Rolling two dice...

Question

Let $A_n$ be the number of fives, $B_n$ the number of sixes and $C_n$ the number of eights in $n$ rolls of two dices. For which n do we have: $E(A_n) < E(min(B_n,C_n))$ ?

Answer 1

I don't know what you mean with E(5) and E(min(6,8)) but probably you can find the expectation by using the whole table with probabilities of throwing a sum, so if you define X to be the sum of the two dices then P(X=2)=1/36, P(X=3)=2/36,... and if you need a certain combination of dices with the E(.) as definition, just make a table with all combinations.

Answer 2

I run a program with PARI/GP :

? n=0;gef=0;while(gef==0,s=0;su=0;n=n+1;for(a=0,n,for(b=0,n,for(c=0,n,for(d=0,n,
if(a+b+c+d==n,su=su+n!/a!/b!/c!/d!*p1^a*p2^b*p3^c*p4^d*min(b,c);s=s+n!/a!/b!/c!/
d!*p1^a*p2^b*p3^c*p4^d)))));if(su>n/9,gef=1);print(n,"    ",s,"  ",su*1.0,"    "
,n/9*1.0,"      ",(su-n/9)*1.0))

The end of the output is

57    1  6.338345777056660390204361588    6.333333333333333333333333333      0.0
05012443723327056871028254579

So, the answer should be $n=57$.