# Is it better to make two separate contested rolls or one with advantage?

I'm trying to model the benefit of enemies working together in contests with my PCs.

Assume I have a pair of enemies that are attempting to shove a PC. They are smart enough for one to use the Help action if it is better to do so. (Shoving is covered on pg 195-196 of the Player's Handbook and is a contest and the Help action is on pg 192)

Intuitively, it seems better for the enemies to gang up and roll a Strength check once (with advantage) against a single Strength (or Dexterity) check made by the PC. However, this surprising AnyDice result suggests it's actually better for each enemy to make their own contest against the PC, with the PC rolling separately against each (what I'm calling "double opposed" in the AnyDice result). The double opposed result appears to be correct since it is the square of the "single opposed" result as it should be for independent die rolls.

Can someone help me to understand why two independent rolls are better in this situation than two rolls taking the higher result (or point out where I've made a mistake)?

• Are we assuming here that there are no modifiers, just clean 1d20 rolls? Mar 28 at 7:23
• Also note that two shoves can achieve knockdown+move 5ft whereas using the help action oy achieves one. Likewise a 'double grappled' creatures theoretically has to escape grapple twice. Mar 28 at 10:38
• For completeness you should probably also compare the benefits of helping to either counter disadvantage, or provide advantage against a target that itself has advantage (like a raging barbarian with strength checks) Those might lend themselves to helping. Mar 28 at 13:49
• To generalize what @AncientSwordRage said: An additional benefit of two individual shoves is also that if the first PC succeeds in the shove, the second PC can use their action to do whatever they want (attack, heal, shove, grapple).
– RHS
Mar 28 at 14:44
• Double opposed gives enemy #2 freedom to act in case enemy #1 wins their test. That's a tactical advantage that, while not relevant for the mathematical exercise, is still important in the game. Mar 28 at 16:05

Assuming that all involved bonuses are equal:

If you have a single roll with advantage, the attackers roll 2 dice, the defender rolls one die, and the attackers win if the highest of those three die rolls is one of their two die rolls, so they win 2/3 of the time.

If you have two separate attempts, each individual contest is a 50/50 chance, so the attackers win if they win either of two coin flips, so a 3/4 chance.

• That's a really simple (read: awesomely simple) explanation. Mar 28 at 9:11
• It's also a mathematically incorrect explanation. In both cases, the defender's roll result(s) have a 50% chance of beating each of the attackers' die rolls Mar 28 at 16:34
• @TheoBrinkman Right. If you have two rolls each with a 50/50 chance of winning, and you only need to win one roll, then you have a 3/4 chance of success Mar 28 at 18:07
• The difference is the player rolls only once when they have advantage. If she rolls high, she gets to keep that roll for both dice, so is more likely to win. I'm not sure the 2/3rds is correct, but advantage is not just 50%x50%. Mar 28 at 19:04
• Just to be clear, this explanation is correct enough if the rolls cannot tie. Someone has to win ties, and that is where the slight differences away from 50/50, 2/3, and 3/4 come from in AnyDice. Mar 28 at 19:23

## In the double-opposed scenario, your opponent needs to not roll badly multiple times, versus only once against advantage.

To help visualize this intuitively, lets simplify the scenario slightly to a d2 (essentially a 50-50 coin flip), and you win if your result strictly beats your opponent's result. When expanded out to a d20, the visualization is effectively the same, just with many more values and a more complicated calculation of the actual probabilities.

### Single-Opposed - 1d2 versus 1d2

In your single-opposed scenario, only 2d2 is rolled with 4 possible combinations, and you win 25% (1 in 4) of the time based on the following results:

You Opponent Winner
1 1 Opponent
1 2 Opponent
2 1 You
2 2 Opponent

If your opponent rolls their max, you always lose. Otherwise, it's a 50-50.

### With Advantage - [highest of 2d2] versus 2d2

When rolling with advantage, you roll 2d2 against your opponent's 1d2. As before, you still lose if your opponent rolls a 2 (in 50% of the scenarios), but when they don't, you now have a much better chance of rolling a 2 to beat them. The final probabilities in this scenario has you winning 37.5% (3 in 8) of the time.

1 1 1 Opponent
1 1 2 Opponent
1 2 1 You
1 2 2 Opponent
2 1 1 You
2 1 2 Opponent
2 2 1 You
2 2 2 Opponent

### Double-Opposed - 1d2 versus 1d2, twice

However, now let's consider your double-opposed scenario. Here, you're essentially taking two copies of the single-opposed table above. You consider the final outcome "successful" by winning just one of the two attempts, but your opponent only considers the final outcome successful if neither of your individual attempts are successful. As a result, in the first attempt's table, any case where your opponent would win is instead replaced by a second attempt.

This causes the probabilities to break down as such:

• On the first attempt, you win 25% of the time and make a second attempt 75% of the time.
• On the second attempt, you win 25% of the time and lose 75% of the time, but only if this attempt needed to be made (if you won on the first result, this doesn't matter)

In this scenario, because the outcomes are conditional on each other (that is, the results of the second attempt are conditional on the outcome of the first attempt - you only make the second attempt if the first attempt fails), you can find the odds of each possible outcome happening by multiplying together the probabilities of all the individual events leading up to that result. Here, a win on the first attempt and a win on the second attempt are treated as two separate outcomes, statistically speaking.

