No, 666 has a **much lower normal probability** to occur than 1111, so do 1000.

I disagree. It is possible to get 666 posts without ever reaching 1111, but not vice versa.

Where is the disagreement? However, "1111" isn't less likely to happen just because the frequancy of its appearence with its equivalents is ten time less than the frequency of "666", that is it.

Bold = mine. If you get the number 1111, you are guaranteed to have gotten then number 666 at least once: there is a pattern which is followed on this site. You're also saying, apparently, than 1111 appears 1/10 as many times as 666 in the natural progression of numbers. What you MEAN to say, I assume, is that any number with three repeating digits (100x+10x+x) is more likely to occur than any number with 4 repeating digits (1000x+100x+10x+x) Which in the sense that Astreja mentioned makes total sense (because you can get 666 and not get 1111, but not the other way around). However, that's the only sense that's being made here.

What you've said is that it is more likely to get the number sequence 'xxx' in a set of all integers between 1 and 10,000 than it is to get 'yyyy' in a set of all integers between 1 and 10,000. Which is only true if you allow for variations of the codes to be counted among them, i.e. 1666 or 2221 (zxxx or xxxz), which simply makes it obvious and rigs it to show a certain outcome. 1111, 2222, etc. would all be

*included* in the count of the number of integers with 3 repeating digits. In fact, depending on how you wanted to argue it, they would be counted

*twice*, with both the triple digit sequence at the beginning and the end of the number counted. They could serve as both xxxz and zxxx. So in essence, by doing this, you are essentially stating that you're more likely to have a number with at least 3 digits than you are to have a number with at least 4 digits.

I'll say that again, in bold-underline-italics so you can understand this.

*You're saying that it's more probable to get a number with at least 3 digits than it is to get a number with at least 4 digits.*This thread now = epic fail.