A simple-minded but very conservative estimate of this buffer size is to treat the SCA as a 3 microsec long pipeline (60 cells long, so there are 36 free cells) If for a given Lev-1, a (Lev-1)(LCT) coincidence is found, 16 cells are set aside for 26 microsec during digitization. There are 96-16 = 80 cells left and 60 of these continue to function as the pipeline so there are 20 free cells left, i.e. room for 1 more event.
What is the probability of another (Lev-1)(LCT) coincidence occurring during the 26 microsec digitization time? Take (Lev-1)(LCT) coincidence rate of 0.6 KHz, or 1 in 1.7 ms in a chamber area covered by an FEB. If these events are evenly distributed in time, the probability of 1 in 26 microsec is 26x10**-6/1.7x10**-3 = 0.015. The Poisson probability for 2 or more events in 26 microsec is 1-P(0;0.015)-P(1;0.015) = 1.3x10**-4. (Where P(n,x)=x**n*exp(-x)/n! ; x=0.015 and n is an integer)
The above estimate is VERY conservative. The readout controller does not treat the SCA as a pipeline. Most cells are re-used after 400 ns (not 3 microsec). Given the LCT rate, only 2 sets of 16 cells are occupied in the sampling operation more than 99% of the time. The number of free cells is 96-32 = 64, so the buffer size is able to handle 1 + 3 additional tracks during digitization. The Poisson probability for 4 or more (Lev-1)(LCT) coincidences during digitization is 1-SUM[P(n;0.015), n=0,1,2,3] = 2x10**-9.