The Monty Hall problem is a puzzle in probability that is loosely based on the American game show Let's Make a Deal; the name comes from the show's host Monty Hall. In this puzzle a contestant is shown three closed doors; behind one is a car, and behind each of the other two is a goat. The contestant is allowed to open one door, and will win whatever is behind the door he opens; however, after the contestant has selected a door but before he actually opens it, the host (who knows what is behind each door) opens one of the other doors to show that there is a goat behind it, and asks the contestant whether they want to change their mind and switch to the other closed door. Does the contestant improve their chance of winning the car by switching or does it make no difference?
Think.
Think hard.
Harder.
HARDER.
Eww what were you thinking?
Anyway, the answer is yes, as surprising as it may seem.
I read it and at first I just thought that since you already know that host's door has a goat, then yours has either a goat or a car. Regardless of which one you choose, you'll have a 50/50 chance of winning . . . DUHHHH
Apparently not.
The solution to the problem is yes: the chance of winning the car is doubled when the contestant switches to another door rather than sticking with the original choice. When the contestant chooses a door, there is a probability of 1/3 that they choose the door with the car: there is a probability of 2/3 that they do not choose the door with the car. When the host opens a door to reveal a goat, there is still a probability of 2/3 that the contestant has not chosen the door with the car (because when the host reveals a goat it does not affect this probability). Therefore if the contestant switches their choice, there is now a probability of 2/3 that they have chosen the door with the car.
If it still seems illogical, try thinking of it like this:
It may be easier for the reader to appreciate the result by considering a hundred doors instead of just three. In this case there are 99 doors with goats behind them and 1 door with a prize. The contestant picks a door; 99 out of 100 times the contestant will pick a door with a goat. Monty then opens 98 of the other doors revealing 98 goats and offers the contestant the chance to switch to the other unopened door. On 99 out of 100 occasions the door the contestant can switch to will contain the prize as 99 out of 100 times the contestant first picked a door with a goat. At this point a rational contestant should always switch.
There's more info on this problem (it's a pretty famous maths problem) here: http://www.economicexpert.com/a/Monty:Hall:problem.html
No Fifi (if you haven't been bored to sleep yet), it is not a rickroll =P
I'd laugh if they were feeling uber evil and put that in the maths yearly. And made it 109 marks out of 110 =)
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