As well as "Jew" or "gay" id on /pol/. I mean it's a three-letter word where second letter can be written as a number (k0t) and first and last letter aren't the same. 

Now let's talk about IDs. They are 8-character sequences consisting of base64 characters (numbers, lowercase, uppercase letters, / and +). So, there can be 

\eqn{10 + 26 + 26 + 2 = 64} 

possible characters for one position in id and thus

\eqn{K_0 = 64^8 = 2^{48} = 281\ 474\ 976\ 710\ 656}

possible IDs (BIG number :o).

Now back, to KOT. 
You can write letters k, o and t in uppercase or lowercase and you can also write o as zero. That means you have \math{2\times3\times2 = 12} combinations of base64 characters that can make up the word "k0T". 

Suppose that first three letters of your ID is one of 12 possible K0Ts. We have 5 letters of ID left and they can be anything from base64 alphabet. 
So, there are

\eqn{K_1 = 12 \times 64^5 = 12 \times 2^{30} = 12\ 884\ 901\ 888}

So, it's also many, many ids. More than all cats, all humans, but less than all chickens in the world. So, maybe kot ID isn't really that rare. 

Also there are 6 positions on which KoT id can happen:
KOTxxxxx
xKOTxxxx
xxKOTxxx
xxxKOTxx
xxxxKOTx
xxxxxKOT

We can assume that there are the same number of ids with KoT on n-th position where \math{n = 1 ... 6 }.  

Remember, that KOT id can happen twice, so we must exclude duplicates:
KOTKOTxx
KOTxKOTx
KOTxxKOT
xKOTKOTx
xKOTxKOT
xxKOTKOT

Each of these duplicate positions exists in
\eqn{K_2 = 12 \times 12 \times 64^2 = 144 \times 2^{12} = 589824}
configurations. 

Because one duplicate can belong to two sets of IDs with at least one kot id (KOTxxKOT can belong to KOTxxxxx and xxxxxKOT), our equation for number of all KOT ids must be:

\eqn{6 \times K_1 - 6 \times K_2}

Let's calculate it:

\eqn{6 \times K_1 - 6 \times K_2 = }
\eqn{6 \times 12 \times 2^{30} - 6 \times 144 \times 2^{12} = }
\eqn{72 \times 2^{30} - 864 \times 2^{12} = }
\eqn{77\ 305\ 872\ 384}

For every human on this planet there are approximately 10 KOT ids. :o

Now we can calculate probablility of having koT id:

\eqn{\frac{77305872384}{K_0} = \frac{77305872384}{281474976710656} \approx 0.00027464563 \approx \frac{1}{3641}}

Thus, k0t id happens approximately one time for every 3641 ids. 
Now you can calculate daily probability, because I don't know if ids on /bant/ are boardwise or just one for every poster and thread. x)