>>126

Assuming that you can write second 'a' of 'jak' as 4 it's the same probability.

Thoughever, you can write it also as 'jaq' or 'jac', so there are 6 possibilities of writing the last letter there are:

\eqn{2 \times 3 \times 6 = 36}

Possible variation of previous constants for JaK:

\eqn{K_0 = the\ same}

\eqn{K_1 = 36 \times 64^5 = 36 \times 2^{30} = 38\ 654\ 705\ 664}

\eqn{K_2 = 36 \times 36 \times 64^2 = 1296 \times 2^{12} = 5\ 308\ 416}


The final equation for the number of jAc IDs:

\eqn{6 \times K_1 - 6 \times K_2 = }
\eqn{6 \times 36 \times 2^{30} - 6 \times 1296 \times 2^{12} = }
\eqn{216 \times 2^{30} - 7776 \times 2^{12} = }
\eqn{231\ 896\ 383\ 488}

And the probability:

\eqn{\frac{231896383488}{K_0} = \frac{231896383488}{281474976710656} \approx 0.000823861453 \approx \frac{1}{1214}}

So it's just pretty much 3 times bigger than "KoT". :o


/qa/ is dead tohough :(