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Well, I have a new theorem about random number. Just before publish it, I wish to know whether any others know about this, so please, those brilliant guys and girls, try your best to answer it.

Given a situation that, for a total choices of n, each choice has a same probability to be played, ie 1/n. After t times of plays, which t is a very large value, if not infinity, is it true to says that, ALL choices of n will have same frequency of occurance, ie t/n (t * 1/n) after t plays?

Please takes note that, the t is a very very large value, and we predict it in a way of mathematically, not through experiments or computer programs.

2006-07-20 05:18:41 · 11 answers · asked by wyeechen 2 in Science & Mathematics Mathematics

Alright. let says, t is a value greater than 1000000000000000000000.

2006-07-20 05:25:08 · update #1

What happen if I say no? Do you think it is worth to publish?

2006-07-20 05:26:45 · update #2

11 answers

How is this a new theorem? And how does it regard "random numbers?"

All you're asking is about probability... Nothing terribly "new" here. Sorry.

What happens if you say no to what?? If you say "no, it won't be evenly distributed?" It wouldn't be extraordinarily surprising. Since probability is only that. As the number of samples increases, the distribution APPROACHES even distribution, but it will rarely if ever ACTUALLY reach even distribution, simply due to the random nature of the tests. There's no way to FORCE a specific result in a random system to make it exactly even, I think.

So, say you have 10,000 results, and 10 possible outcomes with equal probability (say a 10-sided theoretical die):

You may end up with

Result 1: 1000
Result 2: 1000
Result 3: 999
Result 4: 1002
Result 5: 998
Result 6: 1001
Result 7: 995
Result 8: 1001
Result 9: 1004
Result 10: 1000

Now, if you take the percentage of each:


Result 1: 10%
Result 2: 10%
Result 3: 9.99%
Result 4: 10.02%
Result 5: 9.98%
Result 6: 10.01%
Result 7: 9.95%
Result 8: 10.01%
Result 9: 10.04%
Result 10: 10

So, we're approaching even distribution... But it's still not "evenly" distributed. As we get higher numbers we'll maybe get slightly closer to even. But random result will give random answers. You can't force a non-random result out of random numbers, I think. So, you really can't "force" an even result.

And you can't divorce the theory from practical application. If you can't PRACTICALLY apply the theory to test it, it's not science. Science requires falsifiability. So, you must be able to test it or it's worthless. Otherwise you're basically just asserting the world is flat without any proof or investigation of whether the assertion is accurate. So, to say that you can divorce a "theorem" from "reality" doesn't make sense scientifically. A theorem should describe some "real" cause and effect relationship, and be applicable to some real situation or it's basically not a useful model because you're not describing anything that exists. if that makes sense?

My opinion, of course.

2006-07-20 05:38:10 · answer #1 · answered by Michael Gmirkin 3 · 2 0

I don't see anything new for the theorem about random number. (or I don't get your meaning)

Given a situation that, for a total choices of n, each choice has a same probability to be played, ie 1/n.

True, because it is uniform distribution.

After t times of plays, which t is a very large value, if not infinity, is it true to says that, ALL choices of n will have same frequency of occurance, ie t/n (t * 1/n) after t plays?

Since it is uniform distribution, the chance of each choices (1/n) are same. If you put frequency of occurance (t) * chance of each choices (1/n). You will the number that the choices will the occurance in t times of plays.

It is a very basic common sense in applied statistics, isn't it ?

2006-07-20 06:46:46 · answer #2 · answered by Can You Keep A Secret 1 · 0 0

Well, of course, in binomial (Bernoulli) distributions, we see that this "t" could be any number, but is probably relatively large, in order for the distribution to more closely approach a normal distribution. For instance, the number of times until a fair two-sided coin is flipped enough so that the heads to tails ratio is 1 could be a lot. However, we must confess that there is some finite "t" that satisfies this condition, because we would not be flipping the coin for an infinite time period. I doubt, however, that this "t" is the same in every situation; the chances of this seem slim. (Then again, I may have messed up somewhere in my interpretation of your question.)

2006-07-20 09:34:12 · answer #3 · answered by Captain Hero 4 · 0 0

It is *false* that all the choices will have frequency t/n. In fact for many values of t and n this is impossible, since t/n would not be an integer.

What is *true* is that with "high probability" the measured frequencies will be "close" to this number. (There are equations quantifying what "high probability" and "close" mean.)

Not to dissapoint you, but this is an absolutely standard fact in Probability/Statistics. There's no possibility a journal would publish it.

2006-07-20 07:43:49 · answer #4 · answered by Aaron 3 · 0 0

Even if t is arbitrarily huge (but still finite), there's no guarantee whatsoever that each choice will have been chosen an equal number of times (for example, if t is not divisible by n, it's impossible).

But I think it's safe to say that the limit, as t approaches infinity, of (number of times object x is chosen)/t is 1/n.

2006-07-20 05:35:36 · answer #5 · answered by Matt E 2 · 0 0

What is your theorem?


"ALL choices of n will have same frequency of occurrence"
This is not true. They will all have the same probability of having t/n occurrences, and if n divides t, then an equal frequency would be the most probable outcome, but would be highly improbable in general.

2006-07-20 05:27:22 · answer #6 · answered by Eulercrosser 4 · 0 0

Given that t represents any number greater than one, and since you say it is a rather large number, it could in fact represent ANY number greater than one, unless you gave it a parameter of say, more than a given number. This is an excellent question!

2006-07-20 05:22:59 · answer #7 · answered by Anonymous · 0 0

answer: The risk of commute time more desirable than 50 minutes = 21.19% (a million-0.7881). Why??? time-honored DISTRIBUTION, STANDARDIZED VARIABLE z, risk "look-UP" STANDARDIZED VARIABLE: z = (x - µ)/(?) = (50 - 40 six)/(5)) = 0.8 pattern mean: x = 50 inhabitants mean: µ = 40 six inhabitants popular DEVIATION: ? = 5 pattern length: n = a million significant DIGITS = 4 The table for popular time-honored Distribution is prepared as a cummulative 'section' from the LEFT similar to the STANDARDIZED VARIABLE z. the classic time-honored Distribution is likewise symmetric (referred to as a 'Bell Curve') meaning its an interpretive technique to look-Up the 'section' from the table. For STANDARDIZED VARIABLE z = 0.8 the corresponding LEFT 'section' = 0.7881. and with the help of table's cummulative nature, the corresponding excellent 'section' = a million - 0.7881

2016-11-06 21:24:47 · answer #8 · answered by ? 4 · 0 0

i think you need a few more courses in probability and applied statistics before coming up with some theorem.. what you are saying doesnt make much sense.

2006-07-20 05:51:07 · answer #9 · answered by Annie 4 · 0 0

the law of the big numbers garantees that
the limit of
(number of occurences of i_th choice) / (number of
trials)
as number of trials -> inf = 1/n
It does not garantee that the frequence of all events becomes the same.

2006-07-24 00:47:48 · answer #10 · answered by gjmb1960 7 · 0 0

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