I haven't thought about it for a while, but a few days ago the issue sort of slapped me in the face again. I was watching "21" on our national TV-station and in this movie a math problem is presented like this:
You participate in a game show where you must choose one of three doors. Behind one of these doors is a major prize and behind the other two there is nothing.
You choose one of the doors and now the game host presents you with an opportunity to change your mind. He opens a door behind which there is nothing and he asks you if you want to hold on to the door you already chose or you want to switch to the other remaining door.
What is the best choice? Well, the best choice is to change your mind, because the probabilities have changed. I never got my head around this. It doesn't make any sense, does it?
Actually, I was thinking about writing a post on how this math problem has a flaw in it.
Never the less, after thinking about it for a few days, it suddenly occurred to me why this statement is not a flaw. But the only reason, I suddenly understood it was the fact that I can now explain it in words. So, if you would like to understand it as well, hold on and read on:
The math problem presented here states that the first choice holds a 33.33% chance of winning the prize because there are three doors and only one of them has the prize behind it. Like this:
Probability(door1) = 1/3
Probability(door2) = 1/3
Probability(door3) = 1/3
But everything happening after your choice is completely different. This is where all probabilities changes and this is where it gets a bit nerd like, so try to hold on.
The host knows which door has the prize behind it.
The game host will not open the door you selected.
Neither will he open the door that has the prize behind it.
This means that if you chose the right door, you also have chosen to grant the host a free choice between the remaining doors AND that if you change your mind you will loose. In other words: there is a 1/3 probability that you have chosen the wright door AND that changing your mind will make you loose.
However, if you chose one of the wrong doors, you have also chosen to take away the free will of the game host. If you chose the wrong door, the game host can only open one of the other two doors: the one without the prize!
So, the probability that you've made the wrong choice is 2/3 and if this is true the game host could not have chosen any other door to open than the one he opened. In other words: there is a 2/3 probability that you have made the wrong choice in which case the prize MUST be behind the door that neither you or the game host chose. By changing your mind you have grabbed that 2/3 probability of getting the prize.
Still confused? Try flipping the scenario bottom up: instead of thinking about the first round as choosing a door, you should think of it as disregarding a door. If you disregarded one of the wright doors (that is: disregarding one of the doors with no prize behind) you have actually disregarded BOTH of the doors with no prize, because your disregarding of one of the doors forces the game host to disregard the other wrong doors. Having a 2/3 probability of disregarding ONE of these wrong doors, you have actually a 2/3 probability of disregarding BOTH of the two wrong doors. The choice is simple after that: there's a 2/3 probability that you have disregarded BOTH of the wrong doors in collaboration with the game host and a 1/3 probability that you have disregarded the wrong door, in which case it doesn't matter which of the other doors the host has disregarded.
Still confused? Then I'm sorry for not being able to explain this...
Probability Math will never be my strong side, but if the above helped you to understand this math problem, I'd love to hear about it... Maybe you could explain it to me and the other readers of this blog in a better way and we would all have a laugh?
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