What would be the equation for calculating the chance of getting dubs on an image board?

What would be the equation for calculating the chance of getting dubs on an image board?

99*(user post rate/total post rate)/100

I fucked up on basic math, it's actually
9*(user post rate/total post rate)/100

Math is for losers lmao. Check these dubs.

For any random string of digits, the odds of the last two digits being the same is b^-1, where b is the base of the number system being used. In the case of decimal numbers, it's 1/10 or 10% chance.

You're forgetting about trips and higher, which -according to most- don't count as dubs.
There's a 1/100 change of the three last digits being equal, so it's 1/10 - 1/100 = 9/100.
Also, sage shit threads.

There are 10, not 9, possible dubs combinations: 00, 11, 22, 33, 44, 55, 66, 77, 88, 99 AND on top of that you have to check if third digit from the last does not match the dubs. So basically you have 10 combinations multiplied by 10 because you have to consider three last digits and that amounts to a 1000 posts. So you have 10 sets of dubs in each hundred, but sometimes they become trips, once per each hundred. so 000, 111, 222, 333, 444, 555, 666, 777, 888, 999 are out. So you have 9 actual dubs combination in each hundred. In a set of 1000 posts 90 are true dubs. Therefore is correct.

1, as your number of posts approaches infinity.

"When in doubt, bruteforce it." -- Ken Thompson

Then why stop at trips? By the same reasoning you should also throw out quads, pents, sexts, on up to the length of the string in question.
Also, don't tell me what to sage.