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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.

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.

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

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.

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...

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

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?

quote:

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

Completely arbitrarily. (Equivalently, he is omniscient, and completely maliciously). As for the set problem, I cannot believe what I did. It was originally listed as [1, 100], and then I edited it to be [1, 1000]. I've fixed it. :-P

quote:

Originally posted by Flyboy217 Completely arbitrarily. (Equivalently, he is omniscient, and completely maliciously). As for the set problem, I cannot believe what I did. It was originally listed as [1, 100], and then I edited it to be [1, 1000]. I've fixed it. :-P

Ahh okay. So for part (b) I think the answer should be 113 (highest elements sum to 1085, 1085 - 55 = 1030, which is greater than 1024).

I still can't get the genie one, gah.

quote:

Originally posted by DigiNut Ahh okay. So for part (b) I think the answer should be 113 (highest elements sum to 1085, 1085 - 55 = 1030, which is greater than 1024). I still can't get the genie one, gah.

Nope . What your reasoning shows is why Dirichlet's principle ("the pigeonhole principle") cannot be used to prove part (a) with the interval [1,113]. If it were this simple, I wouldn't have suggested 2 hours .

As for the genie one, here's a hint. Try for the case n=2. You can do it in 3 steps worst-case. That should get you started.

Re: Smart?

quote:

Originally posted by Flyboy217 3a) You are blindfolded with a round table in front of you. It is marked with 4 positions (1, 2, 3, and 4), one in each quadrant. There are 4 cups on the table, one per quadrant, each initially randomly either face up or face down (which you cannot see). On each turn, you may instruct the genie to flip the cups in whichever numbered positions you wish. He will oblige, and then rotate the cups around the table as he wishes, so that the cups are (potentially) all in new positions, but in the same rotational order. Your goal is for all the cups to be face up. The genie will tell you if you've won. Can you give a solution that will always win the game, regardless of his sneaky rotations? [30 min]

following algorithm:

flip all
flip any random cup
flip all
flip any random cup

you are BOUND to win, because you are exhausting the space of possibilities. Explanation -somewhat shitty-
if all cups are down and you flip them all, you win. - easy
if they are NOT all down and you flip them all, you ELIMINATE that possibility. By flipping a SINGLE cup, you are eliminating the possibility that THIS one cup was down and the others were up. But that elimination is only true for this one cup in that one spot. It does not matter which cup you flip, since the genie can arbitrarily rotate them, but since the ORDER does not change, you will -eventually- test every cup.
Next you flip all cups again. If you don't win, flip a random single cup again. then flip all, flip single.

at leas that is what i came up with. Eventually you should win, unless the genie HAPPENS to rotate the cups so that you always pick the same cup to flip as a single cup. SO i guess this isn't right, but i think i am on the right way to getting there. Maybe you need to alternate somewhat more like:
flip all
flip first
flip all
flip 2-3
flip all
flip 4-2
flip all
flip 1-4
flip all

so that eventually you will have flipped all pairs and triplets of cups.

Re: Re: Re: Smart?

quote:

Originally posted by DJ-Fuq imo, these r badly worded mostly

dude i totally agree

Re: Re: Smart?

quote:

Originally posted by drizzt81 so that eventually you will have flipped all pairs and triplets of cups.

thought some more:

you definetely need to flip two adjacient ones and then two 'across' ones like 1-2, 1-3, between flipping all four so that you are _guaranteed_ to flip at least one different cup every time, since they are in he same rotational order.

edit:

so I am getting there:

1. flip all -either we win, or we know that there are 1-3 down cups
options:
DUUU
DUDU
DDUU
DDDU

2. now we flip an ADJACENT pair which gives us the following options for each:
DUUU -> DUUU or DDDU (or rotation thereof)
DUDU -> UDDU
DDUU -> DDDD or UUUU (win) or DUDU
DDDU -> DDDU or UUUD (or rotation)

3. now we flip all again, which eliminates the 'red' option above and leaves us with these:
DUUU
DDUU
DUDU
UUUD

4. because we do not know how the geenie has rotated them, we cannot just continue repeating step 2+3, which would always kill two opportunities, so we need to be a little inventive and now flip a NON-adjacient pair to eliminate another oppotunity:
DUUU -> DDUD or UUDU
DDUU -> UDDU or DUDU
DUDU -> UUUU(we win) or DDDD
DDDU -> UDUD or DUDD

5. flip all, to once again eliminate the red option. this COMPLETELY eliminates that ROW! that means that we have 'killed' of this brach of our 'options tree'.
If we keep operating with this routine, we will win. Each time that we complete step 2-5 we will reduce the problem size, since we will eliminate one branch. Therefore, by induction, this algorithm will terminate by us winning.

