quote:

Originally posted by DigiNut Ugh, I'll let Flyboy argue with you over the egg thing, because that question's been answered and you seem to neither understand what "worst case" means, nor the nature of random numbers. No, a computer does not generate random numbers, in fact, if you read any textbook, it will refer to them as "pseudo-random" numbers because it's based on an algorithm. Very little in this world is actually 100% random, independent of outside sources. But lots of things can be modelled that way, like the movement/position of subatomic particles.

very little in this world is 100% random? name ONE case thats 100% random!!

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quote:

Originally posted by Resnick very little in this world is 100% random? name ONE case thats 100% random!!

Argh... aren't you studying engineering?

You're going to have a lotta trouble if you have so much trouble grasping the concept of a random number. I'm not going to get into a whole discussion of random number theory and stochastic processes, you can read that in a textbook.

Human guesses are not useful as random numbers. Period.

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quote:

Originally posted by DigiNut Argh... aren't you studying engineering? You're going to have a lotta trouble if you have so much trouble grasping the concept of a random number. I'm not going to get into a whole discussion of random number theory and stochastic processes, you can read that in a textbook. Human guesses are not useful as random numbers. Period.

hmm it seems that ur not getting my point either, let me make it nice and simple...probabilityes are basically x/y+x so like x = heads and y = tails.. in a basic flip of a coin, the probability of heads coming up is x / x+y so its 1/1+1 = 1/2 ...thats a really simple concept

so now if u take a range, say [1,10] and pick a random number from there, then x = rational #'s, y = irrational numbers, no i know for a fact that there are rational numbers between 1 and 10, so lets just say there is one..even tho there are infinite but forget that.

so 1/1+y is the probability, which means no matter what y is, the fraction CANT be zero (mind u y cant be infinite cuz thats not a number)

and of course this is a really shitty proof cuz both x and y are infinity and u cant really evaluate it, BUT its the probability is not zero

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Last edited by Resnick on Oct-11-2003 at 20:10

quote:

Originally posted by Resnick hmm it seems that ur not getting my point either, let me make it nice and simple...probabilityes are basically x/y+x so like x = heads and y = tails.. in a basic flip of a coin, the probability of heads coming up is x / x+y so its 1/1+1 = 1/2 ...thats a really simple concept so now if u take a range, say [1,10] and pick a random number from there, then x = rational #'s, y = irrational numbers, no i know for a fact that there are rational numbers between 1 and 10, so lets just say there is one..even tho there are infinite but forget that.

Wrong.

Flipping a coin is at least somewhat random because you're leaving a great deal of the "picking" to gravity, and to the furniture surrounding you. Since the gravity is not exactly the same in every location, and the motion of your fingers is similar each time but beyond your fine control to an extent. One might argue that even the coin flip isn't totally random because you could predict the outcome if you had a perfect physical model for the flip, but that's stretching it a bit. The point is that even if you wanted to, you could probably not repeat the exact same flip each time.

"Picking" a number between 1 and 10 is not random. If you wanted to, you could pick the number 5 every single time. If you want to, you could pick from 1 to 10 counting up. Your choice does not constitute a random number.

Let's think of it this way. I will make you bet for $100. I am going to pick a random number between 1 and 10. If that number is 7, you have to pay me the $100. If it's any other number, I'll pay you the $100.

Would you take that bet?

Please, I urge you to read on random number theory before continuing this argument, what you're saying is incorrect and it's not helping answer any of the questions in the thread.

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quote:

Originally posted by DigiNut If the main set has two nondistinct values, are two subsets using the "other" nondistinct value considered distinct? i.e. if the main set is {2, 2, 3} then can {2, 3} and {2, 3} be considered distinct? I'm assuming the answer has to be yes, because otherwise you could just pick your main set as {2, 2, 2, 2, 2, 2, 2, 2, 2, 2} which would essentially disprove your statement... but I'm asking anyway.

A set, by definition, has distinct elements .

