What is the likelihood that my players can succeed in this escape the city scene?

Question

First-time DM running Curse of Strahd for 6 first-time players. In our upcoming session, I'm going to have Strahd appear in the sky and demand that the citizens of a town bring him an NPC he has been chasing. If my players decide not to hand her over, they will have to try to escape while the whole town chases after them trying to get this NPC.

One option they have to escape is to use a teleportation circle that has already been drawn but incorrectly, so it will kill anyone who uses it. My players know this and I've thought up some mechanics for how they could still use this circle.

I was curious about the likelihood my players can succeed.

Fixing the circle

• Doing so requires succeeding on a DC 15 Arcana check.
• Only the bard and sorcerer (both level 5) can attempt to fix the circle because they are the only ones in the party who have this spell on their spell lists, so they would be the only ones with the actual magic power to use it.
• The bard and sorcerer both have a +1 to Arcana, and I would allow another character to use the Help action to give advantage.
• Other than Bardic Inspiration, I don't think they have any spells or features that give them an edge.

The two of them have to complete their own skill challenge to teleport somewhere. No matter what the spell will teleport them away, but depending on how they rolled they will either go somewhere really friendly or really dangerous.

Possible results

I'm thinking that they should have to get 5 successes before 3 failures. The results would be something like:

• 5 successes with no failures = they go somewhere with strong potential allies like the Wizards of the Winery.
• 5 successes with 1 or 2 failures = they are safe but they are in a remote dangerous environment like on top of Mt. Baratok or Argynvostholt.
• 3 failures = they teleport into Strahd's living room.

Really curious what people think about this and what the likelihood is that the party can succeed. How common is it to get three failures? We've done skill challenges before and there was rarely more than one failure, but that was with the whole party rolling.

Answer 1

You've asked: [W]hat the likelihood is that the party can succeed. How common is it to get three failures?

The answer is: it depends on what DC you set, and on your group's skill bonus. If you set the DC to 25 then they will fail; if you set it to 5 then they will succeed. The best way to get a feel for this is to actually do the die rolls, remembering to account for any bonuses they could have, such as advantage or bardic inspiration or guidance.

If (as I suspect) you have set the DC to 15, and if your sorcerer and bard have a +3 to Arcana, then they have a 45% chance to succeed at any given roll, so failure is very likely for them. If they are helping each other and they have advantage, then they have a 70% chance to succeed at any given roll, but in my simulations they still mostly failed at the skill test.

(I simulated this by rolling seven ten-sided dice and counting any result from 8-10 as success. This guaranteed that the party would get either five successes or three failures, but not both. There are various dice simulator programs that you can use for this.)

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Here is a thing I have learned from being a DM: you can use randomness if you want, but you should never put the group in a situation where one of the random outcomes isn't fun.

In particular, if one of the possible random outcomes is that the group winds up in Strahd's living room, then you as DM need to think about what will happen next. What monsters will the group face? How quickly will Strahd learn of the intrusion and arrive on the scene? How will they flee before that happens? Will this be fun for the group?

If another of your random outcomes is that the group winds up somewhere completely safe, then you need to think about what happens after that. Will that be fun?

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You've asked: Really curious what people think about this.

I think you should spend more time thinking about what threats the group will encounter as the bard and sorcerer work to enable the circle. I also think you should spend more time thinking about what will happen if the group tries something other than repairing the circle. (What happens if they flee on foot? What happens if they leave the circle unrepaired, but they tell Strahd that his girlfriend went into the circle and they try to convince him to follow her into it?)

I also want to point out that it's not super realistic to screw up a teleport spell so badly that it puts you in a very specific place where you don't want to be. A more realistic outcome for repairing a magic circle would look like this:

• Bad result: the circle teleports you into a random part of the wilderness and you take a bunch of damage. Now you have to get to safety, even though you're in some monster's territory.
• Average result: the circle teleports you into a random part of the wilderness and you don't take damage. You still have to get to safety even though you're in a monster's territory.
• Good result: the circle teleports you into a part of the wilderness that's close to safety.

