"Let's Catch all the Shiny Spindas! (a fun math exercise)"

EDIT: My formatting should all be fixed but please let me know if there are any more errors.Earlier, for whatever reason, I was thinking about the pokemon Spinda. And how many forms it has. And how rare it is to find a shiny one. It got me wondering - how long would it take to catch a shiny form of every single Spinda pattern?THE SETUP:This question can be answered fairly easily with math, but first we'll need to outline our problem completely. We know that there are 4,294,967,295 unique Spinda patterns, and we know that, as of Gen VI, there is a 1/4096 chance that any given Spinda will be shiny. Now we need to consider how quickly we're going to be catching these Spindas. I don't want this to take forever, so let's say we have one million people, each catching one Spinda every minute.STEP ONE: SHINY CATCH RATEFirst, we need to figure out how frequently our one million helpers are catching a shiny Spinda. We're getting 1,000,000/60, or 16666.67 Spindas per second. That means we're getting 16666.67/4096 = 4.06901 shiny Spindas per second. So, our shiny Spinda catch rate is 1/4.06901 = 0.24576 seconds. Essentially, we'll catch ourselves a shiny Spinda approximately every quarter second.STEP TWO: HOW MANY SHINIES?Now, we need to figure out how many shiny Spindas we need to collect before we can expect to have one of every pattern. This is a perfect example of the coupon collector's problem of probability theory. It essentially asks, "If there are n different coupons being collected with replacement, with each coupon being equally likely, how many trials are needed to collect each coupon?" In our case each "coupon" is a shiny Spinda pattern, and n = 4,294,967,295. This convenient solution gives us a formula for E(T), the expected number of trials it takes to collect all 4,294,967,295. For us, E(T) = n*H = 4,294,967,295*H, where H is the 4,294,967,295th harmonic number.To calculate H by hand, we would need to calculate the sum of 4,294,967,295 reciprocals. Fortunately, Wolfram Alpha gives us an easy way to approximate H (you'll have to scroll down a bit - it's in section 14). H = 1/(2*4,294,967,295) + ln(4,294,967,295) + gamma, where gamma is the Euler-Mascheroni constant. This gives us H = 22.75792. This is technically an overestimate, but it's accurate up to those 5 decimal places, so I'm fine with it. We can finally calculate E(T) now. E(T) = 4,294,967,295*22.75792 = 9.77445 × 1010. That means we need to catch nearly 100 billion shiny Spindas before we can hope to have every pattern!STEP THREE: PUTTING IT ALL TOGETHERNow, we have our shiny Spinda catch rate of 0.24576 seconds, and our expected number of shiny Spindas, 9.77445 × 1010. To figure out how long this will take, we simply need to multiply the two figures. Our final answer is 0.24576*(9.77445 × 1010) = 2.40217 × 1010 seconds, or 761.2178 years. If you had one million trainers all diligently catching one Spinda a minute, it would take them over seven and a half centuries to find a shiny Spinda in every pattern. I don't know about you, but just thinking about that makes me dizzy! via /r/pokemon http://ift.tt/2yZYmnK