[Shiraz (Ray)]

that's entirely consistent with earlier posts.

Binning occurs in 2 steps.

1. First, the signal from pairs of adjacent pixel is added by putting the charge from 2 pixels (rather than one) into each cell of the shift register - for a saturated 8300, this results in trying to stuff 51,000 electrons in each SR cell. But, the shift register cells cannot hold that much and saturate at 37,000 electrons - so you lose ~1/3 the available full charge in this step.
2. The (37,000 electron) charge packets in the SR cells are then transferred into the output register in pairs, so 74,000 electrons turn up at the output register for each binned pixel. The output register cannot handle more than 55,000 electrons, so some of the available electrons are again lost to saturation and you are again down by about 1/3.

ie, in the two processes, you lose about half of the total number of electrons available from the 4 saturated pixels - you start out with 25,500x4 and end up with ~55,000, simply because you cannot squeeze all of the available electrons from 4 saturated pixels through the transfer electronics.

The final step is that the gain of the output stage is normally set to give 64000 ADU with only one pixel-full of electrons (25,500). With binning, it is presented with a bit more than double that many electrons per pixel, so the gain must be reduced by a roughly a half to keep the maximum reading to 64,000 ADU.