so let me see if I get this right:to you, the mode of the distribution depends on N,but you already agreed that the average does not.However if you look at your graph, it is obvious that the average of your distribution depends on N: accordingly to your earlier posts the peak will move like Log N so therefore the expected value will also move !!–** contradiction **.(by the way, I was also wondering, how exactly do you define "mode" of a random variable that does not assume descrete values?...)AlexPSQuoteOriginally posted by: zerdnaYou are mixed up two different things. No Zerdna, I am not, I am simply trying to understand what you are computing. QuoteOriginally posted by: zerdnaIf you want any fixed value, you need to calculate an average of the first average, or some other characteristic of the distribution of average of N numbers - mode, median, etc. This first average over N number has a distribution, like any other random variable, you know? ….Zerdna,…… what exactly do you think I meant when I explicitly wrote “ E[(x1+x2+x3+x4...+xn)/n] “some posts ago? Is that not the average of the average?...

QuoteOriginally posted by: mikebell The problem is fairly complex and there is a LOT of assumptions that one can make. I dont think the problem is that complex:we are just computing the average of a function given by x/(1-x), for 0<x<1/2 (1-x)/x for 1/2<x<1where the measure of integration is not dx but f(x)dx.It has a unique solution given by There are an infinite number of assumptions one can make. Each one corresponds to a diferent measure, or probability density, f(x).And they will lead to diferent answers, in the interval [1, infinity[.Alex

Quoteso let me see if I get this right:to you, the mode of the distribution depends on N,but you already agreed that the average does not.However if you look at your graph, it is obvious that the average of your distribution depends on N: accordingly to your earlier posts the peak will move like Log N so therefore the expected value will also move !!–** contradiction **.What? Another contradiction? Tell me, how do you know that average depends on N from looking at the graph? Average of the sample average is infinite because its distribution function decays slow enough, say like 1/x^2. Maybe you could see that it does or doesn't from the graph, but i cannot. You should teach me how you do it.Quote(by the way, I was wondering, how do you define "mode" of a variable that does not assume descrete values?!!! )AlexMode is the value of random variable corresponding to the maximum of the density of its probability distribution. This Q and A session is getting boring. I am not sure i could convince you that sample average is a random variable, or that continious distribution has a mode, or any of these other things that seem "contradictory" to you and that i have said numerous times. Let's not argue about probability theory, just do a numerical experiment with drawings from the uniform distribution.

QuoteOriginally posted by: alexandreC….Zerdna,…… what exactly do you think I meant when I explicitly wrote “ E[(x1+x2+x3+x4...+xn)/n] “some posts ago? Is that not the average of the average?...dude, that was me, my point was, the average of a sample of 100,000 matches' ratios is itself a randomly distributed variable thats expecation, through the elementary identities, reduces to the expectation of a single match. Thus it suffices to consider one match, and whatever MB wrote about multichoose was irrelevant.(incidentally, i've not done any prob/stats for a loo~ng time - is this the situation where you can apply CLT and say that the average of 100,000 matches is normally distributed about the mean a single one (assuming it exists) etcetc??)I too don't think it's a particularly tough problem - pick a distribution and off you go...

I have just started looking at this problem. I agree insticntively that one should assume that breaks are uniformly distributed. What I find interesting is that the inverse ratio short/long is well behaved and one can calulate an expected valueof 2ln(2) - 1. I tried to use this result to estimate the original problem but have been unsuccessful thus far.

QuoteOriginally posted by: TripitakaQuoteOriginally posted by: alexandreC….Zerdna,…… what exactly do you think I meant when I explicitly wrote “ E[(x1+x2+x3+x4...+xn)/n] “some posts ago? Is that not the average of the average?...dude, that was me, my point was, the average of a sample of 100,000 matches' ratios is itself a randomly distributed variable thats expecation, through the elementary identities, reduces to the expectation of a single match. Thus it suffices to consider one match, and whatever MB wrote about multichoose was irrelevant.[…]Indeed, absolutely right.And you were the first one writing it- I did give you all the credit for it from the very start,read my post from Wed Jul 28, 04 06:13 PM.QuoteOriginally posted by: TripitakaI too don't think it's a particularly tough problem - pick a distribution and off you go...True. And, to be fair, I don’t know why am I wasting my time trying to explain the meaning of an average.There are people on here that think that “[refering to the number 24] This has to be understood as a mean of sample averages ” Zerdna, (Wed Jul 28, 04 05:02 PM)But then, when they realise they are wrong, and that the average is indeed infinite, they prefer to change the name of whatever they estimated,“What i estimated analytically is […] the value corresponding to the mode of the distribution of the average ratios” (zerdna, Wed Jul 28, 04 11:08 PM)but look, the author of this thread explained - more than once - that he is interested on expected values, or averages, not modes!!...QuoteOriginally posted by: TripitakaI too don't think it's a particularly tough problem - pick a distribution and off you go...Well, It boils down to the computation of an integral, so,No, its not a tough problem – at all!!

whoops - missed a couple of posts in the melee - understand now...

QuoteOriginally posted by: Tripitakawhoops - missed a couple of posts in the melee - understand now...ehehehe no worries! Alex

Quotesorry, I am not going to teach you that, why dont you ask your high school teacher instead?In my school there were no teachers who could do that, we went to different schools. Quote(by the way zerdna, to choose a picture of Richard Feynman, Nobel prize in Physics, absolute genious and father of Quantum Electrodynamics, to represent you... very humble of yours no?)I like Feynman, and like this picture, that's the reason for my choice of the icon. I am neither humble nor arrogant. However, i am somewhat busy and get a bit annoyed with combinations of ignorance and agressiveness. Therefore, i suggest you just leave me alone.

“Without a computer, and giving it about 5 secs thought”:Average length of the long piece =3/4, short piece = 1/4. As a rough estimation: The average ratio of the length of the longer piece to the length of a shorter piece = (Average length of the long piece)/(Average length of the short piece) = (3/4)/(1/4) = 3.

I make it 2.588

QuoteOriginally posted by: zerdnaHowever, i am somewhat busy and get a bit annoyed with combinations of ignorance and agressiveness. Therefore, i suggest you just leave me alone.{edited}yeah I suppose.

QuoteOriginally posted by: daveangelI make it 2.588i.e. 1/(2ln2-1)?if so, i think that's a very dodgy calculation to make:outcome 3 6prob 0.5 0.5 has an expectation of 4.5and 1/outcome has an expectation of 1/4

QuoteOriginally posted by: EStealth“Without a computer, and giving it about 5 secs thought”:Average length of the long piece =3/4, short piece = 1/4. As a rough estimation: The average ratio of the length of the longer piece to the length of a shorter piece = (Average length of the long piece)/(Average length of the short piece) = (3/4)/(1/4) = 3.good call for a first attempt but...I think that that is the ratio of the averages, not the average (of the averages) of the ratios.Alex

QuoteOriginally posted by: alexandreCgood call for a first attempt but...I think that that is the ratio of the averages, not the average (of the averages) of the ratios.AlexAs I said "a rough estimation".