Confession time : I must admit I've just been arguing the toss (incorrectly) with someone about the Monty Hall Paradox
Finalists in a tv game show are invited up onto the stage, where there are three closed doors. The host explains that behind one of the doors is the star prize - a car. Behind each of the other two doors is just a goat.
The host invites the contestant to choose one of the three doors. Now, the host does not initially open the door chosen by the contestant. Instead he opens one of the other doors - let us say it is door number 1. The door that the host opens will always reveal a goat. Remember the host knows what is behind every door!
The contestant is now asked if they want to stick with their original choice, or if they want to change their mind, and choose the other remaining door that has not yet been opened. Does it make any difference whether they change their mind or stick with the original choice?
:oops: :oops: :oops:
Thicko SB
The monty hall paradox
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The monty hall paradox
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This is incredibly hard to explain, but on 2 out of 3 occassions you would do well to swap.
The only way to demonstrate this is to keep playing it and see how it works out.
To explain the mathematical reasons behind it would give Pythagoras a headache.
The only way to demonstrate this is to keep playing it and see how it works out.
To explain the mathematical reasons behind it would give Pythagoras a headache.
Kicks and scrums and ruck and roll.....Is all my brain and body need!
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I was utterly sure that the first decision was a formality and had no bearing on the second swap or stick decision.... :oops:Kinoulton wrote:This is incredibly hard to explain, but on 2 out of 3 occassions you would do well to swap.
It kind of makes sense 'cause it is a 50/50 choice but ultimately if you use your head (and that head contains more brains than mine does) you've got the advantage.
SB
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It depends on whether or not the host knows what is in each box.
If the host DOES know, it is in your best interests to swap.
The odds of you picking right first time were 1/3, therefore 2/3 chance of it being in a box you haven't picked.
If the host knows what is in each box he will always open a box that is incorrect. Therefore he knows at least one of the boxes you didn't choes is a losing one and, by opening it, leaves the 2/3 probability resting in the unopened box you haven't chosen, with the one you DID choose having only a 1/3 chance of being the correct one.
However, if the host DOESAN'T know which box is the winner (a la Noel Edmonds/Deal or no Deal), there is no statistical advantage to be gained in swapping boxes as, by the time you are down to 2 boxes, "knowledge" of the contents is identical to both the host & the player, therefore the odds are 50:50 for each box.
If the host DOES know, it is in your best interests to swap.
The odds of you picking right first time were 1/3, therefore 2/3 chance of it being in a box you haven't picked.
If the host knows what is in each box he will always open a box that is incorrect. Therefore he knows at least one of the boxes you didn't choes is a losing one and, by opening it, leaves the 2/3 probability resting in the unopened box you haven't chosen, with the one you DID choose having only a 1/3 chance of being the correct one.
However, if the host DOESAN'T know which box is the winner (a la Noel Edmonds/Deal or no Deal), there is no statistical advantage to be gained in swapping boxes as, by the time you are down to 2 boxes, "knowledge" of the contents is identical to both the host & the player, therefore the odds are 50:50 for each box.