math probleme

Anoymous wanted dubs.

She posted checkem and did not get dubs yesterday.

She posted checkem today, what is the probability she gets dubs?

(For the duration of this discussion I will drop my RedCream mannerisms.)

Out of a three-digit space, including the outlier four-digit 1000, what is the probability of attaining dubs without regard to looking behind?

There are three sorts of dubs in that space: Simple, Zeroes, and Embedded. People commonly accept grouping the simple and zero dubs together.

Simple dubs appear in every 11 postings, including the zeroes (one for each 100ths): {11, 22, 33, ..., 100, 111, 122, ..., 977, 988, 999, 1000}

That probability is 1/11 + 1/100 =~ 0.1009

Embedded dubs are like so: {110, 111, 112, ..., 220, 221, 223, ..., 997, 998, 999, 1000}

Excluding the 1000 (which was already accounted for as a "Zero" dubs) that probability is 10x(1/110) =~ 0.0909

There are no other dubs to be found in the three-digit space, hence adding the simple dubs to the embedded dubs is:

1/11 + 1/100 + 10x(1/110) =~ 0.1918

So to attain dubs randomly, you have an overall ~19.2% chance of success, or about 5 postings in every 26 attempts.