Your previous solution was nearly right. There are 1024 subsets of the initial 10-element set. Each must sum to a number between 1 and 1000 (more precisely, 55 and <1000). The 1024 subsets can therefore not all have distinct sums.

quote:

Originally posted by DigiNut Wrong. Flipping a coin is at least somewhat random because you're leaving a great deal of the "picking" to gravity, and to the furniture surrounding you. Since the gravity is not exactly the same in every location, and the motion of your fingers is similar each time but beyond your fine control to an extent. One might argue that even the coin flip isn't totally random because you could predict the outcome if you had a perfect physical model for the flip, but that's stretching it a bit. The point is that even if you wanted to, you could probably not repeat the exact same flip each time. "Picking" a number between 1 and 10 is not random. If you wanted to, you could pick the number 5 every single time. If you want to, you could pick from 1 to 10 counting up. Your choice does not constitute a random number. Let's think of it this way. I will make you bet for $100. I am going to pick a random number between 1 and 10. If that number is 7, you have to pay me the $100. If it's any other number, I'll pay you the $100. Would you take that bet? Please, I urge you to read on random number theory before continuing this argument, what you're saying is incorrect and it's not helping answer any of the questions in the thread.

lol

ok i can see this is going nowhere, i know a coin flip isnt random , BUT its pretty close too, and when dealing with probabilities, u have to assume it is random (for instance a finite course wouldnt exist in high school if we assume what u assume, which again IS correct, but we cant just go around saying that, cuz nothing would get done)

now, im trying to prove to u its true that its 100% prob by giving u examples and all u do is is say its not random so what u say is not true, then i say , ok give me something that is random (knowing full well that there is no such thing) and u say ok u dont know what u talking about so i win this arguement.

so fine, be that way, why dont u prove to me why it is 100% that u will pick an irrational number? and not just say that its true...believe me, i can argue any case that u can bring up

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quote:

Originally posted by Resnick hmm it seems that ur not getting my point either, let me make it nice and simple...probabilityes are basically x/y+x so like x = heads and y = tails.. in a basic flip of a coin, the probability of heads coming up is x / x+y so its 1/1+1 = 1/2 ...thats a really simple concept so now if u take a range, say [1,10] and pick a random number from there, then x = rational #'s, y = irrational numbers, no i know for a fact that there are rational numbers between 1 and 10, so lets just say there is one..even tho there are infinite but forget that. so 1/1+y is the probability, which means no matter what y is, the fraction CANT be zero (mind u y cant be infinite cuz thats not a number) and of course this is a really shitty proof cuz both x and y are infinity and u cant really evaluate it, BUT its the probability is not zero

Please try to understand what we are saying instead of being so contentious. Let me try to explain in another way. If this fails, I'll leave you to ask someone you trust (perhaps a math professor).

Suppose we try a different game. I have a number 1-10, and you try to guess it. You get no feedback from me except "you've won!" What's a winning algorithm? The optimal algorithm in this case is to guess each number from 1-10 once (the order won't matter). In the worst case, you'll win in 10 trials. If you decided to guess sequentially, then the number 10 would be your worst case. If you guessed in reverse order, then 1 would be.

Now instead, let's say you picked a random algorithm. Let's allow you to use a computer, even though it's at best pseudo-random. Would you be willing to bet your life that the computer will *necessarily* come up with the right answer? How many rounds must you let it go through before you're sure it will give the right answer? At stage N, it will have guessed right with probability 1-.9^N. There is no *finite* value of N for which this probability is equal to 1.

After infinitely many rounds, it will win. This is fully equivalent to saying "it will, at worst, win at the end of eternity," which is in turn equal to "it may *never* win."

quote:

Originally posted by Flyboy217 A set, by definition, has distinct elements . Your previous solution was nearly right. There are 1024 subsets of the initial 10-element set. Each must sum to a number between 1 and 1000 (more precisely, 55 and <1000). The 1024 subsets can therefore not all have distinct sums.

Oops... heh, what a dumb question that was.

Okay, so did you actually mean by the question that they sum to 1000 or less? Because I was interpreting it as having elements from 1 - 1000, which would include the set {991, 992, ... 1000 } which would add up to 9945 (that number is slightly off, too lazy to do the math).

So yeah, I'm second guessing myself here, it seems as though I answered the question but I'm not sure what question I answered anymore.

