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<blockquote data-quote="babomb" data-source="post: 2634409" data-attributes="member: 1316">I can give you the C++ code I used to figure it out. Will that help?[code]#include &lt;cmath&gt;#include &lt;cstdlib&gt;#include &lt;iostream&gt;using namespace std;const int DIE_TYPE = 10;const double PROB_NOT_ONE = (double)(DIE_TYPE-1)/DIE_TYPE;const double PROB_ONE = 1/(double)DIE_TYPE;const double CUTOFF_PROB = 0.50;int main(){&nbsp; &nbsp; double cum_prob = 0;&nbsp; &nbsp; double cum_prob_not_one = 1;&nbsp; &nbsp; double prob = 0;&nbsp; &nbsp; double last_cum_prob = 0;&nbsp; &nbsp; int i = 0;&nbsp; &nbsp; double median = 0;&nbsp; &nbsp; &nbsp; &nbsp; cout &lt;&lt; &quot;d&quot; &lt;&lt; DIE_TYPE &lt;&lt; endl;&nbsp; &nbsp; &nbsp; &nbsp; for(; cum_prob &lt; CUTOFF_PROB; ++i)&nbsp; &nbsp; {&nbsp; &nbsp; &nbsp; &nbsp; prob = cum_prob_not_one * PROB_ONE;&nbsp; &nbsp; &nbsp; &nbsp; last_cum_prob = cum_prob;&nbsp; &nbsp; &nbsp; &nbsp; cum_prob += prob;&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; cout &lt;&lt; i + 1 &lt;&lt; &quot;: &quot; &lt;&lt; cum_prob &lt;&lt; endl;&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; cum_prob_not_one *= PROB_NOT_ONE;&nbsp; &nbsp; }&nbsp; &nbsp; &nbsp; &nbsp; median = (0.50 - last_cum_prob)/(cum_prob - last_cum_prob) + i - 1;&nbsp; &nbsp; cout &lt;&lt; &quot;median: &quot; &lt;&lt; median &lt;&lt;endl;&nbsp; &nbsp; &nbsp; &nbsp; system(&quot;PAUSE&quot;);&nbsp; &nbsp; return 0;}[/code]The output it gives me is[code]d101: 0.12: 0.193: 0.2714: 0.34395: 0.409516: 0.4685597: 0.521703median: 6.59162Press any key to continue . . .[/code]If you don't understand C++, here's the math class version:First, find the cumulative probability that a one is rolled. You can stop when you get a cumulative probability of greater than 50%.The cumulative probability for the nth roll is as follows:c<img src="https://cdn.jsdelivr.net/joypixels/assets/8.0/png/unicode/64/1f44e.png" class="smilie smilie--emoji" loading="lazy" width="64" height="64" alt="(n)" title="Thumbs down&nbsp; &nbsp; (n)"&nbsp; data-smilie="23"data-shortname="(n)" /> = p(1), n = 1&nbsp; &nbsp;&nbsp; = p(not 1)^(n-1) * p(1) + c(n-1), n &gt; 1I'll use m for the lowest roll with a cumulative property &lt;=.5. For the d10, m=6.Now find the following:A = c(m); For a d10, that's 0.468559 (the 6th roll).B = c(m+1); For a d10, that's 0.521703 (the 7th roll).deltac = .5 - A; For the d10, that's .031441I = the distance between intervals. This is 1 for any die, but if we could find the cumulative probability at half-rolls, we could use 1/2.R = B-A; For a d10, 0.053144.deltax = deltac * I/R; For a d10, that's .59162.median = m + deltax; in this case, 6.59162If geometry helps, you plot the points (m, c(m)) and (m+1, c(m+1)). Then draw a line between them. Find the slope of that line (R/I). Since you know a point on the line (namely, (m, c(m)), you can write the equation of the line as&nbsp;c - c(m) = (R/I) * (median-m).5 - A = R/I * (median - m); c = .5, A = c(m) (as above)deltac = R/I * (median - m); deltac = .5 - A (as above)deltac * I/R = median - mdeltac * I/R + m = mediandeltax + m = median; deltax = deltac * I/RQED.</blockquote>

[QUOTE="babomb, post: 2634409, member: 1316"] I can give you the C++ code I used to figure it out. Will that help? [code] #include <cmath> #include <cstdlib> #include <iostream> using namespace std; const int DIE_TYPE = 10; const double PROB_NOT_ONE = (double)(DIE_TYPE-1)/DIE_TYPE; const double PROB_ONE = 1/(double)DIE_TYPE; const double CUTOFF_PROB = 0.50; int main() { double cum_prob = 0; double cum_prob_not_one = 1; double prob = 0; double last_cum_prob = 0; int i = 0; double median = 0; cout << "d" << DIE_TYPE << endl; for(; cum_prob < CUTOFF_PROB; ++i) { prob = cum_prob_not_one * PROB_ONE; last_cum_prob = cum_prob; cum_prob += prob; cout << i + 1 << ": " << cum_prob << endl; cum_prob_not_one *= PROB_NOT_ONE; } median = (0.50 - last_cum_prob)/(cum_prob - last_cum_prob) + i - 1; cout << "median: " << median <<endl; system("PAUSE"); return 0; } [/code] The output it gives me is [code] d10 1: 0.1 2: 0.19 3: 0.271 4: 0.3439 5: 0.40951 6: 0.468559 7: 0.521703 median: 6.59162 Press any key to continue . . .[/code] If you don't understand C++, here's the math class version: First, find the cumulative probability that a one is rolled. You can stop when you get a cumulative probability of greater than 50%. The cumulative probability for the nth roll is as follows: c(n) = p(1), n = 1 = p(not 1)^(n-1) * p(1) + c(n-1), n > 1 I'll use m for the lowest roll with a cumulative property <=.5. For the d10, m=6. Now find the following: A = c(m); For a d10, that's 0.468559 (the 6th roll). B = c(m+1); For a d10, that's 0.521703 (the 7th roll). deltac = .5 - A; For the d10, that's .031441 I = the distance between intervals. This is 1 for any die, but if we could find the cumulative probability at half-rolls, we could use 1/2. R = B-A; For a d10, 0.053144. deltax = deltac * I/R; For a d10, that's .59162. median = m + deltax; in this case, 6.59162 If geometry helps, you plot the points (m, c(m)) and (m+1, c(m+1)). Then draw a line between them. Find the slope of that line (R/I). Since you know a point on the line (namely, (m, c(m)), you can write the equation of the line as c - c(m) = (R/I) * (median-m) .5 - A = R/I * (median - m); c = .5, A = c(m) (as above) deltac = R/I * (median - m); deltac = .5 - A (as above) deltac * I/R = median - m deltac * I/R + m = median deltax + m = median; deltax = deltac * I/R QED. [/QUOTE]

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