D&D 5E Introducing the COUNTDOWN DICE Mechanic!

Morrus

Well, that was fun
Staff member
Countdown dice pools are a mechanic which I initially used in my game What's OLD is NEW, and later included in Level Up: Advanced 5E. It's a mechanic which can be used in any game system as long as that system uses (a) dice and (b) turns or rounds. With a heist-themed anthology on the horizon, countdown dice pools could be a useful trick to add tension to encounters or scenes. I've used these in most every game I've run in the last decade or so.

A countdown dice pool is used when there is a time limit on an event, but you (including the DM/GM!) don't know when that time will run out. A ticking bomb, the time until the guard rounds the corner, a dungeon collapsing as you try to escape it--anything where the expiry time is unknown.

Here's how it works: you form a dice pool of d6s. The size of the dice pool can vary. Each round you roll the dice pool, and remove any 6s. When the dice pool is depleted, the countdown ends -- the bomb goes off, the guard comes round the corner, the dungeon collapses.

It's super simple and is great at adding tension to the game. You can even mess with the pool: change the time interval to minutes, hours, even days or more; or certain actions might add dice or remove dice, speeding up or slowing down the timer. Maybe you're trying to disable a trap before it goes off, and you fail a check, cutting the wrong wire, which removes a die from the pool immediately! Or perhaps your pool represents a disease, rolling daily rather than each round, with dire outcomes when the pool reaches zero dice, and an excellent medicine check adds dice to the pool while time depletes it.


 

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aco175

Legend
I like that it is simple and the fact that the DM does not know as well. This may make it appear more fair to the game and players.

It may make it a bit less dramatic though if the party jumps out of the tomb landing as a pile of bodies covered in spiderweb only to have enough time to get up and clean themselves off, rearrange their gear, and take a few potions and a 10 minute heal before the dungeon finally collapses.
 

Aldarc

Legend
ICRPG also uses a countdown mechanic, but it's usually just rolling a 1d4 or 1d6 and counting down from there each round, and then the bad thing happens at the end of the final round. Sure, the players know that something bad will happen, but that meta-knowledge actually contributes to the tension IME.
 

Vendral

Explorer
Has anyone done the math on this and can share the result?
What is the average amount of rolls with different amount of dice and how does the distribution look?
It's been roughly 30 years since I last did this kind of calculations...
 

dave2008

Legend
I don't even know if I realized this was in A5e! Thank you for reminding me / point it out. It is such a simple idea to add some tension.

Anyway, it seems like this could be used in conjunction with the Suspicion mechanic introduced in the "Prisoner 13" heist from the Golden Vault and Kyle Shire's "12 Ways to Add Tension to Your D&D Heist" article on D&D Beyond. Particularly #6 (A Ticking Clock) as he doesn't provide any mechanics for it.
 

Morrus

Well, that was fun
Staff member
I like that it is simple and the fact that the DM does not know as well. This may make it appear more fair to the game and players.

It may make it a bit less dramatic though if the party jumps out of the tomb landing as a pile of bodies covered in spiderweb only to have enough time to get up and clean themselves off, rearrange their gear, and take a few potions and a 10 minute heal before the dungeon finally collapses.
I mean, if you're rolling every round and it's taking 10 minutes of in-game time, you have a lot more dice than I have! That's 60+ rounds! :D
 


Cadence

Legend
Supporter
Has anyone done the math on this and can share the result?
What is the average amount of rolls with different amount of dice and how does the distribution look?
It's been roughly 30 years since I last did this kind of calculations...

--------------------
Edit: See post #23 below for a much shorter way to do it.
---------------------

I think the attached R code using a transition matrix is correct for giving you what turn you run out on.
#Code using transition matrix

#Set the number of starting dice
dat1<-6

#Set how many turns to check for
nturn<-50

#Set the number of sides on the dice
nside<-6

#Make the initial state vector
s1<-c(1,rep(0,dat1))

#Transition matrix
tmat<-matrix(0,ncol=dat1+1,nrow=dat1+1)
for (i in 1: (dat1+1)){
temp<-dbinom(((dat1-i+1):0),
dat1-i+1,(nside-1)/nside)
tmat[i,(i: (dat1+1))]<-temp
}

#Get probability it has run out by that turn
current<-s1%*%tmat
pendby<-current[dat1+1]
for (i in 1: (nturn-1)){
current<-current%*%tmat
pendby<-c(pendby,current[dat1+1])}

#Get probability it has run out on that turn
pendon<-pendby[1]
for (i in 2:nturn){
pendon<-c(pendon,pendby[ i] -pendby[i-1])}

pendon<-c(pendon,1-sum(pendon))
x<-cbind(1: (nturn+1),round(pendon,4),round(c(pendby,1),4))
colnames(x)=c("Turn","Prob On","Prob By")
x

The first value of pendon is the probability of running out of dice on turn 1, etc... The last value is the probability it takes more than whatever you set the max number of turns to check for. pendby will give you the probability of running out on that turn or before.

