Fair question. It doesn't have to be three days in a row. It's easier to visualize & solve it that way, though.
Here's my work out, for a 6 drawing set. Now, remember I was answering RJOh's question of the odds of a "threepeat". He didn't specify a certain number.
Let's call our target number 000. That comes out on the first day.
On the second day, 000 has 1 in 1000 chance of falling again. Overall odds so far: 1/1000
On the third day, 000 has a 1 in 1000 chance of falling again. Overall odds so far: 1/1,000,000
At this point, I'll stop calculating overall odds, for a minute
On day four (this is where it gets interesting) we want 000 to NOT fall. The odds of that are 999/1000 : slightly less that one
On day five, we want, again for 000 to NOT fall. The odds, again are 999/1000 : slightly less than one
(Day 5 could be a repeat of day 4, though)
On day six, we want to avoid 000, and we may want to avoid whatever fell on days 4 and 5, if 5 is a repeat of day 4. We're looking for a single threepeat, not two of them. Using the worst-case scenario (don't I always? ) of a repeat - that means we drop 2 numbers from consideration: 000 and whatever fell on days 4 & 5. Leaving us 998/1000 : slightly less than one. BTW, if day 4 is different than day 5, the odds go back to 999/1000
So, putting them all together: 1st three days [1/1,000,000] * 2nd three days [999/1000 * 999/1000 * 998/1000]
Because the second three days calculate to slightly less than one, the total product is reduced from 1 in 1,000,000 (10 in 10 million) to j_u_s_t under that. We can say our target 000 is a symbol (instead of a number), we can calculate the odds of that symbol showing-vs-not showing a certain number of times, given a certain number of trials. Once you look at it that way, you can see it doesn't matter the days the target number falls on.
My calculation was for any one and only one pick 3 number showing up straight, exactly three times (no more) in a six drawing period.
Now, if you want a particular number to show up or want certain days, then that is a more complex operation.