Branching ratio of two particles decaying to the same channel When looking at the numbers for the $WW$ decay's branching ratios, I have the impression that a factor $2$ is taken into account only when the $W$s decay to different channels.
For example:
$Br(WW \to qql\nu) = 2* Br(W \to qq) \cdot Br(W \to l\nu)$
$= 2 * 0.685 * 0.316 = 0.432$
but
$Br(WW \to l\nu l\nu) = Br(W \to l\nu) \cdot Br(W \to l\nu)$
$= 0.316 * 0.316 = 0.099$
where the numbers are taken from here (a 20 years old reference, so the best numbers might be slightly different now)
Is this the case ? If it is, why is so ?
 A: You have two $W$.
For the final state $qql\nu$, one $W$ decays to $qq$, the other to $l\nu$. But the decay to $qq$ could come either from the first W or the second W, thus you have the following possibilities :
First possibility
-first $W$ decays to $qq$, thus the second $W$ decays to $l\nu$.
-second $W$ decays to $l\nu$, thus the first $W$ decays to $qq$.
So you need to consider the two possibilities. Since the two possibilities have exactly the same branching ratio, it is equivalent to have twice the product of branching ratio for final state $qq$ and $l\nu$.
For the second final state $l\nu l\nu$, there is a unique solution, because one cannot distinguish first $W\rightarrow l \nu$ and second $W\rightarrow l\nu$, thus there is no factor 2.
If the $W$ could be distinguished (which is not the case), for example, if the first particle would be $W'$ and the second $W$, then there would be the factor 2.
Whenever the two $W$ are of same nature, there is no factor 2.
A: As an analogy, consider a toss of two coins. The outcome of each coin toss can be heads (H) or tails (T):
HH
HT
TH
TT

Since we don't care about the order of the coins (just as we don't care which $W$ actually decayed to quarks/leptons), we can write think of this as:
HH
2*HT
TT

