We define the probability of a given linear binary code C [n,k] as:
$$P_{corr}(C)=\sum_{i=0}^n\alpha_i(p)^i(1-p)^{n-i}$$
Where $\alpha_i$ is the number of coset leaders of weight $i$.
I am asked to show that, for the ternary Golay codes:
$$P_{corr}(G_{11})=P_{corr}(G_{12})$$
First of all, both codes are ternary so I believe we have to replace $p$ with $p/2$ (as the akphabet has 3 letters) Since $G_{11}$ is a perfect code, I can tell right away that: $\alpha_0=1, \alpha_1=22, \alpha_2=220$. $G_{12}$ has minimum distance from a given vector at 3, and thus we have: $\alpha_0=1, \alpha_1=24, \alpha_2=264, \alpha_3=440$ Where the last was computed as to complete the right number of coset leaders.
Putting this values into theformula above, however, I get:
$$P_{corr}(G_{11})=(1-p/2)^9(199/4*p^2+10p+1)$$ $$P_{corr}(G_{12})=(1-p/2)^9(1+p/2)(199/4*p^2+10p+1)$$
And that means they are obviously not equal. I went through my algebra again and again, and also double-checked with wolframalpha. My coset computation seems to be correct. Am I mistaken regarding the $p/2$ (not that it will help), or is it someplace else? Where am I mistaken?
I got (or rather, Mathematica got) the same polynomial for both codes. I also got the same numbers of coset leaders of various weight that you did. For comparison:
$$ \begin{aligned} P(G_{11},p)&=(1-p)^{11}+22(\frac p2)(1-p)^{10}+220(\frac p2)^2(1-p)^9\\ &=1 - 165 p^3 + 990 p^4 - 2772 p^5 + 4620 p^6 - 4950 p^7 + 3465 p^8 - 1540 p^9 + 396 p^{10} - 45 p^{11} \end{aligned} $$ and $$ \begin{aligned} P(G_{12},p)&=(1-p)^{12}+24(\frac p2)(1-p)^{11}+264(\frac p2)^2(1-p)^{10}+440(\frac p2)^3(1-p)^9\\ &=1 - 165 p^3 + 990 p^4 - 2772 p^5 + 4620 p^6 - 4950 p^7 + 3465 p^8 - 1540 p^9 + 396 p^{10} - 45 p^{11}. \end{aligned} $$