Golliwog Numbers


Golliwog Numbers - lower left is G(1,1)
00000000001120543840
0000000001106286406668781166080
0000000011026576628308907776303443622431870976
00000001109584724589337638149548322406417810567950611972096
000000113068104587344673781602752387886492069401621006340945438768128
00000117641942416174483590414134702901761083688832185344806595068762689536
00001274480766998824929081776117550462624145008661029761765130436471424
00015016604676012177763048080074749792018139003520437786795776
001118557536612263113745833375501942130174551
013715316312725551110232047
111111111111

This differs from the table in [DKB] only by the correction of an obvious typo. We note that the non-zero entries in row k (from the bottom) have as rational generating function

                    k
                _________
                  |   |
                  |   |             1
                  |   |     _________________    
                  |   |              
                  |   |              k! 
                  |   |       1 -  ______ z  
                  |   |                        
                  |   |              p                    
                  
                   p=1
 

Pez probabilities

Table of golliwog Numbers - g(m,k) - G(m,k) divided by k!m-k+1 [to get Pez Probabilities - (k-1)!*g(m,k)]
000000000000.000000002
00000000000.0000000250.000000076
0000000000.000000280.00000080.0000014
000000000.0000030.00000780.0000130.0000175
00000000.0000250.0000670.000110.0001440.000167
0000000.0001980.0005140.0008270.001040.001190.00128
000000.001390.003400.005200.006490.0073050.007780.00804
00000.008330.019030.0278220.033750.037340.039370.040470.04105
0000.041670.086810.120080.140940.152940.159500.162980.164790.16571
000.166670.305560.393520.443670.470810.485060.492420.496170.498070.49903
00.500000.750000.875000.937500.968750.984380.992190.996090.998050.999020.99951
1.01.01.01.01.01.01.01.01.01.01.01.0

This table is perhaps more fun to read than the more extensive Table 4.5.2 in [DKB]. In the context of the Pez problem we see, of course, that for N distinct flavors, the probability that there is a flavor missed by the sharer (or victim!) goes to zero as the number of dispensers increases (trace the probabilities up a diagonal), while if we consider the number of dispensers fixed at k, and the number of flavors (hence the height of the dispensers) to increase, the probability of missing a flavor [(k-1)!*g(m,k)] seems to approach 1 (This limit is easily verified using the iterated difference formula for g(m,k)).


Len Smiley
Last modified: Wed Nov 21 01:17:16 AKST 2001