I just read the mechanics for dice rolling in Vampire: the Masquerade 5th edition and something is bothering me. It seems to me that the larger your dicepool is, the more likely it is to get a messy critical.

Just in case somebody wants to answer without knowing the rules, here is a very brief rundown:

A character rolls a number of d10s called a dicepool. For the vast majority of rolls, at least one of those dice would be a Hunger die. Let's just assume it's a single one for simplicity. If at least two dice come up as 10s, that's considered a critical. If at least one of those 10s is on a Hunger die, then it's a messy critical where the character succeeds spectacularly but in the most direct and brutal way possible. Picking a lock with a messy critical can lead to the character ripping the door off the hinges - grants passage but it's not subtle.

These are the relevant rules here.

Comments asked for clarification, so here it goes:


A result of 10 on two regular dice (00) is a critical success. [...] A winning roll containing at least one critical success is called a critical win, or sometimes just a critical.

(page 120 of VtM 5th Edition Core Rulebook)

Messy Critical

A critical win in which one or more 10s appears on a Hunger die is a messy critical.

(page 207 of VtM 5th Edition Core Rulebook)

It doesn't matter how the results of Hunger and non-Hunger dice are combined. If there are any two 10s that's a critical win for the roll. If there is any 10 on a Hunger die on a critical win, that's a messy critical. So, if somebody scored two 10s on non-Hunger dice and a 10 on a Hunger die, that's still a messy critical.

It seems to me that the larger the dicepool is, the more of a chance for a messy critical. My intuition is the following:

  • with a dicepool of 3, if the Hunger die comes up at 10, then you have 2 chances to roll a 10 on the other dice.
  • with a dicepool of 7, if the Hunger die comes up at 10, then you have 6 chances to roll a 10 on the other dice.

Is my intuition here correct? Is a master at picking locks would be more likely to let the Beast do his job than somebody who's just average at locks? I am not sure how to properly calculate the odds of messy criticals.

There is a mechanic that allows you to re-roll some dice but I'm not interested in this, only the chance of scoring a messy critical from the initial roll.

  • \$\begingroup\$ Does it matter how many 10s are rolled from the non-hunger pool, if the hunger die shows a 10 result? \$\endgroup\$
    – Upper_Case
    Commented Feb 12, 2021 at 18:11
  • \$\begingroup\$ @Upper_Case not sure what you mean. You have a dicepool that is (let's assume) always the same Dexterity + Larceny. So, if you have Dex 3, Larceny 4, that's 7 dice. Of those dice one (again, assume one) is a hunger die. So this dicepool is always 7 dice - 1 hunger, 6 normal. If you were rolling, say, a dicepool of 3 (Dex 2 + Larceny 1) then you have 1 hunger die + 2 normal ones. \$\endgroup\$
    – VLAZ
    Commented Feb 12, 2021 at 18:16
  • \$\begingroup\$ What I mean is, if I roll three 10s with non-hunger dice, will a 10 on the hunger die still cause a messy critical? \$\endgroup\$
    – Upper_Case
    Commented Feb 12, 2021 at 18:18
  • 1
    \$\begingroup\$ @VLAZ I think Upper_Case is asking if you can "ignore" a 10 on the hunger dice if the non-hunger dice already have 2 10s. As an extreme example, what happens if you roll 7 dice and they're all 10s? Is that a regular critical because you have 2 10s on non-hunger dice, or is it a messy critical because you have at least 2 10s and one of them is on the hunger die? Basically, what are the rules for when you roll more than 2 10s? \$\endgroup\$ Commented Feb 12, 2021 at 18:53
  • 1
    \$\begingroup\$ So, the answer to @Upper_Case's question is that you can't ignore the tens that come up on the hunger dice, and the only way to "ignore" tens on non-hunger dice is by rerolling them. \$\endgroup\$
    – Jadasc
    Commented Feb 12, 2021 at 19:35

2 Answers 2


Messy criticals become more likely overall, but less likely as a fraction of all criticals

The short answer is: as you add more (non-hunger) dice to the dicepool, messy criticals become a smaller fraction of all criticals, but the overall probability of getting a critical goes up even faster, which means the overall probability of a messy critical goes up along with it. This is perhaps best illustrated in graphs. First, let's look at the probability of a critical being messy as a function of dicepool size (with 1 hunger die):

Fraction of messy criticals for one hunger die

As you can see, the probability of a messy critical starts out at 100% (since all criticals are messy for 2 dice) and then decays exponentially down to 10%. This makes sense, since as the dicepool size approaches infinity, you will always get a critical, so the only thing that matters is whether the hunger die rolls a 10, which gives a 10% chance of a messy critical. Hence, with one hunger die the chance of any given critical being messy can never be less than 10%. Meanwhile, as we said, adding more dice increases your overall probability of getting any kind of critical, and because the chance of a critical being messy is always at least 10% , this means that the overall probability of a messy critical must also increase:

Critical probabilities for one hunger die

Adding more hunger dice changes the "floor" for the exponential decay: the general rule is that the probability of a critical being messy can never be less than the probability of rolling a 10 on a hunger die. For example, with 2 hunger dice, there is a 19% chance that at least one of them will be a 10, so the chance of a critical being messy can never go below 19%. Meanwhile, the trend for increasing the total dicepool remains the same: criticals become overall more likely, and messy criticals become more likely along with them.

