Poisson process for which we throw a die at each jump. I have to solve the following exercise,

a) Let $X(t):t≥0$ be poisson process of rate 1 starting at 0. Each time $X$
jumps we roll a (fair) die. What is the distribution of the number of
odd numbers rolled in interval $[0,4]$ ?
b) Given that on interval $[0,12]$
there were 13 jumps, what is the probability that exactly 3 6’s were
rolled by the die ?

a) So the number of jumps in the interval [0,4] is $N(4) \sim Poisson(4) $ and since the probability of having a odd number is $0.5$ then the distribution of the odd numbers $\sim Poisson(4\times 0.5)$.
b) But if I do the same for the next question, since the probability of getting a 6 is $\frac{1}{6}$, then on $[0,12]$ the distribution of 6 should $Z \sim Poisson(12\times\frac{1}{6})$ so the the probability of having 3 6's should be $P(Z=3)=e^{-2}\frac{2^3}{3!}$.
But the answer is $P(3 $ $ $ $ 6's)={13\choose 3}(\frac{1}{6})^3(\frac{5}{6})^{10}$. Which I understand. But what is wrong about the previous interpretation? And why is it wrong?
 A: Let $\{X(t):t\geqslant 0\}$ be a Poisson process with rate $\lambda>0$  and let $T>0$ be fixed. Then the distribution of  the odd numbers rolled in $[0,T]$, call it $O_T$, is given by
\begin{align}
\mathbb P(O_T=k) &= \sum_{n=k}^\infty \mathbb P(O_T=k\mid X(t)=n)\mathbb P(X(t)=n)\\
&=\sum_{n=k}^\infty\binom nk \left(\frac12\right)^n e^{-\lambda T}\frac{(\lambda T)^n}{n!}\\
&=\frac{e^{-\lambda T}}{k!} \sum_{n=k}^\infty \frac1{(n-k)!}\left(\frac{\lambda T}2\right)^n\\
&=\frac{e^{-\lambda T}}{k!}\left(\frac{\lambda T}2\right)^k\sum_{n=k}^\infty \frac1{(n-k)!} \left(\frac{\lambda T}2\right)^{n-k}\\
&=\frac{e^{-\lambda T}}{k!}\left(\frac{\lambda T}2\right)^k \sum_{n=0}^\infty \frac{(\lambda T/2)^n}{n!}\\
&= \frac{e^{-\lambda T}}{k!}\left(\frac{\lambda T}2\right)^k e^{(\lambda T)/2}\\
&= \frac{e^{-(\lambda  T)/2} ((\lambda  T)/2)^k}{k!}.
\end{align}
Substituting in $\lambda=1$ and $T=4$, we get
$$
\mathbb P(O_4=k) = e^{-2}\frac{2^k}{k!},
$$
so that $O_4\sim\mathrm{Pois}(2)$.
For b), let $Z_6$ be the number of $6$'s rolled, $J$ the number of jumps in $[0,T]$ where $T>0$ is fixed, and $n_6$ a nonnegative integer. Then conditioned on $\{N(T)=J\}$, $Z_6$ has $\mathrm{Bin}(J,1/6)$ distribution, independent of $T$. Hence
$$
\mathbb P(Z_6 = n_6\mid N(T)=J) = \binom{J}{n_6} \left(\frac16\right)^{n_6}\left(\frac56\right)^{J-n_6}.
$$
Substituting $n_6=3$, $T=12$, and $J=13$ we have
\begin{align}
\mathbb P(Z_6 = 3\mid N(12)=13) &= \binom{13}{3} \left(\frac16\right)^{3}\left(\frac56\right)^{13-3}\\ &= \binom{13}3 \left(\frac{5^{10}}{6^{13}}\right)\\
&= \frac{1396484375}{6530347008} \approx0.2138454.
\end{align}
