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[Banjax]

You are taking part in a game show......you have won through to the final round!!!!!

The host shows you 3 doors

behind 1 of the doors is a brand new Ferrari
behind the 2 other doors is a bicycle  

you choose 1 door, before the host opens the door he opens one of the other doors, revealing a bicycle.

Now, armed with this information - he asks if you want to change your mind.

The question is this:

is it better to:-
a) change your mind,
b) stick with your original choice or
c) makes no difference

 

[Martin_1962]

Depends if you were correct or not

[Banjax]

Depends if you were correct or not

yep - but you don't know yet.........

[STMO123]

      DOOR 1  DOOR 2 DOOR 3  RESULT
GAME 1 AUTO  BIKE BIKE Switch and you lose.
GAME 2 BIKE AUTO BIKE Switch and you win.
GAME 3  BIKE BIKE AUTO Switch and you win.
 
GAME 4 AUTO BIKE BIKE Stay and you win.
GAME 5 BIKE AUTO BIKE Stay and you lose.
GAME 6 BIKE BIKE AUTO Stay and you lose.


Switch.
    

[STMO123]

Wisely Choosing A Door
The "three door puzzle" is an interesting and unusual probability question. Here's how it works. You are a contestant in a game show, and the game show host tells you there is a prize behind one of the three doors you face. You have to guess which door to open.

But when you make your guess, instead of opening the door you picked, the game show host opens a different door...one that he knows has nothing behind it. So now you're down to two doors. And the game show host says, "I'll let you change your choice, if you want to."

And the question is, do you change your guess? Or keep your original choice?

The natural assumption is that it makes no difference, and so people tend to keep their original guess - reasoning being that we all know you should never "second guess" yourself. But is that really a wise choice?

Here's the math behind the question.

When you pick your door, the probability that you picked the correct on is one-third (1/3). Thus, the probability that you chose incorrectly is two-thirds (2/3). Nothing that the game show host does at this point has ANY effect on that probability.


Want To Keep Your Guess?
Let's suppose that you guessed correctly. Then it makes no difference what the game show host does, the other door is always the wrong door. So in that case, by keeping your choice, the probability that you win is 1/3 x 1 = 1/3.

But let's suppose you guessed incorrectly. In that case, the remaining door is guaranteed to be the correct door. Thus, by keeping your choice, the probability of winning is 2/3 x 0 = 0.

Your total chances of winning by keeping your guess is: 1/3 + 0 = 1/3.


Want To Change Your Guess?
Again, let's suppose that you guessed correctly. By changing your guess the probability that you win is 1/3 x 0 = 0.

But let's suppose you guessed incorrectly. Again, this means that the remaining door must be the correct one. Therefore by changing your choice, the probability of winning is 2/3 x 1 = 2/3.

Your total chances of winning by changing your guess is: 2/3 + 0 = 2/3.

[woodsy_mv6]

Turned a 33.33% game into a 50% game.


Makes no difference

[Banjax]

Wisely Choosing A Door
The "three door puzzle" is an interesting and unusual probability question. Here's how it works. You are a contestant in a game show, and the game show host tells you there is a prize behind one of the three doors you face. You have to guess which door to open.

But when you make your guess, instead of opening the door you picked, the game show host opens a different door...one that he knows has nothing behind it. So now you're down to two doors. And the game show host says, "I'll let you change your choice, if you want to."

And the question is, do you change your guess? Or keep your original choice?

The natural assumption is that it makes no difference, and so people tend to keep their original guess - reasoning being that we all know you should never "second guess" yourself. But is that really a wise choice?

Here's the math behind the question.

When you pick your door, the probability that you picked the correct on is one-third (1/3). Thus, the probability that you chose incorrectly is two-thirds (2/3). Nothing that the game show host does at this point has ANY effect on that probability.


Want To Keep Your Guess?
Let's suppose that you guessed correctly. Then it makes no difference what the game show host does, the other door is always the wrong door. So in that case, by keeping your choice, the probability that you win is 1/3 x 1 = 1/3.

But let's suppose you guessed incorrectly. In that case, the remaining door is guaranteed to be the correct door. Thus, by keeping your choice, the probability of winning is 2/3 x 0 = 0.

