The short of it is that, mathematically:

Whether attacker or defender rolls against a flat target makes no difference. None.

An opposed roll changes the probability distribution, making it more swingy.

About the second point: In d20 you normally replace the target of *10 + mods* with *1d20 + mods*. So you had a 50% chance of succeeding before hand, that's basically unchanged. If you had a very small chance of succeeding, now you have better odds. If you had a large chance of succeeding, now you have worse odds.

You can see that in this anydice output, which represents the two d20 methods. Click on *view:graph* and *data:at least* to best contrast the two distributions. Note how if you needed a high number on your original d20 roll, now you have a significantly better chance. And with opposed rolls, it's possible to succeed or fail on checks which would be automatic failure/success on a single die.

Now, these different options can definitely change the feel and flow of the game, but those things are pretty subjective. So I'm limiting my answer just to the math.

**To preserve the probabilities exactly, the new DC should be "14 + monster defense."**

# How I got that number

So, you want to convert this:

d20 + monster_save vs. 8 + caster_modifiers

Into this:

d20 + caster_modifiers vs. ?? + monster_save

Here's how to figure out the "??" using a bit of intuition about probability:

- Ignore the modifiers for a second, since you'll be keeping those the same anyway. What's the probability of making a DC 8 check on a straight d20 roll? There are 7 values on a d20 (1 through 7) that fail, and 13 values (8 through 20) that succeed. So 13/20, or 65%. (Try
`output d20 >= 8`

in AnyDice.)
- Now, flip the percentages. You know you want the monster to succeed 65% of the time and fail 35% of the time (before mods). So that means the caster should succeed 35% of the time (7/20) and fail 65% of the time (13/20).
- In other words, you want the lowest 13 numbers on the d20 (conveniently, those are 1 through 13) to be failures. What DC is that? It's actually 14 (because a roll
*equal* to the DC means you've beaten it).
- Thus, to preserve the probabilities exactly, the new formula is:
d20 + caster_modifiers vs. **14** + monster_save

- Try a few examples to check your work.

# System caveats

(This is a bit speculative because the game is unreleased, so we haven't necessarily seen all the rules, and some of them are likely in flux.)

Note that just being the one who gets to make a roll is sometimes a big benefit. For example, D&D Next playtest stuff had the concept of *advantage*, which lets you roll two dice and keep the highest. If defenders were previously able to claim advantage on saves, then shifting rolls to the attacker's side can make spells and monster abilities more powerful even though the raw math is the same.

Likewise, if the system features any kind of "bennies" or "action points" that you can use to enhance a roll, then shifting rolls from defenders to attackers generally makes combat faster and attacks harder to resist. Whether this favors PCs or enemies really depends on who's forcing more saving throws.

## Best Answer

## It doesn't change much

Using the formulas I give further down we can compare the odds of hitting using opposed rolls to normal attack rolls across a wide range of

Attack - (AC - 10)values:For most attack bonus and AC combinations, both methods give very similar results (within 5% of each other). At the extremes, the difference between both methods grows to 6-10%, with one particular combination (e.g. attack +8 vs 10 AC) producing a difference of 11.5%

However, these extreme cases should be rare. Attack bonuses for players usually range from +3 (2 prof bonus + 1 ability mod) to +13 (6 prof bonus + 5 ability mod + magic weapon or fighting style bonuses). For monsters, attack bonuses usually range from +3 at CR 0 to +10 at CR 20 (

Dungeon Master's Guidep.275).Player AC usually ranges from 10 (no armor, 0 Dex mod) to 20 (full plate + shield) and monster AC from 13 to 19 (

ibid.) All the creatures in the Monster Manual with an AC higher than 20 also have CR higher than 20, and even the CR 30 Tarrasque maxes out at 25 AC.By the time characters have attack bonuses as high as +8, they're probably not fighting 10 AC monsters with any regularity. Additionally, players with extremely high attack bonuses will probably invest in Great Weapon Fighting or Sharpshooter for extra damage, which brings their attack bonus back down.

Likewise, it's unlikely that a monster will have much higher attack bonuses than player AC because monsters don't have as many ways of boosting their attack rolls and players tend to invest into AC at least a little.

The red line in the graph can also be shifted up slightly by using

AC - 11instead ofAC - 10.On the topic of randomness...Since dice rolls are independent events, using two dice instead of one has no effect on the outcome of future attacks. Additionally, since there's only two possible outcomes to an attack roll (hit or miss), it doesn't matter how many dice are used to determine the result as long as the probability of hitting stays the same, or close enough for your purposes.

