Problem 1067. The Dark Side of the Die

Created by @bmtran (Bryant Tran)

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{"relationships":[{"relationshipType":"http://schemas.mathworks.com/matlab/code/2013/relationships/document","relationshipId":"rId1","target":"/matlab/document.xml"},{"relationshipType":"http://schemas.mathworks.com/matlab/code/2013/relationships/output","relationshipId":"rId2","target":"/matlab/output.xml"}],"parts":[{"partUri":"/matlab/document.xml","relationship":[],"contentType":"application/vnd.mathworks.matlab.code.document+xml","content":"<?xml version=\"1.0\" encoding=\"UTF-8\"?><w:document xmlns:w=\"http://schemas.openxmlformats.org/wordprocessingml/2006/main\"><w:body><w:p><w:pPr><w:pStyle w:val=\"text\"/></w:pPr><w:r><w:t>It is well-known that opposite sides of a classic hexahedral die add to 7. Given a vector of dice rolls, calculate the sum of the hidden face. That is, the sum of the values opposite the rolled value.</w:t></w:r></w:p><w:p><w:pPr><w:pStyle w:val=\"text\"/><w:jc w:val=\"left\"/></w:pPr><w:r><w:t>For example, if we roll a 2 and then a 6, we calculate that the opposite side of the dice were 5 and 1. Therefore, the answer is 6. This can also be calculated another way: since we know we rolled 2 dice, the total sum of the front and back faces of the dice must be 14 (7x2). If we then subtract the sum of the rolled values (2+6=8) from our total sum, we get the correct final answer (14-8=6)!</w:t></w:r></w:p></w:body></w:document>"},{"partUri":"/matlab/output.xml","contentType":"text/xml","content":"<?xml version=\"1.0\" encoding=\"UTF-8\" standalone=\"no\" ?><embeddedOutputs><metaData><evaluationState>manual</evaluationState><layoutState>code</layoutState><outputStatus>ready</outputStatus></metaData><outputArray type=\"array\"/><regionArray type=\"array\"/></embeddedOutputs>"}]}

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Last Solution submitted on Sep 16, 2020

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dice plus15 reverse

Problem 1067. The Dark Side of the Die

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Problem 1067. The Dark Side of the Die

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