https://www.youtube.com/watch?v=cXqDIFUB7YU
Apologies, don't know how to upload actual video. If someone could PM me how, it would be appreciated. Thanks.
https://www.youtube.com/watch?v=cXqDIFUB7YU
Apologies, don't know how to upload actual video. If someone could PM me how, it would be appreciated. Thanks.
isthereanybodyoutthere (04-08-14)
http://en.wikipedia.org/wiki/Monty_Hall_problem
The solution: Always switch
"The mass of men live lives of quiet desperation" - Henry David Thoreau.
Brilliant love this thread! Best in I don't know how long....
Empirical (03-08-14)
Initially I agreed with Dr Ivory until I read the Wikipedia link
You have to think about this from the point of view of the car, not the player.
When the host removes one of the "wrong" doors, the car remains behind the same door. So you can really think of the right door and one of the wrong doors as being the same door, but with a 2/3 probability of having the car behind....
Empirical (04-08-14)
If I asked you to pick the ace of hearts out of a face down pack and you picked one card and left it aside unchecked...................what are the chances of it being the ace of hearts?
1 chance in 52................correct?
Now if I take the other fifty one and say............."I'll remove 50 of these cards and if the ace of hearts is here then I'll not remove it."
Now the chances of the ace of hearts being in that bundle are 51 out of 52 so there is a better chance of the card I leave being the ace of hearts than the original card you choose.While there are only two cards it is not a 50/50 chance as the two sides of the equation haven't changed
your card is still a 1 in 52 chance while the new card offered instead of 51 cards is almost a certain winner.
IF the two cards were to be mixed after the cards were removed then it would be a 50/50 chance but that is not the case.
Should you be unable to understand the above scenario then I refer you to this site for further elucidation
http://betterexplained.com/articles/...-hall-problem/
Magicman