My crazy die has one side with one pip on it, two with two pips , and three on each of three remaining sides.
I tossed it 7 times in a row and wrote down the result of those tosses as a 7 digit number.

What is the probability that this number
a. Is bigger than 1231111?
b. Contains at least one triplet of identical digits?
c. Is divisible by 3?
d. Exhibits all the 3 features mentioned above?

   10       for D1=1 to 3
   20         P1=D1//6
   30       for D2=1 to 3
   40         P2=P1*D2//6
   50       for D3=1 to 3
   60         P3=P2*D3//6
   70       for D4=1 to 3
   80         P4=P3*D4//6
   90       for D5=1 to 3
  100         P5=P4*D5//6
  110       for D6=1 to 3
  120         P6=P5*D6//6
  130       for D7=1 to 3
  140         P7=P6*D7//6
  150         Num=1000000*D1+100000*D2+10000*D3+1000*D4+100*D5+10*D6+D7
  160         Flag=1
  170         if Num>1231111 then Aprob=Aprob+P7:else Flag=0
  180         if D2=D3 and (D1=D2 or D3=D4) or D4=D5 and (D3=D4 or D5=D6) or D5=D6 and D6=D7 then
  190           :Bprob=Bprob+P7:else Flag=0
  200         if Num@3=0 then Cprob=Cprob+P7:else Flag=0
  210         if Flag then Dprob=Dprob+P7
  220       next
  230       next
  240       next
  250       next
  260       next
  270       next
  280       next
  290       print Aprob;Bprob;Cprob;Dprob
  300       print Aprob/1;Bprob/1;Cprob/1;Dprob/1
 
  finds
 
 44063/46656  11629/23328  3455/10368  1819/11664
 0.9444230109739368998  0.4984996570644718792  0.3332368827160493827  0.1559499314128943758 
 
 as the rational, and decimally approximate, solutions to parts a, b, c and d respectively, assuming a triplet must be successive within the number.

Posted by Charlie on 2014-02-03 18:47:16