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JollyJoker
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posted June 25, 2010 05:35 PM |
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Ohfor, that's simple (or maybe not so simple) mathematic.
With n coins we habe 2>n possibilities
For even-numbered n the number of even-split possibilities is
n!/[(n/2)!]>2
Consequently,
{n!/[(n/2)!]>2}/2>n will deliver a series: (you can substitute 2n for n which will allow you to count off all n)
1/2 for n=2
3/8 for n=4
5/16 for n=6
35/128 for n=8
and so on.
If you look at this series, you'll see that the limes of this series is 0 for n->infinite. You can prove that with an Epsilon-surrounding, I'm sure, but an easier way is, to insert (n+2) for n in the equation above (going from the assumption that n is even- numbered) and dividing the n equation by the n+2 equation:
((n+2)!/[((n+2)/2)!]>2}/2>n+2) / {n!/[(n/2)!]>2}/2>n
This can quite easily be simplified to
(n>2 + 3n + 2)/(n>2 + 4n + 1)
This term is smaller than 1 which means that the chances indeed get smaller with n getting higher, until , for n-> infinite it's ONE split against an enddless number of possible other splits and probability is virtually non-existant.
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ohforfsake
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posted June 25, 2010 05:38 PM |
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I thought continous as well as discrete distributions approximated gauss curves better and better for higher amount of measurements and gauss curves width, or uncertainty, or standard deviation or what it is called, approaches zero for an infinite amount of measurements hitting the expected value at infinity (which here is 50/50). Though maybe I've misunderstood something.
@JJ Maybe it also depends on how one understands infinity. I have always understood it as a "place/point" where all combinations have happened at the frequency expected.
I understand your formulas and I agree with the way you divide the binomial coefficients (or what they're called), but I just don't think you can let the limit go to infinite and conclude it's zero, I don't think your function is defined in that area, I don't think so because of the way I understand infinity.
Edit: If I should guess, I'd say when you let n go to infinte, you simply evaluate over which function increases the fastets (like one normally do) see that the demoninator increases faster and determine at infinite it's zero chance, but I think your function only works for finite series, because it's based on finite series. As an example, your function at infinite, can't be any combination, because the chance for any given combinations becomes zero, because there are infinite many possible combinations. That at least seem not right to me, and rather I'd conclude it means that there's no infinite finite series.
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mvassilev
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posted June 25, 2010 05:39 PM |
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JJ:
Price collusion? A basic look at game theory will tell you why that fails. Each company has an incentive to undersell, and so they do. These kinds of agreements fall apart quite quickly.
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Of course, GOVERNMENT is a job as well and that job follows the same profit-greed-self-interest rule, so there is no reason why the government shouldn't cheat as well.
And that's why we need an educated electorate that keeps government small and under close supervision.
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Not to mention the fact that each and every business idea and new product would actually have to be tested thoroughly for long-term effects - except that it isn't because no one can affortd (!) to wait that long, meaning, the system is HIGHLY LIKELY to produce MASSIVE hazards.
Everything has a risk. There's much to be lost if the government were to ban risky things.
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JollyJoker
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posted June 25, 2010 05:59 PM |
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@ Ohfor
It's just a limes of a simple series, which is well defined.
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JJ:
Price collusion? A basic look at game theory will tell you why that fails. Each company has an incentive to undersell, and so they do. These kinds of agreements fall apart quite quickly.
Not so. Game theory doesn't work, because it doesn't matter who moves first. All examples about incentives and so on assume that ONLY ONE (the first) will hit the jackpot. This is different in business: If an Oil company drops prices against agreement by 5 cents, 10 minutes later the other drop their prices as well.
So, no, game theory doesn't apply.
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Of course, GOVERNMENT is a job as well and that job follows the same profit-greed-self-interest rule, so there is no reason why the government shouldn't cheat as well.
And that's why we need an educated electorate that keeps government small and under close supervision.
Except that they can and will cheat as well, if it's worth the while.
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Not to mention the fact that each and every business idea and new product would actually have to be tested thoroughly for long-term effects - except that it isn't because no one can affortd (!) to wait that long, meaning, the system is HIGHLY LIKELY to produce MASSIVE hazards.