If that's the proof though, then it answers part (b) of the question too, should be 1024.

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ok for the genie problem, say u have 3 cups face down/up and 1 going the other way, all youd have to do is get that one cup face up and u win, thats the best way i can come up with..but that still doesnt gaurantee

solution: u tell him to flip all the cups between every turn, so that incase u get all cups face down by accident, u end up winning

now, either its 2:2 split or 1:3 split, if its 2:2 then u turn any one cup, which makes it a 3:1 split...

so u gotta 'guess' that one right, if u mess up, u get 2:2 split again and then u just turn one over and ur back to 3:1 split...so its an endless loop until u get that 1 cup by chance

again i know its wrong cuz it doesnt guarantee, but thats all i can get, more later tonight possibly...

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quote:

Originally posted by Flyboy217 Please try to understand what we are saying instead of being so contentious. Let me try to explain in another way. If this fails, I'll leave you to ask someone you trust (perhaps a math professor). Suppose we try a different game. I have a number 1-10, and you try to guess it. You get no feedback from me except "you've won!" What's a winning algorithm? The optimal algorithm in this case is to guess each number from 1-10 once (the order won't matter). In the worst case, you'll win in 10 trials. If you decided to guess sequentially, then the number 10 would be your worst case. If you guessed in reverse order, then 1 would be. Now instead, let's say you picked a random algorithm. Let's allow you to use a computer, even though it's at best pseudo-random. Would you be willing to bet your life that the computer will *necessarily* come up with the right answer? How many rounds must you let it go through before you're sure it will give the right answer? At stage N, it will have guessed right with probability 1-.9^N. There is no *finite* value of N for which this probability is equal to 1. After infinitely many rounds, it will win. This is fully equivalent to saying "it will, at worst, win at the end of eternity," which is in turn equal to "it may *never* win."

sorry flyboy, im not understanding what your saying, again at infinite is a limit, which u cant use...but anyways lets drop it, this is going nowhere, ill try to solve the other problems with a real solution

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quote:

Originally posted by Resnick lol ok i can see this is going nowhere, i know a coin flip isnt random , BUT its pretty close too, and when dealing with probabilities, u have to assume it is random (for instance a finite course wouldnt exist in high school if we assume what u assume, which again IS correct, but we cant just go around saying that, cuz nothing would get done) now, im trying to prove to u its true that its 100% prob by giving u examples and all u do is is say its not random so what u say is not true, then i say , ok give me something that is random (knowing full well that there is no such thing) and u say ok u dont know what u talking about so i win this arguement. so fine, be that way, why dont u prove to me why it is 100% that u will pick an irrational number? and not just say that its true...believe me, i can argue any case that u can bring up

Okay, I can give you a proof. The cardinality of the set of rational numbers is countably infinite. The cardinality of the set of irrational numbers is uncountably infinite. The division of a countable infinity by an uncountable one yields zero.

I promised you a link:
http://mathforum.org/library/drmath/view/52145.html

quote:

What is the probability that, if Wilhelm chooses a random number, it will be 2.07948756? I claim that it's zero: there are an uncountably infinite number of numbers (a set in the real line with infinite measure) that aren't 2.07948756, and only one number that is (a point in the real line, with measure zero). So the probability is Zero/Infinity (pardon my abuse of notation), which is zero. What if you only got to choose from numbers between 2 and 5? Still, the answer would be zero. What's going on in your example is that there are SO many numbers whose decimal expansion is, for all our purposes, infinite. In fact, all the irrational numbers and all the transcendental numbers are like that. It can be proven that the set of irrational numbers are very dense (MUCH more dense that the rational numbers, i.e. there are WAY more irrational numbers than rational) in the real line. So if you choose a random number from the real line, the probability of choosing an irrational number is 1. Now, that doesn't mean choosing a rational number is impossible, but it just means it's VERY unlikely if you're truly choosing at random. So since almost all of these irrational numbers have 987987987 in their expansion, the probability is 1. I hope this helps, and write back if it doesn't. -Ken "Dr." Math

I'm not sure if this was clarified earlier, but does the genie rotate it completely arbitrarily, or one position at a time?

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