No. 50^6 is about 938x bigger than (50 choose 6), which is the correct chance of picking 6 of 50 numbers.
> And wouldn't the odds really be: p = (1/49 * 1/48 * 1/47 * 1/46 * 1/45) for the balls to be drawn in order in one lottery?
This is not correct either. There are 45 sequential possibilities of winning numbers: 1-2-3-4-5-6, 2-3-4-5-6-7, ... 45-46-47-48-49-50. So "45 / (50 choose 6)", about 647 times more likely than that calculation for p.
Of course, if you start adding in other striking patterns like 2-4-6-8-10-12, the chance of a "fishy" draw becomes more and more likely.
It's very, very far, between non-replacement (~1.36x), order (5!=120x), the fact that final ball is out of 20 instead of out of 50 (2.5x), and the number of possible sequences which would be notable (at least 16x, but actually probably a fair bit more because recognizable patterns are pretty broad).
The upshot of which is that estimating at 50^6 is off by multiple orders of magnitude...
And wouldn't the odds really be: p = (1/49 * 1/48 * 1/47 * 1/46 * 1/45) for the balls to be drawn in order in one lottery?
The odds for them not to be drawn in order are then: 1 - p.
If there are 1000 lotteries per day, the odds it doesn't happen in a year are: ynp = (1-p)^(365 * 1000).
The yearly probability would be: 1-ynp.
If the balls can be drawn in any order, but they have to end as 6 ascending balls - then I think the probability is much higher, right?