Everything has a risk. There's much to be lost if the government were to ban risky things.
Oh, PLEASE. No risk no fun, right? You have to break an egg to make an omelette. I'm sure, we can find a couple more empty husks of general clichès, if we look long enough.
You can't seriously think, such a general non-statement is a point.
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mvassilev
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posted June 25, 2010 06:05 PM |
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If an Oil company drops prices against agreement by 5 cents, 10 minutes later the other drop their prices as well.
Exactly, which is why it's good for consumers. And yet, because of game theory, there is still that incentive to cheat. That's just how it works.
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Except that they can and will cheat as well, if it's worth the while.
How can an educated populace of voters "cheat"?
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I'm sure, we can find a couple more empty husks of general clichès, if we look long enough.
It's a cliché, but it's also true. You take a risk every time you get into a car. So should cars be banned?
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ohforfsake
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posted June 25, 2010 06:14 PM |
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Quote:
@ Ohfor
It's just a limes of a simple series, which is well defined.
Yeah, I just don't think you find the chance of an even split when you got infinity.
Here's an analogy that hopefully will make my critique easier/funnier to follow:
I am going to pick out a number between 0 and infinite, what are the chance I pick any given number? Well the chance for any given number is zero, because there are infinite more numbers that could have been picked in stead. In fact, if I ever were to pick a number, it'd in itself be infinite large, because for any given finite sequence of numbers, there's always infinite as many bigger numbers than there are smaller numbers, meaning the number would be infinite.
Now what they I actually do there? Well I took actually infinite many 9-sided dices (since the number picked out must be infinite big) and put them all together in a box that can carry infinite many 9 side dices, then I threw them all out at random producing 1-9. However as I get a single number, I must know them all at once, but since no finite series can be infinite big, I can't get any number via this method.
With finite series, I don't think there's any difference of taking n dices and threw them one at a time or all together as long as they are randomly choose between 1 to 6 with the same frequency.
With an infinite series, I think there's a huge difference of taking infinite many dices/coins and throw them all out at once (like I believe is what you did) and the alternative of going through them one at a time.
So if I were to pick a random number from 0 to infinite that'd not be possible, but if I were to pick out one number at a time, and all numbers have the frequency of 1/9, then at an infinite series, I'd have gone through all possible combinations at the amount of times the frequency tells me, that is, I'd for every nine numbers in average have one of each number for the infinite series.
So I don't think you find the chance of an even split via your method, rather you find the chance of any split at all if you throw all dices out at once and since there's infinite many ways, no decision can be made, however going through it the way of taking one dice at a time, you'll keep on throwing forever, but at infinite all frequences are true.
Well those are my thoughts on the matter.
Edit: Just as another example, anyone can claim the divergence of 1/r^2 is zero, and it is for all points except 0, so a conclusions that includes zero would be wrong, eventhough the method is solid for all other numbers than zero.
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JollyJoker
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posted June 25, 2010 07:26 PM |
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Ohfor, that's not how it is working.
It's
n->infinite
and NOT
n=infinite
What we do here is a series or sequence of numbers, and we ask where this sequence will end. The limes of a sequence or series is defined this way:
1) The sequence won't ever reach the limes (there is no n delivering the value of the limes)
2) If you add a fraction of a value - called Epsilon - that may be as small as you want, you can find an n that delivers a value between limes and lines + epsilon.
Simple example: 1/n
The limes n-> infinite of this series is 0 - even though NO n WILL EVER make 1/n 0.
Same thing here. The LIMES of the series given by the percentage of even-split possibilities to all possibilities is 0. That doesn't mean the percentage will ever BE 0 for any given 0. It just means that it gets ever smaller.
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del_diablo
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posted June 25, 2010 07:35 PM |
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mvassilev: Please answer somebodies statement from last page, explain why its wrong:
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Prices rise as competition drops out and monopolies form though.
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mvassilev
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posted June 25, 2010 07:40 PM |
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del_diablo:
It's not wrong. That statement is quite correct. The less competition there is, the higher prices are.
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JollyJoker
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posted June 25, 2010 07:43 PM |
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Quote:
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If an Oil company drops prices against agreement by 5 cents, 10 minutes later the other drop their prices as well.
