Logic FAIL: Monty Hall Problem – GAH!
Now those of you who know me well will know that I love maths, I can spend hours and hours playing around with equations and figures and formulae, however a friend (http://rocko.co.nr/) told me at the weekend about the Monty Hall Problem and this one got me right in the logic chip. My first reaction to this was ‘that can’t be right, it doesn’t make mathematical sense’. It is a very counter intuitive problem and one that has caused much debate over the years if my googling on the subject is anything to go by.
The problem for those of you who don’t know it:
"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 other 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 take the switch?” Ref: http://math.ucsd.edu/~crypto/Monty/montybg.html
My first thoughts were, well of course it doesn’t matter. Initially you have a 1 in 3 chance of picking the right door and once one door has been removed you have a 1 in 2 chance of the door you’ve chosen being correct, so it doesn’t matter if you switch doors. But my friend insisted that no, you now had a 2 in 3 chance of being correct. So I’ve been playing with the maths and doing some reading up on this to find out why.
It seems that one of the biggest factors in why the odds of 2 in 3 is correct is that the game show host knows which door the car is behind, so he will always accurately remove a door without a prize behind it. If the game show host did NOT know what was behind each door, then my original thinking that you now have a 1 in 2 chance of winning would have been correct, phew, the logic chip isn't as flawed as first thought. I hadn’t been taking into account the game show host. Silly me!
Someone has already tested the problem using a little application they’ve written that knows which goatish door to remove. The application resulted in ~66% win rate for the switched door, his detailed notes can be found here: http://skepticblog.org/2009/03/10/monty-hall-problem-put-to-the-test/. So now the alternate theory of the application NOT knowing which door is a goat door needs to be tested to see what happens to the figures. Feel free to knock up an app to do this, if I get to it first then I’ll post my results on here.
The problem for those of you who don’t know it:
"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 other 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 take the switch?” Ref: http://math.ucsd.edu/~crypto/Monty/montybg.html
My first thoughts were, well of course it doesn’t matter. Initially you have a 1 in 3 chance of picking the right door and once one door has been removed you have a 1 in 2 chance of the door you’ve chosen being correct, so it doesn’t matter if you switch doors. But my friend insisted that no, you now had a 2 in 3 chance of being correct. So I’ve been playing with the maths and doing some reading up on this to find out why.
It seems that one of the biggest factors in why the odds of 2 in 3 is correct is that the game show host knows which door the car is behind, so he will always accurately remove a door without a prize behind it. If the game show host did NOT know what was behind each door, then my original thinking that you now have a 1 in 2 chance of winning would have been correct, phew, the logic chip isn't as flawed as first thought. I hadn’t been taking into account the game show host. Silly me!
Someone has already tested the problem using a little application they’ve written that knows which goatish door to remove. The application resulted in ~66% win rate for the switched door, his detailed notes can be found here: http://skepticblog.org/2009/03/10/monty-hall-problem-put-to-the-test/. So now the alternate theory of the application NOT knowing which door is a goat door needs to be tested to see what happens to the figures. Feel free to knock up an app to do this, if I get to it first then I’ll post my results on here.
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