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

Question

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)?

Answer 1

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.

Answer 2

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.

Your First	Your Second	Opponent	Winner
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).

Answer 3

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.

Answer 4

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.

Answer 5

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.