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<blockquote data-quote="orsal" data-source="post: 3532212" data-attributes="member: 16016">a formulafor &quot;higher of two dn&quot;:The possible results are 1,2,3,...,nThe probability of getting 1 is 1/n^2The probability of getting 2 is 3/n^2The probability of getting 3 is 5/n^2The probability of getting k is (2k-1)/n^2The probability of getting n is (2n-1)/n^2Now the expected value (mathematical jargon for mean average in this context) isthe sum from k=1 to k=n of k(2k-1)/n^2Distribute the 1/n^2 and focus on the sum of k(2k-1). This is the sum of 2k^2-k.The sum of k, k=1 to k=n, is n(n+1)/2The sum of k^2, k=1 to k=n, is n(n+1)(2n+1)/6So the sum of 2k^2-k is 2n(n+1)(2n+1)/6 - n(n+1)/2= [n(n+1)/2][(4n+2)/3 - 1]= [n(n+1)/2][(8n-2)/6]= n(n+1)(8n-2)/12Now remember that 1/n^2 we put aside a little earlier. Then we have(n+1)(8n-2)/12nThis is the formula. Checking it for n=20, to see if it agrees with the people who did that case by itself, we get(21)(158)/240=13.825which agrees with what two people got from calculating this specific case. So I believe my algebra is correct.</blockquote>

[QUOTE="orsal, post: 3532212, member: 16016"] [b]a formula[/b] for "higher of two dn": The possible results are 1,2,3,...,n The probability of getting 1 is 1/n^2 The probability of getting 2 is 3/n^2 The probability of getting 3 is 5/n^2 The probability of getting k is (2k-1)/n^2 The probability of getting n is (2n-1)/n^2 Now the expected value (mathematical jargon for mean average in this context) is the sum from k=1 to k=n of k(2k-1)/n^2 Distribute the 1/n^2 and focus on the sum of k(2k-1). This is the sum of 2k^2-k. The sum of k, k=1 to k=n, is n(n+1)/2 The sum of k^2, k=1 to k=n, is n(n+1)(2n+1)/6 So the sum of 2k^2-k is 2n(n+1)(2n+1)/6 - n(n+1)/2 = [n(n+1)/2][(4n+2)/3 - 1] = [n(n+1)/2][(8n-2)/6] = n(n+1)(8n-2)/12 Now remember that 1/n^2 we put aside a little earlier. Then we have (n+1)(8n-2)/12n This is the formula. Checking it for n=20, to see if it agrees with the people who did that case by itself, we get (21)(158)/240=13.825 which agrees with what two people got from calculating this specific case. So I believe my algebra is correct. [/QUOTE]

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