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<blockquote data-quote="The Crimson Binome" data-source="post: 6563859" data-attributes="member: 6775031">Thanks for the response. I get that there's a 3-in-8 chance of getting at least one success on each d8, but there's also a chance of getting two or more successes on each d8. I'm trying to find out the average expected number of success to get from each die.There's a 37.5% chance that the first roll of a d8 will be a success.The chance that rolling one die would result in more than one success is equal to the chance of rolling an 8 on the first die (12.5%) times the chance of rolling at least one success on one of the two explosion dice.Of the 64 possible outcomes of rolling 2d8, there are:25 combinations which result in no additional successes.20 combinations which result in one more success.10 combinations which result in one more success, and allow for rolling two more dice.4 combinations which result in two more successes.4 combinations which result in two more successes, and allow for rolling two more dice.1 combination which results in two more successes, and allows for rolling four more dice.And that's as far as I got. Maybe I just need to math-hammer this some more, but I was hoping anyone else might see an intuitive solution.</blockquote>

[QUOTE="The Crimson Binome, post: 6563859, member: 6775031"] Thanks for the response. I get that there's a 3-in-8 chance of getting at least one success on each d8, but there's also a chance of getting two or more successes on each d8. I'm trying to find out the average expected number of success to get from each die. There's a 37.5% chance that the first roll of a d8 will be a success. The chance that rolling one die would result in [I]more[/I] than one success is equal to the chance of rolling an 8 on the first die (12.5%) times the chance of rolling at least one success on one of the two explosion dice. Of the 64 possible outcomes of rolling 2d8, there are: 25 combinations which result in no additional successes. 20 combinations which result in one more success. 10 combinations which result in one more success, [I]and[/I] allow for rolling two more dice. 4 combinations which result in two more successes. 4 combinations which result in two more successes, [I]and[/I] allow for rolling two more dice. 1 combination which results in two more successes, and allows for rolling [I]four[/I] more dice. And that's as far as I got. Maybe I just need to math-hammer this some more, but I was hoping anyone else might see an intuitive solution. [/QUOTE]

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