Lateral Thinking XI

[Holmes]

Lateral Thinking XI is also known as " The Monty Hall problem "

HINT!: Probability. So, ayw, you´ll love it.

Here it is:

You are on a game show. You are shown three closed doors. A prize is hidden behind one, and the game show host knows where it is. You are asked to select a door. You do. Before you open it, the host opens one of the other doors, showing that it is empty, then asks you if you'd like to change your guess. Should you, should you not, or doesn't it matter?

[sstimson]

Lateral Thinking XI is also known as " The Monty Hall problem "

HINT!: Probability. So, ayw, you´ll love it.

Here it is:

You are on a game show. You are shown three closed doors. A prize is hidden behind one, and the game show host knows where it is. You are asked to select a door. You do. Before you open it, the host opens one of the other doors, showing that it is empty, then asks you if you'd like to change your guess. Should you, should you not, or doesn't it matter?

My answer

Spoiler:

The question is that normal behavior. If so then it does not mater. ( the 1 in 3 or 33% is now 1 in 2 or 50%)
The Host's behavior is key. He should be misleading you, so you chose the wrong door. Of course he show the the empty door, after all there are two of them. Look for host behavior, my thoughts if he looks startled or acts funny, there is a good chance you won the prize. If he acts normal, then change doors. Doing this should increase your change of winning

[S.H.]

Spoiler:

Yes, it will matter...it is better to change

Spoiler:

Explanation:
Let:E1 = empty door 1
     E2 = empty door 2
    P = prize door

List of Combinations
E1   E2      P
E1   P       E2
E2   E1      P
E2   P       E1
P     E1     E2
P     E2     E1
Assuming you picked the red one

Then remove E1 or E2(Just 1..) which is not the one you selected first..(or just not the red one)
You will be left with:
E1       P
E1       P      
E2       P
E2       P      
P         E2
P         E1

As you can see.. there are 4 P if you change and 2 P if you stayed with your answer..
It means there's a higher chance that you will get the prize IF you changed your answer..

But still....its not 100% or the game show will not earn money...XD

[sstimson]

Spoiler:

Yes, it will matter...it is better to change

Spoiler:

Explanation:
Let:E1 = empty door 1
      E2 = empty door 2
    P = prize door

List of Combinations
E1  E2      P
E1  P      E2
E2  E1      P
E2  P      E1
P    E1    E2
P    E2    E1
Assuming you picked the red one

Then remove E1 or E2(Just 1..) which is not the one you selected first..(or just not the red one)
You will be left with:
E1      P
E1      P     
E2      P
E2      P     
P        E2
P        E1

As you can see.. there are 4 P if you change and 2 P if you stayed with your answer..
It means there's a higher chance that you will get the prize IF you changed your answer..

But still....its not 100% or the game show will not earn money...XD

more you always remove a empty door so your chart should look like this

Spoiler:

where is the door you pick

E1 E2 before E1 after
E1 E2 before E2 after
p E2 before p after ( this case does not have a second cause then the host would open the prize door )
p E1 before p after ( this case does not have a second cause then the host would open the prize door )
so out of 6 cases 2 removed because of prize door would be showed next leaving four
you pick right the prize in two cases and picked one of the empty doors in the other case
so you pick the right door 2 out of 4 cases or 50%

Spoiler:

in the long run it might be true but this is short term. A bad door removed still means there are now only two doors to chose from. You either picked the right first or you do not. It is the shell game, only more legit.Me I would be suspicious if someone tried to change my choice. A few believe the first is normally the best. still it is 50 - 50 you chose the right door the first time.

Later

[Sayumi]

Here I go.

Spoiler:

Okay, I think it would be better to change.

The first door you choose:
possibility a)Wrong1
possibility b)Wrong2
possibility c)Right

after elimination the other door is:
possibility a)Right
possibility b)Right
possibility c)Wrong (1 or 2)

So in two out of three cases it would be more profitable to change. This applies as long as you know that the game host will eliminate one of the empty doors.

it's the same as S.H.'s answer 

Thanks for this Holmes!

[ayw]

I'm late at this show!
Holmes, I know this one already so i'll pass. But it's one of my favourites!
 

[bash7353]

Spoiler:

The point is just that at the beginning there were 2 wrong doors and only one right one, so your first decision is mor likely to be false.
If you are now picking a wrong door - which is more likely - then is would be better to change.

For those interested in math: 2 wrong doors and 1 right makes a propability of 2/3 that you chose and empty door, if that's the case you automaticly win by changing your mind. (Only two doors left, if you got the wrong one, then the other one's right).
That gives you a 66,667% chance of winning if you chance, if you don't you only win in 33,333% of the cases.

Again: There's no guarantee that you'll win, this move just gives you a higher chance of winning.

[Holmes]

Very Nice everyone!

[sstimson]

ok i show why i think that way now to look at it from the other side

Spoiler:

left mean door you first chose and [] the second you chose, p is the prize and e1 and e3 empty rooms
E1 E2 the host opens E1 if you change your vote then you are going form a win to a lost ( p [E2] )
E1 E2 the host opens E2 if you change your vote then you are going form a win to a lost ( p [E1] )
p E2  the host opens E2 changing your vote means you win ( E1 [p] )
E1 p  the host opens E1 changing your vote means you win ( E2 [p] )

Just so you completely understand the other two possibilities are not possible because then the host shows you have lost then and there

these are

p E2 the host opens the Prize door    and
E1 p the host opens the Prize door

that again brings the number of possibilities to four with two winners and two loser by changing your vote. 50 -50
the very same as if you did not.

I do not know about you but I would feel bad it I had chosen the right door and then chose the wrong one.

Later

[sstimson]

more explaining using this quote

Spoiler:

Yes, it will matter...it is better to change

Spoiler:

Explanation:
Let:E1 = empty door 1
      E2 = empty door 2
    P = prize door

List of Combinations
E1  E2      P
E1  P      E2 In both of these cases the Host will open E2 makes these two cases one
E2  E1      P
E2  P      E1 in both of these cases the Host with open E1 making these tow cases one
P    E1    E2
P    E2    E1
Assuming you picked the red one

Then remove E1 or E2(Just 1..) which is not the one you selected first..(or just not the red one)
You will be left with:
E1      P
E1      P      Already pointed out to be the same ( one case )
E2      P
E2      P      Already pointed out to be the same ( one case )
P        E2
P        E1

so that leaves this

E1      P
E2      P
P        E2
P        E1

As you can see.. there are 2  if you change and 2 P if you stayed with your answer..
As you can see it is 50 50

Later