance that the prize is behind that single remaining door that you did not choose.

Still not convinced?

Instead of opening the doors, the host offers you the chance to open the door you chose or open all of the doors you did not choose and you win if the prize is behind any of those doors. The answer should be completely obvious at this point.

Does it matter if the host opens the empty doors before or after you make this decision?

[Fly]

When you make your first choice you have a probablity of 1/3 of choosing the winning door. That means that the probability that one of the other doors is a winning door is 2/3. Since the host knowingly opens a second door which is a losing door their is a 2/3 probability that the final door is a winning door. Thus you should switch.

[jschaud]

After reviewing your suggestion, I went to several other sites and now understand that I was incorrect. My original post was off the top of my head and the first time I ever saw or thought about the monte hall problem. I apologize for the inconvenience.

[laeelin]

The way I finaly convinced my brother:

If you DONT switch --- You win if your first pick is the correct door (1 out of 3)

If you DO switch --- you win if your first pick is either if the wrong doors. (2 out of 3)

Dont see how it works?

Lets pretend the prize is behind door (A). Now lets see what happens when you pick each door.

You pick door (A):
Monty opens door (B) showing the Donkey.
You change to (C) - You lose.
(or he could show door (C), and you change to (B) and still lose)


You pick door (B)
Monty opens doos (C) showing the Donkey.
You change to (A) - You win.

You pick door (C)
Monty opens doos (B) showing the Donkey.
You change to (A) - You win.


Edit: Oops! Sorry about the old post bump. I had several threads open, and forgot this was an old thread. [img]/images/graemlins/frown.gif[/img]