did a wee bit of hacking today, but not much, and some e.V. board duties. it's getting colder (-3C right now) which makes me want to sleep in, and peyton wanted to go out for a bit. while out i grabbed a ticket to vancouver as i'll be there on the 21st speaking at the vancouver linux users group (19:30 at the telus theatre on the bcit campus). i'll just be there for the 21st and then returning home the next day. i was going to spend a couple days there with t. but at the last moment her ailing grandmother scheduled a visit to town so i'm going alone. if you're in the vancouver area, come out and say hi =)
in the novel i'm reading it mentions the infamous monty hall problem which goes something like: you are given the choice between three doors, behind one of which is a prize and behind the other two lies nothing. after you pick one, the person running this little game opens up one of the two doors you didn't pick revealing that it is empty and then gives you the choice to stick with the door you picked or the remaining door that is closed. what do you do?
the answer is to pick the other door as it is twice as likely (66% compared to 33% chance) of having the prize. this works no matter which door you choose first or which door the prize is actually behind. (if you don't believe me, visit google for proofs and what not or simply map out the decision tree to see the possibilities for yourself)
it occurred to me while reading the novel that this is very similar to a number of phenomenon seen in quantum physics. (disclaimer: i am not a physicist, nor do i plya one on t.v.) in the monty hall problem, your choice appears to influence the likeliehood of the car's position (really, it's the host with their knowledge of the situation adding more information to the system that changes the odds); and in quantum physics the act (or lack) of observation tends to change the probabilities of the values of a particle's attributes (or at least seems to). what if, thought i, that was just a more complex version of the monty hall problem and we were observing the effects of probability redistribution through the addition of information as we interact with the "host" (which would be the particle/wave in question which one might assume to be fully self-knowledgeable). i spent part of the night laying awake going through the implications of this and how it would work (or, possible, not), and why this would work at small scales and not larger scales (which i decided was a red herring; the real question would be one of interaction or non-interaction resulting in a flow (or not) of information with interaction being nearly unavoidable at the macro scale (nearly, i say, due to the realities of exotic things like bose-einstein condensates)). unsurprisingly, i'm not the first to consider this concept and there is apparently a whole school of thought in the quantum physics world that suggests exactly such a thing. (i love how the internet puts the world's information at your fingertips)
the concept fascinates me, and to arrive at it was very satisfying. sleep robbing, but satisfying.
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3 comments:
it's probably easier to think of the monty hall problem this way:
when the host reveals the empty door, if the one you had first chosen has the prize, the remaining one will be empty; however, if the door you chose first was empty, the remaining one will have the prize.
as there's originally 3 doors, 2 of which are empty, you have a 66% chance of choosing an empty door the first time -- which means a 66% chance that the remaining door will have the prize.
(I'm saying this more for other readers' benefit, Aaron has no doubt thought it through already...)
and while the subject is mind games: http://www.borrett.id.au/computing/petals-j.htm
it's really quite maddening ;)
The key point with the monty hall problem is that it assumes that monty guarantees to open a wrong door for you every time. That is why you should switch. If he chooses to open a random door, than you still have 1/3 chances of having picked the right door. Of course he may open the correct door, and in fact make you lose.
I've heard another variation of this problem. I have 3 envelopes, and 1 of them has a winning lottery ticket. You choose one envelope, and I take one away, and tell you that it has nothing in it. What do you do?
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