BMAD 65 Posted April 16, 2013 Report Share Posted April 16, 2013 Suppose n fair 6-sided dice are rolled simultaneously. What is the expected value of the score on the highest valued die? Quote Link to post Share on other sites

0 Solution witzar 18 Posted April 29, 2013 Solution Report Share Posted April 29, 2013 7 - (1^n + 2^n + 3^n + 4^n + 5^n + 6^n)/6^n Quote Link to post Share on other sites

0 bonanova 85 Posted April 16, 2013 Report Share Posted April 16, 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. Quote Link to post Share on other sites

0 BMAD 65 Posted April 28, 2013 Author Report Share Posted April 28, 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. Quote Link to post Share on other sites

0 bushindo 14 Posted April 29, 2013 Report Share Posted April 29, 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. Quote Link to post Share on other sites

0 bonanova 85 Posted April 30, 2013 Report Share Posted April 30, 2013 7 - (1^n + 2^n + 3^n + 4^n + 5^n + 6^n)/6^n Agree. Quote Link to post Share on other sites

0 bushindo 14 Posted April 30, 2013 Report Share Posted April 30, 2013 7 - (1^n + 2^n + 3^n + 4^n + 5^n + 6^n)/6^n Nice compact form! Quote Link to post Share on other sites

0 BMAD 65 Posted April 30, 2013 Author Report Share Posted April 30, 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 âˆ’ 1)^{n}. So, k is the highest score in k^{n} âˆ’ (k âˆ’ 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} âˆ’ (k âˆ’ 1)^{n})/6^{n}. The expected value, E(n), of the highest score is the sum, from k = 1 to 6, of k Â· p_{n}(k). Quote Link to post Share on other sites

0 BMAD 65 Posted April 30, 2013 Author Report Share Posted April 30, 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 April 30, 2013 by BMAD Quote Link to post Share on other sites

0 witzar 18 Posted April 30, 2013 Report Share Posted April 30, 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 âˆ’ 1)^{n}. So, k is the highest score in k^{n} âˆ’ (k âˆ’ 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} âˆ’ (k âˆ’ 1)^{n})/6^{n}. The expected value, E(n), of the highest score is the sum, from k = 1 to 6, of k Â· 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? Quote Link to post Share on other sites

0 BMAD 65 Posted April 30, 2013 Author Report Share Posted April 30, 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. Quote Link to post Share on other sites

0 bushindo 14 Posted April 30, 2013 Report Share Posted April 30, 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 Quote Link to post Share on other sites

0 BMAD 65 Posted April 30, 2013 Author Report Share Posted April 30, 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 âˆ’ 1)^{n}. So, k is the highest score in k^{n} âˆ’ (k âˆ’ 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} âˆ’ (k âˆ’ 1)^{n})/6^{n}. The expected value, E(n), of the highest score is the sum, from k = 1 to 6, of k Â· p_{n}(k). Hence E(n) = [6(6^{n}âˆ’5^{n}) + 5(5^{n}âˆ’4^{n}) + 4(4^{n}âˆ’3^{n}) + 3(3^{n}âˆ’2^{n}) + 2(2^{n}âˆ’1^{n}) + 1(1^{n}âˆ’0^{n})]/6^{n}. = 6 âˆ’ (1^{n} + 2^{n} + 3^{n} + 4^{n} + 5^{n})/6^{n}. Quote Link to post Share on other sites

0 bushindo 14 Posted April 30, 2013 Report Share Posted April 30, 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 âˆ’ 1)^{n}. So, k is the highest score in k^{n} âˆ’ (k âˆ’ 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} âˆ’ (k âˆ’ 1)^{n})/6^{n}. The expected value, E(n), of the highest score is the sum, from k = 1 to 6, of k Â· p_{n}(k). Hence E(n) = [6(6^{n}âˆ’5^{n}) + 5(5^{n}âˆ’4^{n}) + 4(4^{n}âˆ’3^{n}) + 3(3^{n}âˆ’2^{n}) + 2(2^{n}âˆ’1^{n}) + 1(1^{n}âˆ’0^{n})]/6^{n}. = 6 âˆ’ (1^{n} + 2^{n} + 3^{n} + 4^{n} + 5^{n})/6^{n}. Nevermind, they are equivalent. See witzar's Quote Link to post Share on other sites

