Posted 16 Apr 2013 Suppose n fair 6-sided dice are rolled simultaneously. What is the expected value of the score on the highest valued die? 0 Share this post Link to post Share on other sites

0 Posted 29 Apr 2013 7 - (1^n + 2^n + 3^n + 4^n + 5^n + 6^n)/6^n 0 Share this post Link to post Share on other sites

0 Posted 16 Apr 2013 I'll be a smart a** and say it's seven minus the expected value of the lowest valued die. Seriously, I'll work on it. 0 Share this post Link to post Share on other sites

0 Posted 28 Apr 2013 We seek the expected value of the highest individual score when n dice are thrown. We first find p_{n}(k), the probability that the highest score is k. There are k^{n} ways in which n dice can each show k or less. 0 Share this post Link to post Share on other sites

0 Posted 29 Apr 2013 Suppose n fair 6-sided dice are rolled simultaneously. What is the expected value of the score on the highest valued die? Here's my answer Let P_{N}( i ) be the chance of having i as the highest score on simultaneous rolls of N dice. Obviously, P_{N}( 0 ) = 0. The general expression for P_{N}( i ) is Now that we have P_{N}( i ), the expected value is simply E = sum P_{N}( i )* i for i from 1 to 6. 0 Share this post Link to post Share on other sites

0 Posted 30 Apr 2013 7 - (1^n + 2^n + 3^n + 4^n + 5^n + 6^n)/6^n Agree. 0 Share this post Link to post Share on other sites

0 Posted 30 Apr 2013 7 - (1^n + 2^n + 3^n + 4^n + 5^n + 6^n)/6^n Nice compact form! 0 Share this post Link to post Share on other sites

0 Posted 30 Apr 2013 7 - (1^n + 2^n + 3^n + 4^n + 5^n + 6^n)/6^n For the highest score to equal k, we must subtract those cases for which each die shows less than k; these number (k − 1)^{n}. So, k is the highest score in k^{n} − (k − 1)^{n} cases out of 6^{n}. In other words, p_{n}(k), the probability that the highest individual score is k, is (k^{n} − (k − 1)^{n})/6^{n}. The expected value, E(n), of the highest score is the sum, from k = 1 to 6, of k · p_{n}(k). 0 Share this post Link to post Share on other sites

0 Posted 30 Apr 2013 (edited) The table below shows E(n), correct to four decimal places, for various values of n, from 1 to 50. As expected intuitively, E(n) approaches a number, as n increases. Expected values n E(n) 1 3.5 2 4.4722 3 4.9583 4 5.2446 5 5.4309 6 5.5603 7 5.6541 8 5.7244 9 5.7782 10 5.8202 20 5.9736 30 5.9958 40 5.9993 50 5.9999 Edited 30 Apr 2013 by BMAD 0 Share this post Link to post Share on other sites

0 Posted 30 Apr 2013 7 - (1^n + 2^n + 3^n + 4^n + 5^n + 6^n)/6^n For the highest score to equal k, we must subtract those cases for which each die shows less than k; these number (k − 1)^{n}. So, k is the highest score in k^{n} − (k − 1)^{n} cases out of 6^{n}. In other words, p_{n}(k), the probability that the highest individual score is k, is (k^{n} − (k − 1)^{n})/6^{n}. The expected value, E(n), of the highest score is the sum, from k = 1 to 6, of k · p_{n}(k). That's the obvious and natural approach. And that's exactly how I got the formula. When you say "close but", you mean that the formula is wrong? 0 Share this post Link to post Share on other sites

0 Posted 30 Apr 2013 it wouldn't be the first time I made a mistake on one of my own problems , but yes, i derived a slightly different answer. 0 Share this post Link to post Share on other sites

0 Posted 30 Apr 2013 it wouldn't be the first time I made a mistake on one of my own problems , but yes, i derived a slightly different answer. I applied witzar's formula to various different N, and I reproduced the same exact table that you gave in here The table below shows E(n), correct to four decimal places, for various values of n, from 1 to 50. As expected intuitively, E(n) approaches a number, as n increases. Expected values n E(n) 1 3.5 2 4.4722 3 4.9583 4 5.2446 5 5.4309 6 5.5603 7 5.6541 8 5.7244 9 5.7782 10 5.8202 20 5.9736 30 5.9958 40 5.9993 50 5.9999 0 Share this post Link to post Share on other sites

