 All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars  perplexus dot info  High Rollers (Posted on 2004-12-08) If I rolled n fair six-sided dice rolled simultaneously, what's the expected value (in terms of n) of the highest valued die?

(Note: "expected value" refers to the number you would expect to get if you ran this simulation many times and averaged the results.)

 Submitted by Gamer Rating: 2.5000 (4 votes) Solution: (Hide) First, let's get the chances of each of the six "__ is the highest die" probabilities of occurring. There is 1 way to get it so no number is greater than 1; all the dice must say 1. There is 2^n ways to get it so no number is greater than 1 because you can get 2 favorable outcomes, 2 on the die or 1 on the die. Since we want at least 1 die to say 2, we must subtract however many ways there are to get at most 1 on every dice, 1 in this case. So there are (2^n)-1 ways to get a highest number of 2. Similar logic can be used with 2 and 3. There are 3^n ways to get no more than 3 on each die, and 2^n ways to get no more than 2 on each die, so there are (3^n)-(2^n) ways to get 3 as the highest number on any die. This logic also applies for highest numbers of 4, 5, and 6. Since each of the n dice has 6 faces, there are 6^n different outcomes. Weighting the probabilities gives: (1(1)+2((2^n)-1)+3((3^n)-(2^n))+4((4^n)-(3^n))+5((5^n)-(4^n))+6((6^n)-(5^n)))/(6^n) = (6(6^n)-(5^n)-(4^n)-(3^n)-(2^n)-1)/(6^n) = 6-(((5^n)+(4^n)+(3^n)+(2^n)+1))/(6^n)) (Problem from here) Comments: ( You must be logged in to post comments.)
 Subject Author Date re(2): solution by charlie Charlie 2004-12-08 19:48:33 re(2): solution by charlie owl 2004-12-08 19:34:50 re: solution by charlie Ady TZIDON 2004-12-08 19:27:54 Simulation results Charlie 2004-12-08 19:00:05 solution Charlie 2004-12-08 18:37:31 No Subject Old Original Oskar! 2004-12-08 18:04:15 Please log in:

 Search: Search body:
Forums (1)