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A366451 The sum s > n which has the greatest probability of occurring when summing rolls of an n-sided die.

%I A366451 #90 Jun 27 2025 17:27:50
%S A366451 4,6,8,9,11,13,15,16,18,20,21,23,25,27,28,30,32,34,35,37,39,40,42,44,
%T A366451 46,47,49,51,52,54,56,58,59,61,63,64,66,68,70,71,73,75,76,78,80,82,83,
%U A366451 85,87,88,90,92,94,95,97,99,101,102,104,106,107,109,111,113
%N A366451 The sum s > n which has the greatest probability of occurring when summing rolls of an n-sided die.
%C A366451 A player chooses a target t > n and rolls a fair n-sided die repeatedly, keeping a running total of the numbers rolled. The player wins if t occurs in the sequence of running totals. a(n) is the choice of t that maximizes the probability of winning
%C A366451 A fair die with sides numbered 1 through n is assumed.
%C A366451 For given n, sum s occurs by adding one die role to a sum s-n .. s-1; so the probability p(s) of s occurring is p(s) = Sum_{i=1..n} p(s-i)*(1/n), i.e., the mean of the preceding n probabilities, and starting from p(0) = 1 for s=0 always occurring and p(s) = 0 for s < 0 never occurring.
%C A366451 It is immediately apparent upon plotting the probabilities from sum s = n+1 onwards that the highest probability is achieved at s = a(n).
%C A366451 It can be shown that, given the equation of n < s <= 2n, the maximum near a(n) is the only zero of the first derivative, and the second derivative is always negative, therefore any following probability must necessarily be less the maximum of the previous n probabilities, and therefore less than p(a(n)). Therefore any subsequent terms must then be less than p(a(n)). Because this is appending a rolling average of the previous n terms the probabilities dampen in their oscillations about a value of 2/(n+1).
%H A366451 Paolo Xausa, <a href="/A366451/b366451.txt">Table of n, a(n) for n = 2..10000</a>
%H A366451 James Monroe, <a href="https://www.youtube.com/watch?v=zF814SmPY5A">Pi is Evil</a>, Numberphile youtube video. Note the errata by the authors in the comments: a(20) is actually 35. In addition, the more accurate approximation of (e-1)(n+1/2)
%F A366451 For n < s < 2n, the probability of sum s occurring is p(s) = ((n+1)^(s-1) - s*n^n*(n+1)^(s-n-2))/n^s, which means that, from the difference between s and s-1, the sequence value is given by:
%F A366451 a(n) = floor(((n+1)*(n+1)^n - n^(n+1))/n^n).
%F A366451 a(n) = A060644(n) - n.
%F A366451 An empirical observation is that a(n) is well approximated by (e-1)*(n+1/2).
%F A366451 From _Javier Múgica_, May 01 2025: (Start)
%F A366451 Taylor expansion of the maximum of p(s) gives a(n) ~ (e-1)*(n+1/2) - (e-2)/(24n).
%F A366451 The maximum for s > 2n is attained at s ~ (e-sqrt(-e^2+2e+2))*n, approx. 2.5*n. In general, the position of the maximum for s > kn is given by an algebraic equation in s and e of degree k in each of them. (End)
%e A366451 For n=20, the probability of occurrence p(s) of sum s > n begins at p(21) = 0.08266... and rises to p(35) = 0.09767... before decreasing to p(36) = 0.09760... and never reaches p(35) again, so a(20) = 35 is the sum s with greatest probability of occurrence.
%e A366451            s     p(s)
%e A366451           --  ----------
%e A366451           21  0.08266...
%e A366451           ...
%e A366451           34  0.09751...
%e A366451   a(20) = 35  0.09767... <--- maximum probability
%e A366451           36  0.09760...
%e A366451           ...
%e A366451           42  0.09350...
%t A366451 A366451[n_] := Floor[(n+1)^(n+1)/n^n] - n; Array[A366451, 100, 2] (* _Paolo Xausa_, Oct 12 2024 *)
%o A366451 (Python)
%o A366451 from fractions import Fraction
%o A366451 a = lambda n: int(Fraction((n+1)*(n+1)**n-n**(n+1),n**n))
%Y A366451 Cf. A060644, A365443.
%K A366451 nonn
%O A366451 2,1
%A A366451 _Nicholas Stefan Georgescu_, Oct 10 2023