High Rollers (Posted on 2004-12-08)

	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)

	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