# Thread: Varying Dice Rolls in Order

1. ## Varying Dice Rolls in Order

I'm not positive if this is the right place for this(not certain if it's "pre-university" level or not) but my question has to do with rolling dice and getting at least a certain number, in a certain order.
Best way to ask this I think, is to give my primary example. These rolls are made using standard 6 sided dice(in a magical world where no bias exists in the dice).

3d6 roll #1: needs to result in a sum of all rolls equal to 8 or more(83.80% chance)
3d6 roll #2: needs to result in a sum of all rolls equal to 4 or more(99.54% chance)
3d6 roll #3: needs to result in a sum of all rolls equal to 11 or more(50% chance)
3d6 roll #4: needs to result in a sum of all rolls equal to 3 or more(100% chance)
3d6 roll #5: needs to result in a sum of all rolls equal to 3 or more(100% chance)
3d6 roll #6: needs to result in a sum of all rolls equal to 4 or more(99.54% chance)

The order of the rolls IS important. If someone could tell me(and possibly explain) the formula for figuring out the percentage chance for the rolls succeeding in this order, I'd be very thankful.

2. Originally Posted by Arnkel
I'm not positive if this is the right place for this(not certain if it's "pre-university" level or not) but my question has to do with rolling dice and getting at least a certain number, in a certain order.
Best way to ask this I think, is to give my primary example. These rolls are made using standard 6 sided dice(in a magical world where no bias exists in the dice).

3d6 roll #1: needs to result in a sum of all rolls equal to 8 or more(83.80% chance)
3d6 roll #2: needs to result in a sum of all rolls equal to 4 or more(99.54% chance)
3d6 roll #3: needs to result in a sum of all rolls equal to 11 or more(50% chance)
3d6 roll #4: needs to result in a sum of all rolls equal to 3 or more(100% chance)
3d6 roll #5: needs to result in a sum of all rolls equal to 3 or more(100% chance)
3d6 roll #6: needs to result in a sum of all rolls equal to 4 or more(99.54% chance)

The order of the rolls IS important. If someone could tell me(and possibly explain) the formula for figuring out the percentage chance for the rolls succeeding in this order, I'd be very thankful.
Your wording is not very clear. Are you trying to derive the percentages in parentheses?

If so, I don't understand why you included #4 and then #5. They are the exact same thing. Also #2 and #6 are the same thing.

Hopefully you can see that rolling 3 dice, the minimum total is 3. So the probability of getting 3 or more total has to be 1, or 100%.

There are (1/6)^3 ways to roll the dice. Only one of these: (1, 1, 1) results in a total of 3. Therefore the probability of getting total of 4 or more is 1-1*(1/6)^3 = 215/216 $\displaystyle \approx$ 0.99537.

#1 and #3 are a little more involved. I'm not going to derive them unless you confirm that you are indeed trying to derive the values in parentheses.

3. My apologies for being unclear. What I'm trying to do is determine the chance I have of rolling 3d6 6 times, and getting the end results to equal or exceed the numbers listed in the order listed. I realize that a result of 3 would always be 100%, but I do not know if having the 100% chance is useless information or not when the order of rolls is important.

4. Originally Posted by Arnkel
My apologies for being unclear. What I'm trying to do is determine the chance I have of rolling 3d6 6 times, and getting the end results to equal or exceed the numbers listed in the order listed. I realize that a result of 3 would always be 100%, but I do not know if having the 100% chance is useless information or not when the order of rolls is important.
It's still unclear, at least to me. Are you making 18 die rolls total? Define 3d6 and what it means to roll 3d6 6 times, and make sure that "probability of rolling 3d6 6 times" makes sense with your definition.

Edit: Perhaps what you mean is simply the probability that event #1 happens, followed by event #2, etc. Since all events are independent, just multiply

P(desired) = (0.8380)(0.9954)(0.5)(1)(1)(0.9954)

5. Ah, ok, yes, I'm rolling 18 dice, the thing is, that I'm doing it in batches of 3. and each batch of 3 has to be equal to or above a certain number(listed in the initial post). The order of the batch rolls(ie, is this the 1st, 2nd or 3rd set of 3 six sided dice I'm rolling) determines what the target number to equal or beat is. I'm trying to determine the probability that I'll manage to succeed on all the rolls.

6. Originally Posted by Arnkel
Ah, ok, yes, I'm rolling 18 dice, the thing is, that I'm doing it in batches of 3. and each batch of 3 has to be equal to or above a certain number(listed in the initial post). The order of the batch rolls(ie, is this the 1st, 2nd or 3rd set of 3 six sided dice I'm rolling) determines what the target number to equal or beat is. I'm trying to determine the probability that I'll manage to succeed on all the rolls.
Okay, then it's

P(desired) = (0.8380)(0.9954)(0.5)(1)(1)(0.9954)

like I wrote above in my edit.

7. Thank you very much for your help!