Answer 2

Rolling dice is not the exciting part

You can tell this by the amount of times you get your dice bag out and roll dice for excitement. Once you’ve got over the novelty of polyhedral dice it’s basically never. Rolling dice, like stacking dice, is a sign of boredom, not excitement.

Page 6 of the the PHB tells you how to play they game so it’s easy to see where the excitement is: The players describe what they want to do. What’s exciting is deciding what to do, and what happens next, not the rolling of dice.

“What do you do?”

“I try to fix the teleportation circle.”

That’s a decision. That’s exciting. Will I fix it? Won’t I fix it?

Roll.

”You fail. Do you want to keep trying or hack your way out through innocent townspeople?”

”Um … keep trying, I guess?”

That’s not a choice. That’s you getting me to tell you what I already told you again. That’s not fun.

It’s also not in the rules; PHB p.6:

In cases where the outcome of an action is uncertain, the Dungeons & Dragons game relies on rolls of a 20-sided die, a d20, to determine success or failure.

A 20-sided die, not several 20-sided dice.

Once a player declared an action, the DM should resolve it fully, including determining how long it takes.

Alternative solution

Set the DC. If they roll this exactly it takes 6 rounds. For each number higher than this, it takes one round less to a minimum of 1. Or for each 2 numbers higher depending on how long you want the confrontation to last and remembering that combat rarely takes more than 3-4 rounds. If they fail by 5 or less it takes 6 rounds to “fix” and takes them somewhere they don’t want to go. If they fail by more than this they just fail.

This plays out like this:

“What do you do?”

“I try to fix the teleportation circle.”

1. “You’re making progress, you think it will take another 2 or 3 rounds.”

2. “You’re making progress but it’s not going quite as you planned. It’s going to take another 5 rounds and you won’t have any control over the destination.”

3. “Your clumsy efforts break the circle beyond your current level of knowledge to fix. This isn’t going to work.”

Now, the player’s have real decisions to make. Now, they’re having fun.

In addition, because it’s only one roll, there’s no tricky probability calculation involved in working out the chance of success. You can decide if it’s easy, medium or hard and use that number. Or, knowing the PC’s bonus you can set the DC to give a known chance of success/failure.

Helping

The bard can help the sorcerer or vice-versa. No one else can. PHB p.175:

A character can only provide help if the task is one that he or she could attempt alone.”

What does failure look like here?

They try. They fail. What happens next?

• The players chop their way out in a sea of innocent blood?

• The townsfolk overwhelm them and hand the hostage over?

• The players calm the crown and negotiate a compromise?

• Something else?

For the game to be meaningful failure has to be a realistic possibility for both the players and the DM. This is a cool little scenario. Just make sure you can deal with the consequences if it all goes pear shaped.

Answer 3

The odds are not in their favor (unless...)

There are a number of very good "frame challenge" answers to this question: answers that suggest why this method might not be fun for the players, or fit the expectations of the game. With the understanding that those answers are an important part of this conversation, let's also take a look at the mathematical likelihood that your players succeed at the task that you've given them.

Let p= probability of a single success on a single roll. (So 1-p = probability of failure on a single roll).

To calculate the probability of 5 successes before 3 failures, we can calculate the following:

1. Seven rolls, with 2 failures and 5 successes (where the last roll is a success)

The probability of this will be:

p⁴(1-p)²(6C2) p

= p⁵(1-p)²(15)

(Here, 6C2 means "6 choose 2", which accounts for the different places the two failures could go in the first six rolls. It is equal to 6 * 5/(2 * 1) = 15)

2. Six rolls with 1 failure and 5 successes (where the last roll is a success)

The probability of this is: p⁴(1-p)¹(5C1) p

=5 * p⁵(1-p)¹

3. Five rolls, all of which are successes.

Probability of this is: p⁵

If we sum the above probabilities, we get the probability of 5 successes before 3 failures:

p⁵(1-p)²(15)+5 p⁵(1-p)¹ + p⁵

Ok... is that good?

