Strategies for Calculating Dice Roll Probabilities for Specific Sums and Conditions

Question

I'm working on some probability problems involving multiple dice rolls, and I'm uncertain about how to approach them systematically. I hope someone could guide me through the correct method for calculating these probabilities. I'll list a few examples of the type of problems I'm dealing with:

1. What is the probability of getting a sum of 7 or 8 when rolling two dice?
2. What is the probability of getting a sum of 12, with no fives appearing, when rolling three dice?
3. What is the probability of getting a sum of 12, with at least one five, when rolling three dice?
4. What is the probability of rolling an odd total sum with two dice?
5. What is the probability of rolling a total sum greater than 7 with two dice?
6. What is the probability of getting two sixes when rolling three dice?
7. Should one bet on the event "at least one six appearing in three dice rolls"?
8. What is the probability of rolling either the minimum or maximum sum with three dice?

For instance, for the first problem, I started by listing all possible outcomes for two dice rolls and counted the cases that sum up to 7 or 8. However, as the questions become more complex, like questions 2 and 3, I'm not sure how to account for all the cases without extensive enumeration, especially when dealing with restrictions like 'no fives' or 'at least one five'.

Could you please provide some guidance or strategies for tackling these types of questions more effectively? Any help or insights into how one should approach these from a probability standpoint would be greatly appreciated.

BTW I found a useful post, but I can't figure out if it can be use for other problems.

Thank you in advance!

Answer 1

I may have the equations you're looking for. For the 2 dice, the probability of a sum is (||x-7|-6|-|x-7|+6)/72, where x is the sum of the dice. It's made this way so that the whole sums that the dice can't make are at y=0. For the 3 dice, (((||x-10.5|-8.5|-|x-10.5|+8.5)^2)/1728)-3(((||x-10.5|-2.5|-|x-10.5|+2.5)^2)/1728)+(||x-10.5|-8.5|-3||x-10.5|-2.5|+2|x-10.5|+1)/864. However, if the number of sides of any of the dice aren't 6, then the equation for the 2 dice is (||x-((a1+a2)/2)-d1-d2+1|-((a1+a2)/2)|-||x-((a1+a2)/2)-d1-d2+1|-|(a1-a2)/2||+((a1+a2-|a1-a2|)/2))/2a1a2, where a is the number of sides and d is the starting number of either die number 1 or 2 as long as both their numbers go up by 1 from lowest to highest. The equation for 3 dice is too complex for me to type, so for now, you could use the previous equation for 2 of the 3 dice, except it would end with ".../2" to calculate the number of ways to get a sum from those two, and use the numbers of the third one to help find the number of ways to get the sum before dividing the total by the product of each die's number of sides. For example, if you want to know the probability of getting a sum of 6 from rolling 3 4-sided dice, with 1-4 on each die, you'd start by putting in 4 for both a's and 1 for each d in the equation and find that in order from sum-number of ways, the results you could use are: 2-1, 3-2, 4-3, 5-4, because 6-1=5 and 6-4=2, so you could add the y-values of 2-5 together to get 10 total ways of getting a sum of 6 and divide it by 4 cubed, or 64, to get 5/32 as the probability of such. You could rewrite these equations if you'd like as long as they still have the same points. If they're wrong, though, I'm sorry. I'm also sorry for all the parenthesis in the equations. I just wanted to make sure they were easy to follow.

Answer 2

Often one of the most useful options for exploring these sorts of problems is a spreadsheet program - Counting hundreds of outcomes is very tedious, but computers don't care - as long as you can accurately enumerate the probability space and define your criteria, spreadsheet formulae and filters make it very easy to solve such problems.

Let's, as an example, look at your second question. Assuming we're talking about 6-sided dice, we know that the probability space is going to be $6^n$, where $n$ is the number of dice. For 3 dice, that's 216, not too bad for a manual spreadsheet search. I went and put in all 216 outcomes in a spreadsheet, then had the spreadsheet sum the three dice in one column (so I know which ones are equal to 12) and then count the number of 5s in another column. Once I have that data, it's quite easy to filter on those columns which returns... 10 outcomes. So, the probability is $\frac{10}{216}$.

Indeed, we've now set up this, we can get the third question as well for no more effort - We can filter to return all the rows with 1 or 2 5s instead of the rows with 0 5s... and we get 15 outcomes back, so the probability is $\frac{15}{216}$.

Most of your questions should be pretty amenable to a spreadsheet approach, so I'll leave those as an exercise for you, but the last two are actually a little easier to solve with another strategy - by reframing the problem.

Question #7 is asking us what the odds are of any of three dice rolls showing 6. It turns out, it's easier to solve the reverse of this problem - what's the probability of rolling no 6s - and then subtracting that answer from 1. The probability of not rolling 6 is $\frac{5}{6}$, so the probability of all three dice not rolling a 6 is $\left (\frac{5}{6}\right )^3$, which equals $\frac{125}{216}$, and $1-\frac{125}{216}=\frac{91}{216}$. In fact, this generalises for any number of dice - the odds of rolling at least one 6 on $n$ dice is $1-\left (\frac{5}{6}\right )^n$.

Question #8 is actually incredibly easy to solve, because it is the same as asking "what is the probability that all the dice thrown show the highest/lowest value", and on a standard, fair die, that answer is always going to be $\left (\frac{1}{6}\right )^n$, where $n$ is the number of dice thrown, because there is only 1 outcome that is the maximum/minimum outcome on a standard fair die. So, for three dice, the probability of both the highest and lowest sum is $\frac{1}{216}$

My strongest recommendation for further reading would be to explore binomial distributions and their various generalisations,because if you can reduce a probability space to "success" and "fail", binomials can simplify calculations enormously. Looking into combinations, permutations and multi-sets will also often bear a lot of fruit, because these concepts explicitly revolve around counting, and it turns out that a lot of dice probability questions can be reduced down to counting problems.

Answer 3

These problems seem to be carefully curated to be possible to do with a relatively small amount of enumeration.

Consider the probability of rolling a sum of $12$ on three dice without rolling a five. There is only one way to do this without rolling a six: $4+4+4.$ And there is no way to get a sum of $12$ with more than one six, because after two sixes the third die puts you over. So we are looking for rolls that have one six where the other two dice sum to $6.$ There are three possible ways for the single six to appear, and for each of those ways we count the number of ways to get the sum $6$ on two dice with no fives. (Three ways, since $(1,5)$ and $(5,1)$ are both eliminated.)

That’s $3\times3=9$ ways with one six, plus the single way with all fours, total $10$ ways. It takes a lot of words to explain but very little actual calculation.

For at least one five, you have either one five with the sum $7$ on two dice with no five, or some permutation of $5+5+2.$

For an odd sum on two dice, roll the first die. Whatever it shows is either even or odd. The next die has $1/2$ chance to change the sum to the other parity and $1/2$ chance to stay the same parity. One of those results is an odd parity; whichever it is, the probability we get it is $1/2.$