Your total chances of winning by keeping your guess is: 1/3 + 0 = 1/3.


Want To Change Your Guess?
Again, let's suppose that you guessed correctly. By changing your guess the probability that you win is 1/3 x 0 = 0.

But let's suppose you guessed incorrectly. Again, this means that the remaining door must be the correct one. Therefore by changing your choice, the probability of winning is 2/3 x 1 = 2/3.

Your total chances of winning by changing your guess is: 2/3 + 0 = 2/3.

perfect answer and you're right it IS interesting - most mathematicians cant see past the "it makes no difference" argument - but when you examine it - ALWAYS switch and you have twice the chance of winning a Ferrari  

[Banjax]

Turned a 33.33% game into a 50% game.


Makes no difference

always always switch - you have twice the chance of being wrong first time - so switching flips the odds in your favour  

[woodsy_mv6]

Turned a 33.33% game into a 50% game.


Makes no difference

always always switch - you have twice the chance of being wrong first time - so switching flips the odds in your favour  

The odds may change, but the situation does not.

Like pulling cloured marbles from a bag

[psychnurse]

Stick with the bike... Its cheaper and you know you will get fitter  

 

[Banjax]

Stick with the bike... Its cheaper and you know you will get fitter  

 

  

oh - and welcome back stranger  

[Bandit127]

Turned a 33.33% game into a 50% game.


Makes no difference

Woodsy is right.

Any other (clever) argument just appeals to human nature. Like getting a heads on a random coin flip is any more likely after 6 tails in a row than on the first throw. It isn't. It's still 50%. But human nature reasons that it must more likely.

Answer is c - makes no difference.

[Banjax]

Turned a 33.33% game into a 50% game.


Makes no difference

Woodsy is right.

Any other (clever) argument just appeals to human nature. Like getting a heads on a random coin flip is any more likely after 6 tails in a row than on the first throw. It isn't. It's still 50%. But human nature reasons that it must more likely.

Answer is c - makes no difference.

nope it DOES make a difference - as brilliantly explained by STM0123

you have a 2 in 3 chance of picking a bike, and a 1 in 3 chance of picking the Ferrari - so odds are you picked the wrong door - by switching you are doubling the chance of being right (because you've seen the other bike) ALWAYS switch - in the unlikely event of this happening to you  

[Bandit127]

It was a clever explanation. But I maintain that opening the first door had no effect on what is behind the remaining 2. All that has happened is you have changed the chance from 1/3 to 1/2 by doing that because they are independant.

Let's say the game is tossing coins and you need 3 heads in a row.

1st go - your chance of winning is 2³ (2 outcomes to the power of 3 goes) of winning. 1/8.

2nd go. 2 throws left. Your chance has changed. 2². 1/4.

Last go. At this point you win with a head or loose with a tail. 50% chance. The previous turns have no effect on the spinning coin . The physics is the same whether the game was 3 in a row or just 1 go. The chance of you winning the game at this point is still 1/2.

It's the same with the doors. It essentially became a new game. When you choose, you simply face 2 doors. 1/2.

[Bandit127]

It was a clever explanation. But I maintain that opening the first door had no effect on what is behind the remaining 2. All that has happened is you have changed the chance from 1/3 to 1/2 by doing that because they are independant.

Let's say the game is tossing coins and you need 3 heads in a row.

1st go - your chance of winning is 2³ (2 outcomes to the power of 3 goes) of winning. 1/8.

2nd go. 2 throws left. Your chance has changed. 2². 1/4.

Last go. At this point you win with a head or loose with a tail. 50% chance. The previous turns have no effect on the spinning coin . The physics is the same whether the game was 3 in a row or just 1 go. The chance of you winning the game at this point is still 1/2.

It's the same with the doors. It essentially became a new game. When you choose, you simply face 2 doors. 1/2.

I have been reading up on this and it appears that we could both be right.

If the game show host knows where the bikes are (so always shows a bike - not the car) the chance is 2/3 for changing and Bannjaxx &STMO123 are right.

If the host doesnt know, the chance is 1/2. Presumably because 1/3 of the time he removes the car...

Try this. Use a big number of games - I used 2,000.
http://math.ucsd.edu/~anistat/chi-an/MonteHallParadox.html