0 Posted 30 Apr 2013 There are k^{n} ways in which n dice can each show k or less. For the highest score to equal k, we must subtract those cases for which each die shows less than k; these number (k − 1)^{n}. So, k is the highest score in k^{n} − (k − 1)^{n} cases out of 6^{n}. In other words, p_{n}(k), the probability that the highest individual score is k, is (k^{n} − (k − 1)^{n})/6^{n}. The expected value, E(n), of the highest score is the sum, from k = 1 to 6, of k · p_{n}(k). Hence E(n) = [6(6^{n}−5^{n}) + 5(5^{n}−4^{n}) + 4(4^{n}−3^{n}) + 3(3^{n}−2^{n}) + 2(2^{n}−1^{n}) + 1(1^{n}−0^{n})]/6^{n}. = 6 − (1^{n} + 2^{n} + 3^{n} + 4^{n} + 5^{n})/6^{n}. 0 Share this post Link to post Share on other sites

0 Posted 30 Apr 2013 There are k^{n} ways in which n dice can each show k or less. For the highest score to equal k, we must subtract those cases for which each die shows less than k; these number (k − 1)^{n}. So, k is the highest score in k^{n} − (k − 1)^{n} cases out of 6^{n}. In other words, p_{n}(k), the probability that the highest individual score is k, is (k^{n} − (k − 1)^{n})/6^{n}. The expected value, E(n), of the highest score is the sum, from k = 1 to 6, of k · p_{n}(k). Hence E(n) = [6(6^{n}−5^{n}) + 5(5^{n}−4^{n}) + 4(4^{n}−3^{n}) + 3(3^{n}−2^{n}) + 2(2^{n}−1^{n}) + 1(1^{n}−0^{n})]/6^{n}. = 6 − (1^{n} + 2^{n} + 3^{n} + 4^{n} + 5^{n})/6^{n}. Nevermind, they are equivalent. See witzar's 0 Share this post Link to post Share on other sites

0 Posted 30 Apr 2013 (edited) I concede to Witzar. There are k^{n} ways in which n dice can each show k or less. For the highest score to equal k, we must subtract those cases for which each die shows less than k; these number (k − 1)^{n}. So, k is the highest score in k^{n} − (k − 1)^{n} cases out of 6^{n}. In other words, p_{n}(k), the probability that the highest individual score is k, is (k^{n} − (k − 1)^{n})/6^{n}. The expected value, E(n), of the highest score is the sum, from k = 1 to 6, of k · p_{n}(k). Hence E(n) = [6(6^{n}−5^{n}) + 5(5^{n}−4^{n}) + 4(4^{n}−3^{n}) + 3(3^{n}−2^{n}) + 2(2^{n}−1^{n}) + 1(1^{n}−0^{n})]/6^{n}. = 6 − (1^{n} + 2^{n} + 3^{n} + 4^{n} + 5^{n})/6^{n}. Let's take a look at the highlighted equation above Let's play with only 1 die so n = 1. So the expected value of the highest number is simply (1 + 2 + 3 + 4 + 5 + 6)/6 = 3.5 But if you plug it in the above formula, you'll get a different value for the expected face number 6 - (1 + 2 + 3 + 4 + 5 + 6)/6 = 6 - 3.5 = 2.5 Edited 30 Apr 2013 by BMAD 0 Share this post Link to post Share on other sites

0 Posted 30 Apr 2013 There are k^{n} ways in which n dice can each show k or less. For the highest score to equal k, we must subtract those cases for which each die shows less than k; these number (k − 1)^{n}. So, k is the highest score in k^{n} − (k − 1)^{n} cases out of 6^{n}. In other words, p_{n}(k), the probability that the highest individual score is k, is (k^{n} − (k − 1)^{n})/6^{n}. The expected value, E(n), of the highest score is the sum, from k = 1 to 6, of k · p_{n}(k). Hence E(n) = [6(6^{n}−5^{n}) + 5(5^{n}−4^{n}) + 4(4^{n}−3^{n}) + 3(3^{n}−2^{n}) + 2(2^{n}−1^{n}) + 1(1^{n}−0^{n})]/6^{n}. = 6 − (1^{n} + 2^{n} + 3^{n} + 4^{n} + 5^{n})/6^{n}. 7 - (1^n + 2^n + 3^n + 4^n + 5^n + 6^n)/6^n = = 6 + 6^n/6^n - (1^n + 2^n + 3^n + 4^n + 5^n + 6^n)/6^n = = 6 - (1^n + 2^n + 3^n + 4^n + 5^n)/6^n 0 Share this post Link to post Share on other sites

