Reply To: Challenge 10B
One way to answer @cat_robin ‘s question, and obtain the 36:16 split, is to count the frequencies of the individual ‘cards’. Having established they occur in pairs:
AC 2C 3C 4C 5C 6C 7C 8C 9C XC JC QC KC
First 23 83 35 20 47 20 128 123 141 97 64 56 22
Second 0 0 0 0 0 0 0 0 0 0 0 0 0
AD 2D 3D 4D 5D 6D 7D 8D 9D XD JD QD KD
First 4 51 47 11 29 49 50 83 39 41 40 44 56
Second 0 0 0 0 0 0 0 0 0 0 0 0 0
AH 2H 3H 4H 5H 6H 7H 8H 9H XH JH QH KH
First 69 39 31 48 10 16 19 10 12 4 0 0 0
Second 0 0 0 0 0 0 0 0 0 0 197 107 76
AS 2S 3S 4S 5S 6S 7S 8S 9S XS JS QS KS
First 0 0 0 0 0 0 0 0 0 0 0 0 0
Second 144 125 93 109 175 79 43 60 68 155 45 39 146