Odds

Post #214,134 by broomberg 7/9/05 11:42:36 PM Reply	Odds Can someone explain the odds of something to me? Ben had a 100 sided die. He was playing a game that required him to roll a 20, 40, 60, and 80. In any order. He did it. That means the 1st roll he had a 4 in 100 chance. The 2nd was a 3 in 100. The 3rd was a 2 in 100. And finally the last was a 1 in 100. While each roll is independant, there has to be some relationship between each since they are part of the total goal. So what is the formula and what are the odds?
Post #214,135 by ChrisR 7/9/05 11:55:03 PM Reply	Multiply the odds out Forgot the exact formula, but the odds between rolls are multiplied from one to the next.
Post #214,138 by broomberg 7/10/05 12:14:47 AM Reply	So it's 4/100 = 1/25 3/100 = 1/33 2/100 = 1/50 1/100 = 1/100 253350*100 = 1 in 4,125,000
Post #214,142 by ben_tilly 7/10/05 12:57:26 AM Reply	Yup. He was lucky. I have come to believe that idealism without discipline is a quick road to disaster, while discipline without idealism is pointless. -- Aaron Ward (my brother)
Post #214,137 by Another Scott 7/10/05 12:09:52 AM Reply	A decent page on n-sided dice If the 100-sided die is fair, then each roll is independent. The odds of obtaining a particular result is independent of the previous roll, and each face has a 1/100 chance of turning up on top. [link\|http://ken.duisenberg.com/potw/archive/arch99/990111sol.html\|This] page discusses the odds of a particular result using 20-sided dice, and someone extended it to n-sided dice. The odds of rolling a particular value from [1, 100] are: 1/100 The acceptable values for the 4 rolls are: 20, 40, 60, 80 40, 60, 80, 20 60, 80, 20, 40 80, 20, 40, 60 20, 80, 60, 40 20, 80, 40, 60 20, 60, 80, 40 20, 40, 80, 60 40, 80, 60, 20 40, 80, 20, 60 40, 20, 60, 80 40, 20, 80, 60 Etc. Each of these combinations of the 4 values is equally likely. Figuring out the rest of the math is left as an excercise for the reader. ;-) Cheers, Scott.
Post #214,139 by JayMehaffey 7/10/05 12:21:30 AM Reply	Not good. The problem you describe is the simple case. Because there are no rules or other factors that apply across rolls, you can simply divide the possible winning rolls by the total possible rolls. Because the order doesn't matter there are 4! (24) possible combinations that win, out of a 100^4 (100000000) possible combinations. That is a rather slim 0.000024% chance of success, or 1 in 4166667. Jay
Post #214,140 by broomberg 7/10/05 12:33:52 AM Reply	He did it Playing Dungeons and Dragons. The game came to a stop for about 30 minutes as people freaked out about it.
Post #214,144 by Ashton 7/10/05 2:11:58 AM Reply	(Note too) In this event, we may be reasonably assured that.. He did so without 4! or even, a single prayer to a $Deity ummm ;^>

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