# Rsgoldfast is a cheap OSRS gold online store

Posted by Nfkjasfas on May 17th, 2021

Individuals who wants to play RS3 and OSRS at precisely the exact same time do that and pay x2 and OSRS buy gold for it in the 2nd account. Even if this attribute for paying double was discretionary (cuz it would amazingly unfair to charge double for the folks that dont play OSRS and dont even have interest in enjoying with it ), why could jagex squander time and attempt to alter and adapt such system when they'd get nothing out of it?

Jagex is a business, theres no economical incentive at all for them to do that. That would mean that they need to copy the player database, so that they can let 1 accounts login at precisely the exact same moment. It would be almost impossible. Regrettably infrastructure/servers don't work (well) this way.

Whould they will need to replicate the entire database? Login performance is (or should be) separated from the playerstate. There is absolutely no reason to keep the playerstate of both games at precisely the table. If your login funcionality is broken of as an agency, this shouldn't be hard to do. Biggest reason is for host ability. If you are logged on both versions, you use more server resources while you only pay for a single membership.

Kc's needed to complete OSRS bosses, I graphed the completion kc's and simulated 1 countless completions of OSRS bosses. This way it is possible to observe the spread of completions. Since completion can be somewhat vague I mimicked a couple of different options. By default getting 1 of drops unique into the boss is assumed excluding any jars/pets. Calculating average completion is really hard to do when two drops have distinct weighting. Additionally, it does not show the variance, which I guess you should be able to calculate but I'm not sure how.

Indeed for those that don't know this is very hard since it is a generalization of the coupon collector problem that's when all the drops have equal probabilities. But even then forms don't have closed. Therefore the general case (namely this one) is sort of impossible.

On is take all ordered lists of numbers. Assuming you have an easy tactics to generate the lists, this is going to be an computation that is easy enough. (An easy way to do so is to have a list of span l-1 with repeats at the amounts without I where every other amount seems at least once then append I to the end).

It's also worth noting that the complex parts of the coupon collector's problem only arise since the traditional difficulty attempts to calculate the mean, variance, as well as the pmf right from a pmf using analytic procedures, which makes real ugly since there are lots of nested, iterative calculations. Computational calculation going pmf of only thing - cdf of single thing - cdf of all - pmf of all is a great deal of dull impossible to do by hand mathematics, but is computationally very simple. A computer can quickly compute this without approximations. View my comment above to find the computational solution. This computational solution won't get you the mean or variance, but it will get one of the cdf graph to match the picture above and buy old school runescape gold.