• 25%: You win on the first attempt
• 75% x 25% = 18.75%: You lose on the first attempt (75%), but you win on the second attempt (25%)
• 75% x 75% = 56.25%: You lose on the first attempt (75%), and you lose on the second attempt (75%) (where 25% + 18.75% + 56.25% = 100% of all possible outcomes)

Resummarized, in double-opposed, you win 25% (attempt 1) + 18.75% (attempt 2) = 43.75% of the time, and your opponent wins 0% (attempt 1) + 56.25% (attempt 2) == 56.25% of the times.

A tweak of your AnyDice program confirms these numbers in the d2 scenario, and also validates that your original program was written correctly. All changing the size of the dice does is change the specific probabilities of who wins because it changes the probabilities that the rolls tie.

On d2s, a tie will happen 50% of the time, but only 5% of the time on a d20. This 45% difference gets split between the "you win outright" (rolled higher) and "you lose outright" (rolled lower) outcomes evenly 22.5% each. Since a tie otherwise meant your opponent wins, that difference improves your odds of success by 22.5%, which is where the 47.5% win chance comes from in your original AnyDice program.

### Side Note - "Winning twice" in Double-Opposed

What this analysis has overlooked until now is the difference, gameplay wise, that a double-opposed scenario could result in - that you actually win twice by making the second attempt regardless and get to take benefits from both of them. Whether you want that possibility is something you should consider at a gameplay level.

However, for the sake of completeness, you can find the odds of that happening in the same way we can find the odds of each outcome for double-opposed. In the d2 scenario:

• You win both attempts 25% x 25% == 6.25% of the time
• You win the first attempt but lose the second 25% x 75% == 18.75% of the time
• You lose the first attempt but win the second 75% x 25% == 18.75% of the time
• You lose both attempts 75% x 75% == 56.25% of the time

Where both "win once" outcomes are equivalent to you and can be considered one combined 37.5% outcome.

In the d20 scenario, instead of using 25% and 75% as the outcomes, you can substitute your 47.5% and 52.5% numbers from your original AnyDice program for the equivalent exact results (win twice ~22.5% of the time, win once ~49.9% of the time, and lose both ~27.6% of the time, rounding away several decimal places for each of reading).

• There is another point in favor of the Double-Opposed scenario. If enemy #1 succeeds on the shove (first attempt, 25%), attempt, enemy #2 has a free action to attack. Mar 28 at 16:03

why two independent rolls are better in this situation than two rolls taking the higher result

The critical difference isn't that the enemies are making two independent rolls, since with advantage they are also making two rolls. It's that the player isn't making two independent rolls in the advantage case. In the independent rolls case, the player winning the first roll doesn't have any bearing on their chances of winning the second roll, since they are going to be rolling a fresh check total. In the advantage case, conditional on the player beating the first enemy roll, they probably rolled high, which increases their chances of beating the second enemy roll as well since their first check total is effectively reused.

## Helping does not help

There are only a few variables to consider here:

• the net bonus of the attacker vs. the defender $$\b=b_A-b_D\$$.
• and the various d20 results which are uniform random variables from 1 to 20 $$\D\$$ with various subscripts to indicate who is rolling.

To succeed with the help action we are looking for:

$$Pr\left({\lceil D_{A1},D_{A2}\rceil}+b>D_D\right)$$

And to succeed with 2 shoves, we are looking for:

$$Pr\left(((D_{A1}+b>D_D)+(D_{A2}+b>D_D))>0\right)$$

So what is the probability that the former is worse, the same, or better than the latter?

Now, I could continue to do this analytically but, we live in a computer age and its far easier to do it empirically. This anydice gives the answer for net bonuses ranging from -5 to +5:

loop B over {-5..5} {
output (([highest 1 of 2d20]+B)>1d20) - (((1d20+B>1d20)+(1d20+B>1d20))>0) named "Bonus: [B]"
}


In every case, the chance of doing worse by helping is greater than the chance of doing better although the difference gets less the higher the net bonus.

## Where it does help

This assumes that both shovers are equally good at shoving. However, if you have a mismatched pair such that one is good and one is lousy, this changes things and, depending on the difference, it may be better for the weaker to help the stronger.

• I followed your approach of subtracting the two probabilities to see where one was better, the same, or worse and came up with anydice.com/program/2805d which (I think correctly) shows that the situation changes if there are 4 or more attackers that pair up to use advantage (with odd numbers having a single shover trying it on their own). The improvements are there but only marginal. The biggest difference is with lower numbers of attackers (<4). Easiest seen by clicking 'transpose' Apr 3 at 1:48

# Another visual, with no math!

Alright so I don't think I can beat @The Zach Man's answer for simplicity and conciseness... but I do have another way of looking at this that may resonate with some players.

• Help Scenario: Enemies roll Advantage (two dice) to the player's one.
• Two Attempts scenario: Enemies roll two dice to the player's two dice.
• Sound familiar? This is statistically similar to rolling Advantage against Disadvantage.

Of course it's not the same thing (the Enemies could roll high into the player's high roll and low into the player's low roll), but forcing the player to roll more dice gives them more opportunities to fail (a phenomenon sometimes called rolling to fail when taken to the extreme). Of course this applies to GM controlled entities as well, but it's far less impactful, in general, when it happens to them.