Equivalent DnD dice roll for a rational probability
Matt Tearle
2025 年 2 月 9 日
最新アクティビティ Hans Scharler
さんによる返信 (2025 年 2 月 11 日)
I got thoroughly nerd-sniped by this xkcd, leading me to wonder if you can use MATLAB to figure out the dice roll for any given (rational) probability. Well, obviously you can. The question is how. Answer: lots of permutation calculations and convolutions.
In the original xkcd, the situation described by the player has a probability of 2/9. Looking up the plot, row 2 column 9, shows that you need 16 or greater on (from the legend) 1d4+3d6, just as claimed.
If you missed the bit about convolutions, this is a super-neat trick
[v,c] = dicedist([4 6 6 6]);
bar(v,c)
% Probability distribution of dice given by d
function [vals,counts] = dicedist(d)
% d is a vector of number of sides
n = numel(d); % number of dice
% Use convolution to count the number of ways to get each roll value
counts = 1;
for k = 1:n
counts = conv(counts,ones(1,d(k)));
end
% Possible values range from n to sum(d)
maxtot = sum(d);
vals = n:maxtot;
end
1 件のコメント
新しい順
So, you are saying, when it matters, you roll a 1?
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