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a(n) = min{m: Sum_{j=0..m} binomial(n,j)*(1/6)^j*(1-1/6)^(n-i) >= 0.95}.
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%I #13 Apr 30 2023 15:15:28

%S 0,1,1,2,2,2,3,3,3,3,4,4,4,5,5,5,5,6,6,6,6,6,7,7,7,7,8,8,8,8,9,9,9,9,

%T 9,10,10,10,10,10,11,11,11,11,12,12,12,12,12,13,13,13,13,13,14,14,14,

%U 14,14,15,15,15,15,16,16,16,16,16,17,17,17,17,17,18,18,18,18,18,19,19,19

%N a(n) = min{m: Sum_{j=0..m} binomial(n,j)*(1/6)^j*(1-1/6)^(n-i) >= 0.95}.

%C If you toss an idealized (six-sided) die n times, then the probability of obtaining more than a(n) 6's is <= 5 percent.

%t s[m_, n_] := Sum[Binomial[n, i]*(1/6)^i*(1 - 1/6)^(n - i), {i, 0, m}]; a[0] = 0; a[n_] := a[n] = For[m = a[n - 1], True, m++, If[s[m, n] >= 95/100, Return[m]]]; Table[a[n], {n, 0, 80}] (* _Jean-François Alcover_, Oct 11 2012 *)

%Y Cf. A039708 (similar for a coin).

%K easy,nice,nonn

%O 0,4

%A Klaus Strassburger (strass(AT)ddfi.uni-duesseldorf.de), Dec 15 1999