Probably common sense, but this statistically supports the most common roll of the dice. Of course, this assumes physical normalcy of the die, not loaded, etc. \ud83d\ude42<\/p>\n
\r\n[root@cmhlcanlyodb01 ~]# cat rand.py\r\nimport random\r\nd = dict()\r\nfor i in range(1,100001):\r\n j=random.randint(1,6)\r\n k=random.randint(1,6)\r\n l=j+k\r\n if l in d:\r\n d[l] = d[l] + 1\r\n else:\r\n d[l] = 1\r\n\r\nfor a in d:\r\n print str(a).rjust(10,' '),str(d[a]).rjust(10,' '),\r\n str(round(float(d[a])*100\/float(100001),1)).rjust(10,' ')\r\n[root@cmhlcanlyodb01 ~]# python rand.py\r\n 2 2739 2.7\r\n 3 5568 5.6\r\n 4 8320 8.3\r\n 5 11355 11.4\r\n 6 13765 13.8\r\n 7 16830 16.8\r\n 8 13695 13.7\r\n 9 10991 11.0\r\n 10 8269 8.3\r\n 11 5687 5.7\r\n 12 2781 2.8\r\n[root@cmhlcanlyodb01 ~]#\r\n<\/pre>\n","protected":false},"excerpt":{"rendered":"Probably common sense, but this statistically supports the most common roll of the dice. Of course, this assumes physical normalcy of the die, not loaded, etc. \ud83d\ude42 [root@cmhlcanlyodb01 ~]# cat rand.py import random d = dict() for i in range(1,100001):…<\/p>\n