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<blockquote data-quote="nsruf" data-source="post: 764788" data-attributes="member: 872">Here is a matlab function that should do the trick with a time complexity of (p^2)*(n^2) instead of the awful p^n for complete enumeration. So you can calculate the probabilities for a moderate number of dice (e.g. 10d6) in reasonable time. It uses a recursive scheme without actual recursion:function d = roll(n, p)% determine probabilities to roll less than% or equal to any number with n p-sided dice&nbsp; &nbsp;&nbsp; &nbsp; &nbsp;&nbsp; N = n * p;&nbsp; &nbsp;&nbsp; % maximal roll result&nbsp; &nbsp;&nbsp; &nbsp; &nbsp;&nbsp; last = [ones(1, p), zeros(1, N - p)];&nbsp; &nbsp;&nbsp; % contains the number of combinations&nbsp; &nbsp;&nbsp; % for rolling a given number found in the&nbsp; &nbsp;&nbsp; % previous iteration of the outer for-loop;&nbsp; &nbsp;&nbsp; % initialized with the single die roll&nbsp; &nbsp;&nbsp; &nbsp; &nbsp;&nbsp; next = zeros(1, N);&nbsp; &nbsp;&nbsp; % contains the number of combinations&nbsp; &nbsp;&nbsp; % for rolling a given number if you add&nbsp; &nbsp;&nbsp; % one more die&nbsp; &nbsp;&nbsp; &nbsp; &nbsp;&nbsp; for i = 2:n;&nbsp; &nbsp;&nbsp; % number of dice considered for&nbsp; &nbsp;&nbsp; % calculating next&nbsp; &nbsp;&nbsp; &nbsp; &nbsp;&nbsp; mu = ceil((p + 1) * i / 2);&nbsp; &nbsp;&nbsp; % expectation value, rounded up&nbsp; &nbsp;&nbsp; &nbsp; &nbsp;&nbsp; for j = i:mu; for k = 1<img src="https://cdn.jsdelivr.net/joypixels/assets/8.0/png/unicode/64/1f61b.png" class="smilie smilie--emoji" loading="lazy" width="64" height="64" alt=":p" title="Stick out tongue&nbsp; &nbsp; :p"&nbsp; data-smilie="7"data-shortname=":p" />;&nbsp; &nbsp;&nbsp; % determine values in next up to expectation&nbsp; &nbsp;&nbsp; % mu; k is the number rolled on the additional&nbsp; &nbsp;&nbsp; % die&nbsp; &nbsp;&nbsp; &nbsp; &nbsp;&nbsp; if j - k &gt;= i - 1;&nbsp; &nbsp;&nbsp; % if it is possible to roll k on one die and&nbsp; &nbsp;&nbsp; % still get a result of j&nbsp; &nbsp;&nbsp; &nbsp; &nbsp;&nbsp; next(j) = next(j) + last(j - k);&nbsp; &nbsp;&nbsp; % add the number of possibilities to roll&nbsp; &nbsp;&nbsp; % (j - k) with (i - 1) dice&nbsp; &nbsp;&nbsp; &nbsp; &nbsp;&nbsp; end; end; end;&nbsp; &nbsp;&nbsp; &nbsp; &nbsp;&nbsp; I = i * p;&nbsp; &nbsp;&nbsp; % maximal roll on i dice&nbsp; &nbsp;&nbsp; &nbsp; &nbsp;&nbsp; for j = (mu + 1):I;&nbsp; &nbsp;&nbsp; % determine remaining values in next&nbsp; &nbsp;&nbsp; &nbsp; &nbsp;&nbsp; next(j) = next(I + i - j);&nbsp; &nbsp;&nbsp; % find values larger then mu by symmetry&nbsp; &nbsp;&nbsp; &nbsp; &nbsp;&nbsp; end;&nbsp; &nbsp;&nbsp; &nbsp; &nbsp;&nbsp; last = next;&nbsp; &nbsp;&nbsp; &nbsp; &nbsp;&nbsp; next = zeros(1, N);&nbsp; &nbsp;&nbsp; &nbsp; &nbsp;&nbsp; end;&nbsp; &nbsp;&nbsp; &nbsp; &nbsp;&nbsp; d = cumsum(last) / (p ^ n);&nbsp; &nbsp;&nbsp; % probability to roll less is cumulative sum&nbsp; &nbsp;&nbsp; % divided by total number of combinations&nbsp; &nbsp;&nbsp; &nbsp; &nbsp;&nbsp; d = [n:N; d(n:N)];&nbsp; &nbsp;&nbsp; % first row of output contains possible rolls,&nbsp; &nbsp;&nbsp; % second row contains probability to roll&nbsp; &nbsp;&nbsp; % equal or less</blockquote>

[QUOTE="nsruf, post: 764788, member: 872"] Here is a matlab function that should do the trick with a time complexity of (p^2)*(n^2) instead of the awful p^n for complete enumeration. So you can calculate the probabilities for a moderate number of dice (e.g. 10d6) in reasonable time. It uses a recursive scheme without actual recursion: [font=courier new] function d = roll(n, p) % determine probabilities to roll less than % or equal to any number with n p-sided dice N = n * p; % maximal roll result last = [ones(1, p), zeros(1, N - p)]; % contains the number of combinations % for rolling a given number found in the % previous iteration of the outer for-loop; % initialized with the single die roll next = zeros(1, N); % contains the number of combinations % for rolling a given number if you add % one more die for i = 2:n; % number of dice considered for % calculating next mu = ceil((p + 1) * i / 2); % expectation value, rounded up for j = i:mu; for k = 1:p; % determine values in next up to expectation % mu; k is the number rolled on the additional % die if j - k >= i - 1; % if it is possible to roll k on one die and % still get a result of j next(j) = next(j) + last(j - k); % add the number of possibilities to roll % (j - k) with (i - 1) dice end; end; end; I = i * p; % maximal roll on i dice for j = (mu + 1):I; % determine remaining values in next next(j) = next(I + i - j); % find values larger then mu by symmetry end; last = next; next = zeros(1, N); end; d = cumsum(last) / (p ^ n); % probability to roll less is cumulative sum % divided by total number of combinations d = [n:N; d(n:N)]; % first row of output contains possible rolls, % second row contains probability to roll % equal or less[/font] [/QUOTE]

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