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POCalculate exact result of complex throw of two D30
 

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  1. POCalculate exact result of complex throw of two D30
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    <p>Okay, this bugged me for several years, now. If you sucked in statistics and higher math at school, turn away, <em>now</em>. Too late.</p> <p>Okay. Take a deep breath. Here are the rules. Take <em>two</em> thirty sided dice (yes, <a href="http://paizo.com/store/byCompany/k/koplow/dice/d30" rel="nofollow noreferrer">they do exist</a>) and roll them simultaneously.</p> <ul> <li>Add the two numbers</li> <li>If both dice show &lt;= 5 or >= 26, throw again and <em>add</em> the result to what you have</li> <li>If one is &lt;= 5 and the other >= 26, throw again and <em>subtract</em> the result from what you have</li> <li>Repeat until either is > 5 and &lt; 26!</li> </ul> <p>If you write some code (see below), roll those dice a few million times and you count how often you receive each number as the final result, you get a curve that is pretty flat left of 1, around 45° degrees between 1 and 60 and flat above 60. The chance to roll 30.5 or better is greater than 50%, to roll better than 18 is 80% and to roll better than 0 is 97%.</p> <p>Now the question: Is it possible to write a program to <em>calculate</em> the <em>exact</em> value f(x), i.e. the probability to roll a certain value?</p> <p>Background: For our role playing game "Jungle of Stars" we looked for a way to keep random events in check. The rules above guarantee a much more stable outcome for something you try :)</p> <p>For the geeks around, the code in Python:</p> <pre><code>import random import sys def OW60 (): """Do an open throw with a "60" sided dice""" val = 0 sign = 1 while 1: r1 = random.randint (1, 30) r2 = random.randint (1, 30) #print r1,r2 val = val + sign * (r1 + r2) islow = 0 ishigh = 0 if r1 &lt;= 5: islow += 1 elif r1 &gt;= 26: ishigh += 1 if r2 &lt;= 5: islow += 1 elif r2 &gt;= 26: ishigh += 1 if islow == 2 or ishigh == 2: sign = 1 elif islow == 1 and ishigh == 1: sign = -1 else: break #print sign #print val return val result = [0] * 2000 N = 100000 for i in range(N): r = OW60() x = r+1000 if x &lt; 0: print "Too low:",r if i % 1000 == 0: sys.stderr.write('%d\n' % i) result[x] += 1 i = 0 while result[i] == 0: i += 1 j = len(result) - 1 while result[j] == 0: j -= 1 pSum = 0 # Lower Probability: The probability to throw this or less # Higher Probability: The probability to throw this or higher print "Result;Absolut Count;Probability;Lower Probability;Rel. Lower Probability;Higher Probability;Rel. Higher Probability;" while i &lt;= j: pSum += result[i] print '%d;%d;%.10f;%d;%.10f;%d;%.10f' % (i-1000, result[i], (float(result[i])/N), pSum, (float(pSum)/N), N-pSum, (float(N-pSum)/N)) i += 1 </code></pre>
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    1. COThe curve you describe is the cumulative probability. The probability density curve for this data is interesting. It has two clear peaks (one at 26, and one at 36), with a valley at the mid-point (31). That alone makes me think that your answer will be difficult to determine!
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    2. COI know :) I already asked to professors for statistics and they couldn't think of an answer ...
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    3. COthat should have been "two professors"
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