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%I #12 Mar 09 2020 10:32:07
%S 21,161,1071,6797,42231,259421,1582791,9614717,58230711,351922781,
%T 2123580711,12799240637,77074749591,463808234141,2789504205831,
%U 16769733474557,100779708074871,605475935585501,3636808913042151,21840480209276477,131140458175102551,787328413691288861
%N Expected value of the highest die when n six-sided dice are rolled, multiplied by 6^n.
%H Robert Israel, <a href="/A242191/b242191.txt">Table of n, a(n) for n = 1..1282</a>
%F a(n) = 1 + Sum_{k=2..6, j=1..n} k*binomial(n,j)*(k-1)^(n-j).
%F Conjecture: a(n) = -5^n-3^n+6^(n+1)-2^n-1-4^n with generating function -x*(-21 +280*x -1365*x^2 +2954*x^3 -2688*x^4 +720*x^5) / ( (x-1) *(6*x-1) *(4*x-1) *(3*x-1) *(2*x-1) *(5*x-1) ) - _R. J. Mathar_, May 23 2014
%F Conjecture is true, because the probability that the highest is k is (k/6)^n - ((k-1)/6)^n for k = 1..6. - _Robert Israel_, Mar 09 2020
%e a(1) = 21, because when a die is rolled, the possible outcomes are 1,2,3,4,5,6, whose average is 21/6.
%e a(2) = 161 because when two dice are rolled, the expected value of the higher die is 161/36.
%p f:= n -> 6^(n+1)-add(i^n,i=1..5):
%p map(f, [$1..50]); # _Robert Israel_, Mar 09 2020
%K nonn
%O 1,1
%A _Andrew Woods_, May 06 2014
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