I know this is a studied and well known area, I just don't know what it is called.
The general area is statistics, but I assume you knew that
. Maybe "Gaussian distributions" or "Poisson statistics" is what you're looking for. The important effect (for Gaussian distributions) is (roughly) the following:
Let's say you have a bunch (a large number) of buckets, and a N times as many balls as you have buckets, and you randomly drop balls into buckets until all balls are in a bucket.
On average, each bucket will have N balls in it (because of the way you set things up). The statistical effect is that, if both N and the number of buckets are very large, then probability of finding a particular number of balls in a particular bucket will have a "Bell Curve" (Gaussian) distribution centered on N and of width proportional to sqrt(N). In practical terms, this means that most of the time, the number of balls "n" in a particular bucket will be between N-sqrt(N) and N+sqrt(N). I can factor out an N to write this a different way: between N*(1-sqrt(1/N)) and N*(1+sqrt(1/N)). The reason I wrote it the second was is because the +/- bit is the
relative size of the fluctuations with respect to the average value. So the bigger N is, the smaller the relative fluctuations are (even though the absolute fluctuations are getting bigger).
Because of this distribution if you amplify the opportunities to score a hit by 6 fold. The chance of a given hit probability is very minimal, as you noted it is only a .15% increase or some such, but given the ten year span of 120 rolls verses 720 rolls that 0.15% chance builds and I envision a distribution more like this:
I'm not sure what you're trying to say with your table - I think you're saying that you expect the second case to have more promotion "hits" than the first case. But the situation is exactly set up so this is not the case - the average value of hits is exactly the same in each case. For the 6% case, the average number of hits after each roll is increased by 0.06 = 0.06*1hit + 0.94*0hits. So after 120 6% rolls, you will on average have 7.2 hits = (0.06hit/roll)*120rolls. For the 1% case, the average number of hits after each roll is 0.01 = 0.01*1hit + 0.99*0hits. After 720 1% rolls you will on average have 7.2 hits = (0.01hit/roll)*720rolls, i.e. exactly the same. The higher number of hits is exactly balanced by the lower chance of getting a hit for each roll.
Now 7.2 isn't a very big value for N, so we're probably not very far into the large N regime. But the other important thing is to think about the difference between a single 6% roll and 6 1% rolls. For the 6% roll, you've got a 94% chance of no hits and a 6% chance of one hit. For the 6 1% rolls, you've got a 94.148% chance of no hits, a 5.706%chance of 1 hit, a 0.144% chance of 2 hits, a 0.002% chance of 3 hits, and even smaller chances of 4-6 hits. The important take away here is that I can replace a cluster of 6 1% rolls with a single roll that can have 0-6 hits with the probabilities listed above (i.e. an
identical distribution) and then use these probability distribution to interprete a sequence of 120 random numbers either using this clustered distribution or using the normal 6% distribution. If I do that experiment, then there's only (roughly) a 1-in-5 (= 0.148*120) chance that I'll see any difference at all in the roughly 7.2 hits that I expect. And Gaussian statistics says that if you increase the number of rolls the relative fluctuations will go down, not up. Does that help you see why it's a small effect?
John
PS - All this assumes that there isn't a bug in the code and that Steve actually is multiplying by the appropriate factor for longer intervals. He certainly intended to do so, but he could have missed a spot.
