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September 23rd, 2017, 02:30 AM  #1 
Newbie Joined: Sep 2017 From: United States Posts: 3 Thanks: 0  Rerolling sixsided dice
I have the formal education of, basically, an eighth grader. So excuse my ignorance. But a few questions. The basic situation is that I'm rolling two sixsided dice, trying to roll a 9 or higher. From what I understand, this chance is 27.77%. 1) If I was able to make the attempt twice, what is the chance that either attempt will come up 9 or greater? 2) Alternatively, If I was able to keep the higher of two dice, and reroll the lower die, how does this affect my chances? Now, I also reask both questions if my target was an 8 instead of 9. The initial chance now being 41.66%. Thank you for your time. 
September 23rd, 2017, 04:06 AM  #2  
Math Team Joined: Jan 2015 From: Alabama Posts: 2,966 Thanks: 807  Quote:
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You cannot roll enough to get a "9" if the higher first roll was less than 3. If the higher roll was 3, probability 5/36, then you must roll a 6, probability 1/6, the second time to get "9". The probability is (5/36)(1/6)= 5/216= 0.023. If the higher roll was 4, probability 7/36, then you must roll either a 5 or a 6 on the second roll to get 9 or higher. The probability of rolling a 5 or a 6 is 2/6= 1/3 so the probability of this is (7/36)(1/3)= 7/108= 0.065. If the higher roll was 5, probability 9/36= 1/4, then you must roll a 4, 5, or 6 on the second roll to get 9 or higher. The probability of rolling a 4, 5, or 6 is 3/6= 1/2 so the probability of this is (1/4)(1/2)= 1/8. If the higher roll was 6, probability 11/36, then you must roll 3, 4, 5, or 6 the second time to get 9 or higher. The probability of rolling 3, 4, 5, or 6 is 4/6= 2/3 so the probability of getting 9 or higher is (11/36)(2/3)= 22/108. Add all of those to get the probability of rolling 9 or higher in this way. Quote:
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Last edited by Country Boy; September 23rd, 2017 at 04:18 AM.  
September 23rd, 2017, 06:06 AM  #3 
Newbie Joined: Sep 2017 From: United States Posts: 3 Thanks: 0 
Okay, that makes sense. So, if the target is 9, the chance is 27.77%, if I can reroll both dice, it's 47.7%, if I can reroll the lowest instead, it's 57.88%. Because, after adding together all the combinations that would make the reroll work, 41.67% chance of the reroll succeeding. So to see my total chances of success, I would first find out the chances of failure of the initial roll failing (10.2777) and multiply that by the chances of the reroll failing (1.4167) to get the total chance of failure, which is 42.13%. For success, instead, I'd subtract .4212 from 1, to get 57.88% For this target number, it would be better to reroll the lowest die. 
September 23rd, 2017, 06:37 AM  #4 
Newbie Joined: Sep 2017 From: United States Posts: 3 Thanks: 0 
And, I also eventually got: Success rate Target number is 8 41.66% 65.96% with a reroll 75.4% reroll lowest Target number is 7 58.33% 82.63% with a reroll 76.65% reroll lowest Target number is 6 72.22% 92.28% with a reroll 80.7% reroll lowest I really should show my math. But I want to see trends. I'll work on that next. "When is it better to reroll the lowest die compared to reroll both dice? On a target number where you know the better option between those two, would it be better to reroll, or lower the target number by 1?" For me, this is a lot of math, heh. But I do thank you for your reply 

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dice, rerolling, sixsided 
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