I believe your math is good, but assumptions wrong. There are many different concepts in probability and I believe you are choosing the wrong event. The probability you described is the event in which times straight over a [i]specific[/i] time span. What I believe you should be looking at is the probability that the voidfang has been produced 4 weeks straight over a [i]general[/i] time span. Clearly as we go to infinity, the probability of this occurring at least once go to 1, regardless of it is random. In order to do this, the easiest way I can do it is to consider blocks of 4 weeks at a time, and let us assume that each week the item is chosen uniformly random over some set of size 6. Our first case to consider is the one in which the chosen block has all the same elements, this as you have said is 6^{-4}, very small indeed, helped only slightly by the fact there are 6 different possible ways to drop the same element 4 times, giving us a probability of 6^{-3}. But we must now consider drawing a block which matches with a previous block, what do these situations look like? Previous block ends with 3 identical the current block begins with that same element . 2 identical and 2 identical 1 element and 3 identical elements. That should be clear enough, so let us determine these probability that this chosen block will give us a string of four. We just need to add up the previous probabilities, which I will label with (x,y) where x will be the number of the 4 string on the first block and the y is the 4-x (0,4) - 6^{-3} as shown previously (1,3)- Because we do not want to encroach on (2,2) we must only consider solutions which are of the form xxBA there are 6^3*5 total sequences as such, giving us a probability of one of these occurring at 5/6. The probability that the currently chosen block matches it with 3 identical elements of that same type is 6^{-3}. Giving us a probability of 5* 6^{-4} (2,2)- Same thing, form must be xBAA, of which there are 6 * 5 * 6 = 6^2*5, giving us 5*6^{-2} probability of these occurring, and on the other side we have AAxx, which is 6^{-2}, giving us 5*6^{-4} (3,1)- BAAA- gives us 6*5 outcomes 5*6^{-3}, Axxx gives us 6^3 outcomes and 6^{-1} probability, so total is once again 5*6^{-4}. So the total probability of the currently placed block halting this is (6 + 3 * 5)*6^{-4} = 21*6^{-4} = something annoying. Anyway, because this is the probability that the current block chosen will halt the problem, we can now observe this like it is Bernoulli trial with independent events p except at n=1. So in conclusion the probability of this not occurring in n randomly chosen blocks is 1-(1-6^{-3}) * (1-21*6^{-4})^{n-1} . Because each block of n represents 4 weeks, and there has been about 16 weeks since the release of destiny, the actual probability of this having occurred over this time frame is around 1/20. Which by no stretch of the imagination is a large probability, but by no means unheard of. In conclusion, hunters rock.