I checked it by writing one that did a simulation.
#Code using a simulation

#Set the number of starting dice
dat1<-6

#Set how many turns to check for
nturn<-50

#Set the number of sides on the dice
nside<-6

#Set number of simulations
nsims<-100000

simcount<-rep(0,(nturn+1))
for (i in 1:nsims){
curcount<-dat1
turn<-0
while ((curcount>0)&(turn<(nturn+1))){
turn<-turn+1
curcount<-rbinom(1,curcount,(nside-1)/nside)
if (curcount==0){simcount[ i]<-turn}
if ((curcount>0)&(turn==nturn)){simcount[ i]<- (nturn+1)}
}}

table(simcount)/nsims


#Compare them to make sure it worked

cbind(1: (nturn+1),round(pendon,4),
round(table(simcount)/nsims,4))

Here are the probabilities for starting with three dice and going up to 20 turns. (Remember the last entry is "More than 20").

Turn Prob On Prob By
[1,] 1 0.0046 0.0046
[2,] 2 0.0239 0.0285
[3,] 3 0.0462 0.0748
[4,] 4 0.0640 0.1388
[5,] 5 0.0752 0.2140
[6,] 6 0.0802 0.2942
[7,] 7 0.0805 0.3747
[8,] 8 0.0773 0.4520
[9,] 9 0.0720 0.5240
[10,] 10 0.0655 0.5895
[11,] 11 0.0586 0.6481
[12,] 12 0.0517 0.6999
[13,] 13 0.0451 0.7450
[14,] 14 0.0391 0.7841
[15,] 15 0.0336 0.8176
[16,] 16 0.0287 0.8464
[17,] 17 0.0244 0.8708
[18,] 18 0.0207 0.8915
[19,] 19 0.0175 0.9090
[20,] 20 0.0148 0.9238
[21,] 21 0.0762 1.0000

Here it is for six dice

Turn Prob On Prob By
[1,] 1 0.0000 0.0000
[2,] 2 0.0008 0.0008
[3,] 3 0.0048 0.0056
[4,] 4 0.0137 0.0193
[5,] 5 0.0265 0.0458
[6,] 6 0.0408 0.0866
[7,] 7 0.0538 0.1404
[8,] 8 0.0639 0.2043
[9,] 9 0.0703 0.2746
[10,] 10 0.0730 0.3475
[11,] 11 0.0725 0.4201
[12,] 12 0.0697 0.4898
[13,] 13 0.0652 0.5550
[14,] 14 0.0597 0.6148
[15,] 15 0.0538 0.6685
[16,] 16 0.0478 0.7163
[17,] 17 0.0419 0.7583
[18,] 18 0.0365 0.7948
[19,] 19 0.0315 0.8263
[20,] 20 0.0271 0.8534
[21,] 21 0.1466 1.0000

Edit: I think I fixed all the auto-emoji and mark-up making in the code snippets.
 
Last edited:

Dausuul

Legend
Has anyone done the math on this and can share the result?
What is the average amount of rolls with different amount of dice and how does the distribution look?
It's been roughly 30 years since I last did this kind of calculations...
Here's the average number of rounds and standard deviation for dice pools from 1 to 10:

DICEAVG (MEAN) TIMEMEDIAN TIMESTD DEV
16.004.005.46
28.737.006.13
310.569.006.39
411.9311.006.54
513.0212.006.63
613.9413.006.69
714.7213.006.74
815.4114.006.77
916.0215.006.80
1016.5615.006.82

Note that the standard deviation is very high, indicating a very wide spread. If you use this mechanic, be prepared for a lot of variation in your results.

Edit: I realized that median time is more useful than the mean here, so I added a column for that.

If you don't know what the difference is: "Mean" is "add up all your results, then divide by the number of trials." "Median" is "sort all your results from highest to lowest, then pick the one in the middle."

In other words: If you grab 100 people off the street and put them in a room with Jeff Bezos, the mean (average) wealth of the people in that room is over $1 billion. But the median wealth is that of person-off-the-street #51.

Since it can't be skewed by a handful of outliers, median generally gives you a better picture of the typical case. You'll note, for instance, that the 1-die pool has a mean duration of 6 rounds but a median of only 4 -- that's because of a handful of cases where the duration goes super long, pushing the average up.
 
Last edited:

Cadence

Legend
Supporter
Here's the average number of rounds and standard deviation for dice pools from 1 to 10:

DICEAVG ROUNDSSTD DEV
16.005.46
28.736.13
310.566.39
411.936.54
513.026.63
613.946.69
714.726.74
815.416.77
916.026.80
1016.566.82

Note that the standard deviation is very high, indicating a very wide spread. If you use this mechanic, be prepared for a lot of variation in your results.

Those match pretty well what I got for 6 and 10. What did you use to get them? (I'm going to feel silly if I did matrix stuff and/or simulation for nothing :) ).
 

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