0 BMAD 65 Posted April 30, 2013 Author Report Share Posted April 30, 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 âˆ’ 1)^{n}. So, k is the highest score in k^{n} âˆ’ (k âˆ’ 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} âˆ’ (k âˆ’ 1)^{n})/6^{n}. The expected value, E(n), of the highest score is the sum, from k = 1 to 6, of k Â· p_{n}(k). Hence E(n) = [6(6^{n}âˆ’5^{n}) + 5(5^{n}âˆ’4^{n}) + 4(4^{n}âˆ’3^{n}) + 3(3^{n}âˆ’2^{n}) + 2(2^{n}âˆ’1^{n}) + 1(1^{n}âˆ’0^{n})]/6^{n}. = 6 âˆ’ (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 April 30, 2013 by BMAD Quote Link to post Share on other sites

0 witzar 18 Posted April 30, 2013 Report Share Posted April 30, 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 âˆ’ 1)^{n}. So, k is the highest score in k^{n} âˆ’ (k âˆ’ 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} âˆ’ (k âˆ’ 1)^{n})/6^{n}. The expected value, E(n), of the highest score is the sum, from k = 1 to 6, of k Â· p_{n}(k). Hence E(n) = [6(6^{n}âˆ’5^{n}) + 5(5^{n}âˆ’4^{n}) + 4(4^{n}âˆ’3^{n}) + 3(3^{n}âˆ’2^{n}) + 2(2^{n}âˆ’1^{n}) + 1(1^{n}âˆ’0^{n})]/6^{n}. = 6 âˆ’ (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 Quote Link to post Share on other sites

0 BMAD 65 Posted April 30, 2013 Author Report Share Posted April 30, 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 âˆ’ 1)^{n}. So, k is the highest score in k^{n} âˆ’ (k âˆ’ 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} âˆ’ (k âˆ’ 1)^{n})/6^{n}. The expected value, E(n), of the highest score is the sum, from k = 1 to 6, of k Â· p_{n}(k). Hence E(n) = [6(6^{n}âˆ’5^{n}) + 5(5^{n}âˆ’4^{n}) + 4(4^{n}âˆ’3^{n}) + 3(3^{n}âˆ’2^{n}) + 2(2^{n}âˆ’1^{n}) + 1(1^{n}âˆ’0^{n})]/6^{n}. = 6 âˆ’ (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. Quote Link to post Share on other sites

0 bushindo 14 Posted April 30, 2013 Report Share Posted April 30, 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 âˆ’ 1)^{n}. So, k is the highest score in k^{n} âˆ’ (k âˆ’ 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} âˆ’ (k âˆ’ 1)^{n})/6^{n}. The expected value, E(n), of the highest score is the sum, from k = 1 to 6, of k Â· p_{n}(k). Hence E(n) = [6(6^{n}âˆ’5^{n}) + 5(5^{n}âˆ’4^{n}) + 4(4^{n}âˆ’3^{n}) + 3(3^{n}âˆ’2^{n}) + 2(2^{n}âˆ’1^{n}) + 1(1^{n}âˆ’0^{n})]/6^{n}. = 6 âˆ’ (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 âˆ’ (1^{n} + 2^{n} + 3^{n} + 4^{n} + 5^{n})/6^{n}. But I misread it as 6 âˆ’ (1^{n} + 2^{n} + 3^{n} + 4^{n} + 5^{n} + 6^{n}) /6^{n}. Mea culpa =) Quote Link to post Share on other sites

0 witzar 18 Posted April 30, 2013 Report Share Posted April 30, 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. Quote Link to post Share on other sites

0 BMAD 65 Posted April 30, 2013 Author Report Share Posted April 30, 2013 oh good, i can stop pulling my hair out now. Well done witzar. Quote Link to post Share on other sites

0 dark_magician_92 4 Posted May 1, 2013 Report Share Posted May 1, 2013 I have a question. What will be the expected value if only 1 dice is rolled? Quote Link to post Share on other sites

0 BMAD 65 Posted May 1, 2013 Author Report Share Posted May 1, 2013 I have a question. What will be the expected value if only 1 dice is rolled? i believe 3.5 Quote Link to post Share on other sites

0 dark_magician_92 4 Posted May 2, 2013 Report Share Posted May 2, 2013 Ok thanks. I didnt realise expected value may be a fraction. Quote Link to post Share on other sites

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## BMAD 65

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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