Re: Re: Re: Smart?

quote:

Originally posted by drizzt81 thought some more: you definetely need to flip two adjacient ones and then two 'across' ones like 1-2, 1-3, between flipping all four so that you are _guaranteed_ to flip at least one different cup every time, since they are in he same rotational order.

Impressive. You've got all the right ideas. I imagine it wouldn't take you long to formulate a complete winning strategy if you tried. In fact, it's close enough that I'll give one solution.

SPOILER







1 = Flip cups that are across
2 = Flip adjacent cups
3 = Flip any three
4 = Flip all

4, 1, 4, 2, 4, 1, 4, 3, 4, 1, 4, 2, 4, 1, 4

Can you find another?

Re: Re: Re: Re: Smart?

quote:

Originally posted by Flyboy217 Impressive. You've got all the right ideas. I imagine it wouldn't take you long to formulate a complete winning strategy if you tried. In fact, it's close enough that I'll give one solution. SPOILER 1 = Flip cups that are across 2 = Flip adjacent cups 3 = Flip any three 4 = Flip all 4, 1, 4, 2, 4, 1, 4, 3, 4, 1, 4, 2, 4, 1, 4 Can you find another?

oops read this 'after' i devised the post above :/

I was thinkin whether i need to flip three cups, just to make sure that i am not flipping two adjacient cups again. I understand that the problem will be that you can be caught in a loop for a while, if you happen to retiterate the same set of cup flips over and over again, but i though -for some reason- that you would always reduce the problem size.

now I am back to my REAL work.. I 'hate' you for giving me these problems, i should have been coding instead of trying to solve "tea party" problems

j/k

Re: Smart?

quote:

Originally posted by Flyboy217 b) For what values of N is this possible (in part a, N=4)? [No limit.]

only for integers
I think it might actually fail for N=6 already, because 2 and four have a GCF of >1, but i am going on a hunch here.

Re: Re: Smart?

quote:

Originally posted by drizzt81 only for integers I think it might actually fail for N=6 already, because 2 and four have a GCF of >1, but i am going on a hunch here.

It in fact fails for n=3. Not sure where you're going with the GCF of 2 and 4, but since I myself never solved this one, I can't comment . I'm told the solution is non-trivial. I'll get on it soon .

quote:

Originally posted by Flyboy217 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

wow, i cant believe we argued all this time just for this.

ok the very first thing i said in response to this was that ur using limits, and that the answer would tend to be zero, BUT truly it is really really small (which is stated in the article btw).

heres the quote from the proof you gave:

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.

Re: Re: Re: Smart?

quote:

Originally posted by Flyboy217 It in fact fails for n=3. Not sure where you're going with the GCF of 2 and 4, but since I myself never solved this one, I can't comment . I'm told the solution is non-trivial. I'll get on it soon .

it fails for 3, yet works for 4? that blows my mind.

quote:

Originally posted by Resnick heres the quote from the proof you gave: 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.

yes but that is what he said.. and in the limit it approaches 1 (i.e. 100%) it seems quite clear to me

quote:

Originally posted by drizzt81 yes but that is what he said.. and in the limit it approaches 1 (i.e. 100%) it seems quite clear to me

nonononon

heres how this discussion started:

flyboy said that it was 100% to pick irrational, i said no its not, its only a limit, then he said no its 100% everytime no matter what, and diginut and i argued this for a while,

so now im saying yes it is a limit and flyboy and diginut dont know what theyre talking about.

Furthermore, i would like to see someone actually prove this limit (im not saying that its not true, just saying that if ur so confident in ur answer, you should be able to mathematically prove this)

quote:

Originally posted by Flyboy217 Not really sure what you mean. The actual value of the probability of choosing an irrational number is 1. Not just close to 1, but 1. You will, with 100% probability, pick an irrational number. But you still may pick a rational one. (This sounds strange, but read on).

^^^ thats his exact quote in reply to my limit answer originally

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