Exactly, which is why it's good for consumers. And yet, because of game theory, there is still that incentive to cheat. That's just how it works.
It's not, and you demonstrate that you don't know much about game theory. As I said the assumption in gaming theory is, that first is only. Your game theory was valid, if only the first to drop prices would get the deal.This isn't the case here, however.
That's why it is just the other way round. Provided there is a limited and known number of competitors there is ALWAYS the incentive for price agreements - because IT DOESN'T HURT.
In THEORY, the system counters unreasonably high prices with NEW competitors - but of course that doesn't work when the existing comptetitors are big enough.
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Except that they can and will cheat as well, if it's worth the while.
How can an educated populace of voters "cheat"?
Depends on what you mean exactly. If it's a small group they can be bought. If it'ss a big group they can be cheated.
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I'm sure, we can find a couple more empty husks of general clichès, if we look long enough.
It's a cliché, but it's also true. You take a risk every time you get into a car. So should cars be banned?
Counter-question: A guy hands you 1.000.000 bucks and a six-shooter. He says, one chamber is loaded, and you can keep the money, if you rotate the drum, put that gun to your head, pull the trigger and survive.
Now what?
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ohforfsake
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posted June 25, 2010 07:54 PM |
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@JJ
I don't know what I wrote that made you find it relevant to specify limits of a variable going towards infinite is not the same as setting the variable equal to infinite. I know it's about growth rate, that's how you got to your very first conclusion. However following that conclusion, every possible outcome is impossible, because there's always infinite more possible outcomes.
Again, I am convinced what you do is to throw all coins at one and then since there's no finite infinite of course you get the answer of no possible outcome.
The way of taking one coin at a time however always generates a outcome and for the finite serie it generates the same outcomes as throwing them all at one, but going to infinite, you don't remove and rethrow, no you add a new value, either H, frequency 1/2 or T, frequency 1/2 to the previous set, after having done so forever, i.e. at infinite, you have all possible combinations of any number of throws at the frequency expected. Looking at the isolated H contra T, you've both at frequency 1/2, so both are equally present. Looking at HH, contra TT, contra HT, contra TH, again all are at frequency of 1/4th and all HH's and TT's suits the previous claim of equal H's and T's. Continueing, all HHH's, HTH's, etc. and again for every HHH you have a TTT at the same frequency.
I don't mind being wrong, I write here in the hope of I might be wrong and can better my knowledge on this field, but honestly, I don't think you address the problem I'm trying to show here.
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JollyJoker
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posted June 25, 2010 08:10 PM |
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I have no idea where your problem is - you never actually are at "infinite" - you always have a finite number.
Is your problem the idea, that what I wrote is calid only when I flip one coin after another instead of flipping a number of coins at the same time? There is no difference, because you can just number the coins - just because you don't see the individual markings doesn't mean they are not there.
In short, I don't see your problem.
Edit: I read your post again, and sorry, I just don't understand what you describe and what your problem is.
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ohforfsake
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posted June 25, 2010 08:19 PM |
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Actually, I think it's opposite, your method flips all coins at once, I believe.
Anyway, my problem is that I think the value you get for going towards infinite is not correct. I don't think so, because what you get is that any combination is impossible, while when going one throw at a time you get through all possible combinations. That is, I think you use the same rules for finite set on a set you let go to infinite, and I don't think it's correct for the reasons stated.
I don't say anything magical happens at larger and larger values, I think you're completely correct in any finite set the coins will be less and less likely to have a 50/50 distribution, eventhough I'd expect otherwise [I'd actually expects the histogram to become closer and closer to a continous function, forming a gauss curve that gets thinner and thinner the more times you throw. It could be interesting to find out what actually happens though]. However when you keep on throwing, that is letting the set go towards infinite, you get all sets repeating at the expected frequency, when throwing forever you'll as often his H as T, you'll as often hit HH as TT as HT as TH, you'll as often hit HHH as HHT, as HTH, as THH, as TTH, as THT, as HTT, as TTT and all in all that means an even split when going towards infinite.
