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Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat. He then says to you, "Do you want to pick door No. 2?" Is it to your advantage to switch your choice?
Do any of you have any noteworthy experiences where knowledge of math helped you in an unusual way?
https://en.wikipedia.org/wiki/Monty_Hall_problem
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Get Those Brain Cells Working: The Monty Hall Problem
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(Score: 4, Informative) by morpheus on Friday July 29 2016, @10:45AM
When offered to swap doors, you change your odds from 1/3 to 1/2 if you swap the door as totally there are one car and one goat left and there is 50% chance of car in the other door.
I may have misunderstood what you are trying to say here but the odds of winning the car if you always switch are 2/3, not 1/2. Just think of trying to pick the door without the car first. Your chances of doing this are 2/3 and if you are successful (and always switch your choice) you get the only door with the car in the end. If you randomly decide between staying with your original choice and switching, then the probability of picking a car is 1/2.
(Score: 2) by tonyPick on Friday July 29 2016, @11:24AM
Yeah - the trick is that since "Monty" always knows which doors have what (and he's always got an empty/goat door) then he's not changing the selection odds in any way from the initial choice when he does the first part of the reveal, even though the question implies otherwise.
Basically the first choice splits the set into your door ("1/3" chance of having the car) and Monty's two doors ("2/3" chance of having the car), and then you get the chance to swap over to Monty's set: when phrased like that the problem becomes a lot more obvious.
(Score: 2) by Reziac on Saturday July 30 2016, @04:26AM
Quick, James -- move the car!
And there is no Alkibiades to come back and save us from ourselves.