0 Posted 30 Apr 2013 There are k^{n} ways in which n dice can each show k or less. For the highest score to equal k, we must subtract those cases for which each die shows less than k; these number (k − 1)^{n}. So, k is the highest score in k^{n} − (k − 1)^{n} cases out of 6^{n}. In other words, p_{n}(k), the probability that the highest individual score is k, is (k^{n} − (k − 1)^{n})/6^{n}. The expected value, E(n), of the highest score is the sum, from k = 1 to 6, of k · p_{n}(k). Hence E(n) = [6(6^{n}−5^{n}) + 5(5^{n}−4^{n}) + 4(4^{n}−3^{n}) + 3(3^{n}−2^{n}) + 2(2^{n}−1^{n}) + 1(1^{n}−0^{n})]/6^{n}. = 6 − (1^{n} + 2^{n} + 3^{n} + 4^{n} + 5^{n})/6^{n}. 7 - (1^n + 2^n + 3^n + 4^n + 5^n + 6^n)/6^n = = 6 + 6^n/6^n - (1^n + 2^n + 3^n + 4^n + 5^n + 6^n)/6^n = = 6 - (1^n + 2^n + 3^n + 4^n + 5^n)/6^n if they are equivalent then why don't they produce the same expected values as Bushindo pointed out. 0 Share this post Link to post Share on other sites

0 Posted 30 Apr 2013 There are k^{n} ways in which n dice can each show k or less. For the highest score to equal k, we must subtract those cases for which each die shows less than k; these number (k − 1)^{n}. So, k is the highest score in k^{n} − (k − 1)^{n} cases out of 6^{n}. In other words, p_{n}(k), the probability that the highest individual score is k, is (k^{n} − (k − 1)^{n})/6^{n}. The expected value, E(n), of the highest score is the sum, from k = 1 to 6, of k · p_{n}(k). Hence E(n) = [6(6^{n}−5^{n}) + 5(5^{n}−4^{n}) + 4(4^{n}−3^{n}) + 3(3^{n}−2^{n}) + 2(2^{n}−1^{n}) + 1(1^{n}−0^{n})]/6^{n}. = 6 − (1^{n} + 2^{n} + 3^{n} + 4^{n} + 5^{n})/6^{n}. 7 - (1^n + 2^n + 3^n + 4^n + 5^n + 6^n)/6^n = = 6 + 6^n/6^n - (1^n + 2^n + 3^n + 4^n + 5^n + 6^n)/6^n = = 6 - (1^n + 2^n + 3^n + 4^n + 5^n)/6^n if they are equivalent then why don't they produce the same expected values as Bushindo pointed out. I misread the equation. Yours was 6 − (1^{n} + 2^{n} + 3^{n} + 4^{n} + 5^{n})/6^{n}. But I misread it as 6 − (1^{n} + 2^{n} + 3^{n} + 4^{n} + 5^{n} + 6^{n}) /6^{n}. Mea culpa =) 0 Share this post Link to post Share on other sites

0 Posted 30 Apr 2013 if they are equivalent then why don't they produce the same expected values as Bushindo pointed out.They do of course. Bushindo made a silly error evaluating your formula. 0 Share this post Link to post Share on other sites

0 Posted 30 Apr 2013 oh good, i can stop pulling my hair out now. Well done witzar. 0 Share this post Link to post Share on other sites

0 Posted 1 May 2013 I have a question. What will be the expected value if only 1 dice is rolled? 0 Share this post Link to post Share on other sites

0 Posted 1 May 2013 I have a question. What will be the expected value if only 1 dice is rolled? i believe 3.5 0 Share this post Link to post Share on other sites

0 Posted 2 May 2013 Ok thanks. I didnt realise expected value may be a fraction. 0 Share this post Link to post Share on other sites

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Suppose n fair 6-sided dice are rolled simultaneously. What is the expected value of the score on the highest valued die?

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