Depends on how likely they are to succeed on a single roll.

In the comments attached to your question, you said that the two characters who could attempt this have "+1 to Arcana," and that this would be a DC 15 check. Assuming that means their total roll will be 1d20+1 (e.g. the sorcerer has an Intelligence of 12, the bard has an intelligence of 10, and neither is proficient in Arcana [allowing the bard to get +1 from their Jack of All Trades class feature]), then p= 7/20. In this case, their total probability of getting 5 successes before 3 failures is 5.56% (about the same as if this were a single DC 21 check).

Now, perhaps these characters could attempt to improve their odds, and one of the two (either the Bard or the Sorcerer) could use the Help action each turn, giving the other advantage on the checks. In that case, p= 1-(13/20)² (that is, the probability that you don't fail twice on two rolls). In that case the total probability of 5 successes before 3 failures goes up to 37.192%. Still worse than a coin flip, but a lot more likely than before.

_{NOTE: If these characters were proficient in Arcana (bringing the total check up to 1d20+4), their odds of getting 5 successes before 3 failures increase considerably (to 22.7% normally, 75.64% if they are using the Help action each turn). But given your description of them having "+1 to Arcana," I'm assuming this is not the case.}

The exact influence of Bardic inspiration is hard to calculate here, because they will likely run out before the task is finished (and we're not sure how many successes they will still need when they run out). But if they had Bardic Inspiration on every single roll (they won't), that would add an average of 4.5 to each attempt, bringing p up to .575. This will change the probability of 5 successes before 3 failures to 36.67% (or 85.62% using the Help action).

And if you have a Cleric who can cast Guidance every turn (giving an extra 1d4 to one check per turn), that will influence things as well. But if all these things are happening simultaneously (Help Action, Bardic Inspiration, Guidance, and someone making the actual check), then three of the six PCs are using their actions every turn to influence (or make) this check. That means the party is at half strength to deal with whatever else is going on (for example, an angry mob of villagers attempting to bust the door down and take your NPC). That being said, all of these factors together would bring the probability of five successes before 3 failures up to about 98%.

Bottom line, the probability of success here massively changes depending on the resources the PCs bring to bear in this contest, and the strategies they use. And what they have available is very hard to predict, since we're not sure how many resources they will spend on their way to this circle, or if they will even attempt to go there. It's also quite possible they will change their tactics part way through this attempt. If you're trying to balance this check, it will be hard to do ahead of time.

Answer 4

Skill challenges in AnyDice

To address the mathematical side of the question, here is how to compute skill challenges in AnyDice, assuming all rolls are made with the same dice and modifiers:

function: skill challenge S:n successes before F:n failures on D:d {
  result: S@(S+F-1)dD
}

output [skill challenge 5 successes before 1 failures on d20]
output [skill challenge 5 successes before 3 failures on d20]

\ "Normal" is the highest DC that a set of rolls would succeed at.
  "At Least" is the chance of succeeding for a given DC. \

You can replace the d20 with dice and modifiers of your choice.

How it works

The most rolls the challenge can go before ending is S - 1 + F - 1; at this point, the next roll will end the challenge for a total of S + F - 1 rolls. Even if the challenge ends before this, running the challenge out to S + F - 1 rolls will not change the outcome since there's not enough rolls to reach both S successes and F failures at the same time. So we can always consider exactly S + F - 1 rolls.

Now, if the Sth highest roll is a success, then the S-1 rolls above it will be successes as well, so there will be at least S successes overall. On the other hand, if it is a failure, then so are all the rolls below it and there are at most S-1 successes overall. So the Sth highest roll fully determines the result of the skill challenge.

Further reading

Problems of this type fall under order statistics. For a d20, the result is close to a beta distribution.