For example, suppose you were betting on the outcome of a coin toss; there's only two possibilities and each is equally likely, so you have a 50% chance of winning.

Now suppose the game is changed to flipping two coins, and you bet on whether both coins will land same side up. There's 4 possible outcomes, each equally likely: HH, HT, TH, and TT. However, you can win with 2 of those 4 outcomes, so your chance of winning is still 50%, just like when you were flipping one coin.

If you only knew the result (win or lose), someone could change from using 1 coin to 2 coins and you wouldn't be able to notice, because your chances of winning stayed the same.

Every method of resolving attack rolls, regardless of how many dice you use, will boil down to producing hits with certain a probability. Any two methods that produce hits with the same (or close enough) probability are interchangeable. If the differences are small, it would take meticulous tracking of results across multiple game sessions for players to notice a difference.

## Probability for single rolls:

The odds of a

d20 + Bonusbeing greater than or equal toACis\begin{equation} P = \frac{21 + Bonus - AC}{20} \end{equation}

Proof:Since every number on the die is equally likely, we can get the probability of hitting by counting the number of rolls that can hit and dividing that by the number of possible rolls (20).

Since the highest total we can roll is

20 + Bonus, subtracting the highest total that can miss (AC - 1) will leave us with the number of rolls that can hit:20 + Bonus - (AC - 1) = 21 + Bonus - ACExample: With an attack bonus of +2 and an AC of 20, there's only three rolls that result in a total of 20 or higher: 18, 19, or 20. The formula correctly gives(21 + 2 - 20)/20 = 3/20 = 0.15, or15%.## Probability for opposed rolls:

If the attacker has a roll modifier of

Xand the defender has a roll modifier ofY, the odds of the attacker rolling higher is:\begin{equation} P = \begin{cases} (20 + X - Y)(21 + X - Y)/800 & \text{if X $\leq$ Y} \\ 1 - (19 + Y - X)(20 + Y - X)/800 & \text{otherwise} \end{cases} \end{equation}

Proof:Since each number on one die is equally likely to come up, every combination of numbers is equally likely when rolling two dice. Figuring out the number of combinations that result in a hit is tricky, but looking at the table of possible outcomes (taken from this answer in Mathematics Stack Exchange) gives us some insight:

\begin{array} {c|cccccc} &1&2&3&4&5&6 \\ \hline \\ 1&=&<&<&<&<&< \\ 2&>&=&<&<&<&< \\ 3&>&>&=&<&<&< \\ 4&>&>&>&=&<&< \\ 5&>&>&>&>&=&< \\ 6&>&>&>&>&>&= \\ \end{array}

(This table is for a d6, but it's not hard to see that the pattern would be the same for a d20.) When both sides have the same modifiers, the combinations that result in a hit (the

>and=signs) form a triangle whose sides are as long as the table (20 rows and columns).If the attacker's modifier is

xpoints lower than the defender's modifier, the triangle shrinks byxunits:\begin{array} {c|cccccc} &2&3&4&5&6&7 \\ \hline \\ 0&<&<&<&<&<&< \\ 1&<&<&<&<&<&< \\ 2&=&<&<&<&<&< \\ 3&>&=&<&<&<&< \\ 4&>&>&=&<&<&< \\ 5&>&>&>&=&<&< \\ \end{array}

So when

X ≤ Ythe sides of the triangle have length20 + X - Y. Fortunately there's a handy formula for counting the number of characters in this triangle:n(n + 1)/2, wherenis the length of the triangle's sides. Replacingnwith20 + X - Ygives(20 + X - Y)(21 + X - Y)/2. Dividing that the total number of possibilities (20 * 20 = 400) gives us the first part of our function.If the attacker's modifier becomes positive, we get a similar table:

\begin{array} {c|cccccc} &0&1&2&3&4&5 \\ \hline \\ 2&>&>&=&<&<&< \\ 3&>&>&>&=&<&< \\ 4&>&>&>&>&=&< \\ 5&>&>&>&>&>&= \\ 6&>&>&>&>&>&> \\ 7&>&>&>&>&>&> \\ \end{array}

The

>and=signs no longer form a triangle, but we can count the number of<signs (i.e. the roll combinations that miss) and subtract the result from 100% instead. The formula is the same except the triangle length starts at 19 instead of 20 since we're no longer including the=signs.