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January 17th, 2002

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12:58 am - [wins]
stability is more chaotic than chaos, because it is less expected of chaos.

i think i figured out why i'm right about the monty hall problem:
at work, we sell lottery scratch tickets. by law, they have to have the odds of winning written on them. the only one i remember is that one of the 1$ tickets is 1 in 4.72 chance of winning (keep in mind that this is your chance of winning your dollar back. if you want to make a profit, it's slimmer pickings).
now of course, if you buy a bunch of tickets, the chance of at least one of them being a winner goes up. for fun, let's say that if you buy 5 tickets, you have a 1 in 3 chance of having at least one winner. now, say you scratch four of those tickets and they're losers. is the chance of the last ticket being a winner 1 in 3? no, it's still 1 in 4.72, just like always[1]. as you scratch the tickets, each time you get a loser the chance of the remaining tickets containing a winner gets smaller because there are less unscratched tickets left. you can't count the ones you already scratched because you know htey're not winners already: they're out of contention.
similarly, once the wrong door is revealed, the door you 1st picked is no more or less likely to be right than the last door. if this weren't true, he could show both doors you didn't pick to be wrong, and your door would still have only a 1/3 chance of being right.
therefore: dan wins!

[1] this only stops being true if you have bought a significant percentage of the total production run, assuming they don't have some nedlessly complex method of assuring each card's chances are independent of the total run.

(yes, this is essentially teh same argument i was making all along. the difference is then i was saying that both answers seem to be valid, and i was defending mine cuz john said only the other was true. now i think mine is in fact correct)
state: smart
np: violent femmes - american music (live)

(4 shots upside the head | en garde!)


[User Picture]
Date:January 16th, 2002 11:26 pm (UTC)
There are two things at work here, you know. The one for each event, and the one for aggregate of all the events... The first is probability, the second is statistics. IE, each ticket has a 1 in 4.72 chance of winning something, and statistically your chance of having a winner is higher when you have not had one in a while.
[User Picture]
Date:January 16th, 2002 11:43 pm (UTC)


from what i can tell, from dictionary.com, probability seems to be a specific type of statistics relating to chance. they're not seperate things
[User Picture]
Date:January 17th, 2002 12:34 am (UTC)
Being that the chance of winning on the lottery ticket is a specific event, this makes a kind of sense.
[User Picture]
Date:January 18th, 2002 10:56 am (UTC)
you still have a problem.. first, the three doors are your entire lottery run right there, so it's not a direct comparison unless monty gives you a chance at 3 of one billion doors, where you have a good chance that no door wins. you know one of his doors win. also, monty's door uncovering will only be a random choice 1/3 of the time, which is when you actually picked the right door to begin with, and he has two losing doors to uncover. the other 2/3 of the time, he's not making a random choice, he's choosing the only incorrect door he has available to him to open, because you have chosen (but not yet opened) the other loser. which is why your odds are better when you follow him. 1/3 is that you picked the right door, and 2/3 is that you didn't and that monty is telling you which of the other two is the right door by omission. i can't see how this would work any differently..

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