My Math Forum Rerolling six-sided dice

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 September 23rd, 2017, 03:30 AM #1 Newbie   Joined: Sep 2017 From: United States Posts: 3 Thanks: 0 Rerolling six-sided 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 six-sided 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 re-roll the lower die, how does this affect my chances? Now, I also re-ask 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, 05:06 AM   #2
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Quote:
 Originally Posted by Mechanicalpear 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 six-sided dice, trying to roll a 9 or higher. From what I understand, this chance is 27.77%.
Yes, that is correct. Since there are 6 equally likely possible outcomes on each die, from 1 to 6, there are 6*6= 36 possible outcomes with two dice, (1, 1) to (6, 6). Of those, four, (3, 6), (4, 5), (5, 4), and (6, 3), add to 9, three, (4, 6), (5, 5), and (6, 4) add to 10, two, (5, 6) and (6, 5), add to 11, and one, (6, 6), adds to 12. 10 of the 36 possible rolls add to 9 or higher. The probability of rolling 9 or higher is $\displaystyle \frac{10}{36}= \frac{5}{18}= 0.2777$ or 27.77%.

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 1) If I was able to make the attempt twice, what is the chance that either attempt will come up 9 or greater?
The probability one roll does not give 9 or higher is 1- 0.277= 0.723. The probability you do not get 9 or higher on either roll is $\displaystyle (0.723)^2= 0.523$. The probability you do get 9 or higher on at least one of the two rolls is 1- 0.5231= 0.477 or 47.7%.

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 2) Alternatively, If I was able to keep the higher of two dice, and re-roll the lower die, how does this affect my chances?
This is more tedious. Of the 36 possible outcomes for two dice, there is one in which the higher die is 1, ((1, 1)), three in which the higher die is 2 ((1, 2), (2, 2), and (2, 1)), five in which the higher die is 3, ((1, 3), (2, 3), (3, 3), (3, 2), and (3, 1)), seven in which the higher die is 4, ((1, 4), (2, 4), (3, 4), (4, 4), (4, 3), (4, 2), and (4, 1)), nine in which the higher die is 5, ((1, 5), (2, 5), (3, 5), (4, 5), (5, 5), (5, 4), (5, 3), (5, 2), and (5, 2)), and 11 in which the higher die is 6, (1, 6), (2, 6), (3, 6), (4, 6), (5, 6), (6, 6), (6, 5), (6, 4), (6, 3), (6, 2), and (6, 1)).

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.

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 Now, I also re-ask both questions if my target was an 8 instead of 9. The initial chance now being 41.66%.
Yes, that is correct. You can get a total of 8 on two dice by (2, 6), (3, 5), (4, 4), (5, 3), or (6, 2), 5 out of the 36 possible rolls. Adding that to the 10 possible ways of getting 9 or higher, above, there are 5+ 10= 15 ways to get 8 or higher rolling two dice. The probability of rolling 8 or higher with a pair of dice is 15/36= 0.41666... or 41.66%. Do the rest of the problem as above, using 8 instead of 9.

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Last edited by Country Boy; September 23rd, 2017 at 05:18 AM.

 September 23rd, 2017, 07:06 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 re-roll 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 re-roll work, 41.67% chance of the re-roll succeeding. So to see my total chances of success, I would first find out the chances of failure of the initial roll failing (1-0.2777) and multiply that by the chances of the re-roll 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 re-roll the lowest die.
 September 23rd, 2017, 07:37 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 re-roll 75.4% re-roll lowest Target number is 7 58.33% 82.63% with a re-roll 76.65% re-roll lowest Target number is 6 72.22% 92.28% with a re-roll 80.7% re-roll lowest I really should show my math. But I want to see trends. I'll work on that next. "When is it better to re-roll the lowest die compared to re-roll 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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