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del_diablo
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posted June 25, 2010 08:20 PM |
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mvassilev & JollyJoker: So you are unable to clarify what "markeding conditions are", agree on a shared definition to have a conversation over, and both fail to aknowledge and attempt to use absolute arguments.
So how many businesses do you need to compete within a area to have a proper working economy?
How frail is this structure?
How long would it take for the structure to fall apparent if there was no regulations?
How frail is really our current economic variantions?
Etc?
At the least bother to define why the other person is wrong, and how they are wrong within certain parameters.
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JollyJoker
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posted June 25, 2010 09:26 PM |
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@ del diablo
We've both done that.
@ ohfor
I still can't identify your problem. There is no infinite coin flipping, it's always finite, and there is always a finite number of possibilities.
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mvassilev
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posted June 25, 2010 10:02 PM |
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JJ:
Quote:
As I said the assumption in gaming theory is, that first is only. Your game theory was valid, if only the first to drop prices would get the deal.
The first to drop prices does get the deal.
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Depends on what you mean exactly. If it's a small group they can be bought. If it'ss a big group they can be cheated.
That' why they're an informed and educated group of voters. It's better to have no one subsidised than to have everyone subsidised, and they all know that. Thus, if any small group tries to get a subsidy, the rest will vote it down.
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Counter-question: A guy hands you 1.000.000 bucks and a six-shooter. He says, one chamber is loaded, and you can keep the money, if you rotate the drum, put that gun to your head, pull the trigger and survive.
Now what?
This is actually a very interesting question. No, I probably wouldn't pull the trigger unless there was a greater risk that I would die if I didn't get the money. But I'm a risk-averse person. I'm sure there are many adventure-seekers who would pull the trigger. And that's fine by me - I'm not one to tell them that they can't.
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JollyJoker
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posted June 25, 2010 10:40 PM |
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Quote:
JJ:
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As I said the assumption in gaming theory is, that first is only. Your game theory was valid, if only the first to drop prices would get the deal.
The first to drop prices does get the deal.{/quote]I only can say say no to this. This is just not true.
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Depends on what you mean exactly. If it's a small group they can be bought. If it'ss a big group they can be cheated.
That' why they're an informed and educated group of voters. It's better to have no one subsidised than to have everyone subsidised, and they all know that. Thus, if any small group tries to get a subsidy, the rest will vote it down.
It seems you don't know much about cheating. The only important question is whether YOU are subsidised.
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Counter-question: A guy hands you 1.000.000 bucks and a six-shooter. He says, one chamber is loaded, and you can keep the money, if you rotate the drum, put that gun to your head, pull the trigger and survive.
Now what?
This is actually a very interesting question. No, I probably wouldn't pull the trigger unless there was a greater risk that I would die if I didn't get the money. But I'm a risk-averse person. I'm sure there are many adventure-seekers who would pull the trigger. And that's fine by me - I'm not one to tell them that they can't.
This may not be fair, but on the other hand... You missed the point: You accepted the deal and the "risk parameters". If you read again, you'll see that I wrote HE SAYS that one chamber is loaded.
So the answer to NOW WHAT is: I CHECK THE CHAMBERS!
When you risk something, you have to be sure about the odds, which is the problem here (no one klnows the odds). It's like jumping blindly into the darkness, which isn't risk but insanity.
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posted June 26, 2010 05:58 AM |
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Guys this is in 'Statistics 101' back to the topic at hand.
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JollyJoker
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posted June 26, 2010 09:10 AM |
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Last comment on it:
@ Ohfor
Gauß Curve or Normal Distribution is of course valid. Even split is at the apex of the curve. Your expectation would be in agreement with the curve.
The effect of adding always more flips, is (among others) that you add more cases within the standard deviation:
When you flip the coin 1.000.000 times a 498.877/501.123 split isn't so much less likely than the even split, and so are a lot of other splits.
With only a couple flips, like 10, the deviation is bigger for each different split, since the density of the curve is low.
The curve is valid in the infinite as well, though (which was your point, if I'm not wrong: shouldn't, in the infinite, expectation become 50-50).
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shyranis
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posted June 26, 2010 02:16 PM |
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Back to the topic at hand, all of Obama's political decisions so far have been roughly on par with every other President in the last 50 years in terms of Communism.
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