Hello there adventurers! Today I'd like to discuss odds, or the chances you have to roll a particular number. This is, of course, one of the most important topics we can discuss when talking about rolling dice. We want to know what our chances are of hitting that monster or jumping over that ravine, and as gamemasters we want to know how much damage or healing we can expect our monsters and players to do or what the player's chance of success is.
It is fairly simple to determine our odds to roll a particular number on one die. Say we have a 6 sided die and we want to know what our chances are of rolling a 6. Because the die has 6 sides, and (hopefully) our die is fair (the same chance to roll any particular side) we have a 1-in-6 chance of rolling a 6.
The formula for determining probability is:
the number of desired outcomes divided by the number of total possible outcomes
number of desired outcomes
total possible outcomes
What does this mean? Well, we are looking for one specific outcome (6) and there are six possible outcomes (1, 2, 3, 4, 5, and 6). So if we plug in these numbers to our formula
1
6
we get 0.167. We can then multiply this result by 100 to get the percentage chance: 16.7%.
So we have a 16.7% chance to roll a 6 on a six-sided die!
If we check out all our dice we get:
Dice Chance to get any 1 side
d4 25%
d6 16.7%
d10 10%
d12 8.3%
d20 5%
d100 1%
Percentage always totals up to 100%. For example, the chance to roll any one side on a d4 is 25%, and there are 4 sides, so (25 + 25 + 25 + 25) = 100. A way of thinking about this is that when we roll a die, we always get an outcome. We don't have a chance to roll a number that is not on the die. If we roll a d6, we know for certain that we will get an outcome and that outcome will be between 1 and 6!
Range of Numbers
So let's say we want to know the chances of getting a range of results. Maybe we're only interested in our probability of getting a 13 or above on a d20 so we can know our chance to hit a monster. Getting this chance is easy: you simply add together the chances! Since we are interested in getting 8 different outcomes (13, 14, 15, 16, 17, 18, 19, and 20) we can simply multiply our chance of getting each individual result (5%) by the number of outcomes (8) which gives us a total of 40%! We have a 5% chance to roll each outcome, and we are looking for a range of 8 different outcomes.
8 = 0.4 = 40%
20
Keep in mind that when dealing with percentages, the total of all the chances (hits and misses in this case) will always equal 100%. If we make a roll, we will get some result, whether it's the result we want or not. We can therefore always take the inverse of each probability as well. In our above example we have a 40% chance to hit. This also means that we have a 60% chance to miss! It can be very useful to flip the numbers and look at our chances in a different way.
Chances on Multiple Dice
We have so far explored how to calculate our probability of getting the results that we want on one die. Frequently though we are rolling multiple dice in our games however! How do we figure out our chances when multiple dice are involved? It's a little more complicated, but still fairly easy!
The method is similar to calculating with a single die. The formula is still the number of outcomes we want divided by the total number of outcomes. To determine the total number of outcomes with multiple dice, however, we simply multiply the total number of outcomes of each individual die. Remember that the total number of outcomes for a die equals the number of sides of that die. So for example, if we are rolling two d6's, we would have 36 total number of outcomes as 6 x 6 = 36. We can expand this to include any number of any different number-sided dice as well. You simply multiply the number of sides together. The important thing to realize with this though is that this is simply to get any one result on each die. So to see what the chances of getting a 1 on each die if we roll a d6, a d8, and a d20, then we can multiply (6 x 8 x 20) = 960 to get a total of 960 possible outcomes. Keep in mind that that this is not the same as getting a particular result (like, say a 7), but rather the chance to get any one particular side on each of the dice. We will talk about distributions in a future article and discuss the chances of getting a particular result and how that can change (spoiler: some results are much more likely when we roll multiple dice!).
For now though we can use the same formula to determine our chances in the same way. The number of desired outcomes divided by the number of total possible outcomes. So in our above example our chance to roll a 1 on our d6, d8, and d20 in one roll is:
desired outcomes 1
total possible outcomes = 960
Gambits and Longshots
Now that we have figured out how to determine the odds we can make more informed decisions on what actions to take in the game (or inflict upon the characters). We could use this information to guarantee that any actions we take will have a high likelihood to succeed, but where's the fun in that? The characters we play are heroic, and even if they fail catastrophically we don't suffer any harm ourselves, so take the risk!
As an example, I was running a session where the party was going through a mercenary hideout. They knew that the head of the mercenary group was a medusa, so they couldn't look at him (a male medusa!). One of the characters rigged up a mirror to his forehead in the off-chance that the medusa would be affected by his own gaze. The difficulty number to succeed on the saving throw is a 14, but if someone fails the roll by 5 or more, they are instantly petrified, so you need a total result of 9 or less to be instantly turned to stone. The Constitution bonus of the medusa is +3. In order for the medusa to succeed on the saving throw against it's own petrifying gaze, it needs to roll an 11 or above, and it needs to roll a 6 to avoid being turned to stone instantly! If we plug these numbers into our formula, get these results:
chance to make saving throw = 10 / 20 = 0.55 = 55%
chance to be turned to stone instantly = 5 / 20 = 0.25 = 25%
Remember that rolling the difficulty number is a success, so we include that number in our count of desired results. For example, if the medusa needs a result of 11 or better on a d20, then we have (11, 12, 13, 14, 15, 16, 17, 18, 19, 20) = 10 desired results, and if want to roll a 6 or above to avoid being turned to stone instantly we have (1, 2, 3, 4, 5) = 5 desired results, because in this case a roll of a 6 is a success. Also remember that we can swap the percentages around to get a different view of the situation. A 55% chance to make the saving throw also means a 45% chance to fail the saving throw, and a 25% chance to be instantly petrified means that there is a 75% chance that he would not be turned to stone. The total chance always equals 100%; when we roll the dice we always get some result.
To go back to the story, even though there was a 55% chance for the medusa to make the saving throw, and only a 25% chance for him to be petrified instantly, I rolled a 2, for a result of 5 -- below the threshold to be turned to stoned! Even though it was a longshot, it paid off! What would have been a tough and probably memorable fight, it became an epic moment for the group. Their ingenuity and planning paid off.
So we can see from this example, that we would expect that the medusa would probably succeed on their saving throws against their own petrifying gaze, but the longer the fight goes on and the more rolls we have to make the more likely it is that we will get some failures. We also, when planning this encounter, not expect the instant petrification outcome, but we shouldn't be surprised if it does. With a chance of 25%, we should probably plan as if it won't happen, but it makes it more exciting when it does!
Conclusion
We can see that it is not difficult to get more detailed insights into the mechanics so we can make more informed decisions about setting our difficulty numbers. Next time we will look at distributions, which will show us how rolling multiple dice can give us a more consistent and "average" result. See you next time!