Here is the R code I used to generate the above plots. This code exploits the fact that we only care whether each die is a 10 or not, which means we are effectively flipping a biased coin with a 10% probability of heads. With this in mind, we enumerate every possible roll and compute its probability and whether it is a crit or a messy crit. We repeat for each dicepool size from 2 to 20 and plot the results.


## n = total dice; h = hunger dice
summarise_dicepool <- function(n, h = 1) {
    if (h > n) stop("Too many hunger dice")
    dice <- list()
    for (i in seq_len(n)) {
        if (i <= h) {
            die_name = paste0("H", i)
        } else {
            die_name = paste0("D", i)
        dice[[die_name]] <- c(FALSE, TRUE)
    do.call(expand.grid, dice) %>%
        as_tibble %>%
            Num_10s = rowSums(.[seq_len(n)]),
            Hunger_10 = rowSums(.[seq_len(h)]) > 0,
            Prob = 0.1 ^ Num_10s * 0.9 ^ (n - Num_10s),
            Is_Critical = Num_10s >= 2,
            Is_Messy_Critical = Is_Critical & Hunger_10
        ) %>%
        group_by(Is_Critical, Is_Messy_Critical) %>%
        summarise(Prob = sum(Prob), .groups = "drop") %>%
        arrange(Is_Critical, Is_Messy_Critical)

dicepool_sizes <- 2:20
dp_table <- dicepool_sizes %>%
    lapply(function(i) cbind(DicePool = i, summarise_dicepool(i))) %>%
    bind_rows %>%
        Roll_Type = case_when(
            Is_Messy_Critical ~ "Messy_Critical",
            Is_Critical ~ "Critical",
            TRUE ~ "Not_Critical"

dp_table_crits_only <- dp_table %>%
    group_by(DicePool) %>%
    filter(Is_Critical) %>%
    mutate(Prob = Prob / sum(Prob))

p1 <- ggplot(dp_table) +
    aes(x = DicePool, y = Prob, fill = Roll_Type) +
    geom_col(position = "stack") +
    xlab("Dicepool size") +
    ylab("Probability") +
    ggtitle("Critical probabilities for one hunger die")

p2 <- ggplot(dp_table_crits_only) +
    aes(x = DicePool, y = Prob, fill = Roll_Type) +
    geom_col(position = "stack") +
    xlab("Dicepool size") +
    ylab("Fraction") +
    ggtitle("Fraction of messy criticals for one hunger die")

Your intuition is broadly correct. More dice give you a better chance of a critical roll, regardless of other circumstances

The intuition for this situation is that more non-hunger dice means more chances for getting at least two to show tens. As long as that's the metric that matters (meaning we don't care about results that are not tens), then adding more dice always improves your odds of rolling additional tens.

This is completely unrelated to the hunger die (I'm assuming that a hunger die is a specific die marked out before the roll). The results of the hunger die roll are not related to the results of the other dice, and so for this particular scenario it may make things easier to think about the two varieties of dice completely separately (it may be easiest to imagine rolling each set, hunger and non-hunger, separately):

  • If the hunger die is rolled first and shows a 10, more non-hunger dice increases the odds that at least one of them will also show a 10. This matches your intuition.
  • If the non-hunger dice are rolled first, more dice means greater odds that at least one of them shows a ten. Rolling the hunger die afterwards then gives a 1/10 likelihood (or 1:9 odds) that the hunger die will show a ten, yielding a messy critical.

It might be a bit confusing to think in terms of assuming a result (although that is the correct way to calculate the odds you're interested in), because you are removing the randomness of the assumed result. That seems to me to be the cause of confusion in your second-to-last line.

A master at lockpicking is more likely to roll a critical by virtue of their large dice pool-- that's the representation of their lockpicking skills. But their hunger die has an invariant chance of showing a ten. They are very good at picking locks, but have a chance of their hunger complicating the effort, and are not necessarily particularly good at dealing with that hunger just because they pick locks well.*

The rough math

The easiest way I can think of to calculate the chances of a messy critical for a given dice pool is first to calculate the chances of at least one die showing a ten. The easiest way to calculate that is to compute the complementary event (no die shows a ten).

For d10s, this is just (9/10)^N, where N is the number of dice you are rolling. We are computing the 90% probability that each die will show a non-10 number, and multiplying those probabilities to get the likelihood that those outcomes will all happen at once.

With the number of rolls that produce zero tens calculated, we subtract that probability from 1 (here, 1 represents "100%", the set of all possible outcomes). One minus the probability of no ten results equals the probability of one or more tens.

So, with our chances of rolling at least one ten from the regular dice sorted, we now look at our chances of rolling a ten on the hunger die. This is always 1/10 for a single, fair d10. That makes the math pretty easy! Whatever the probability of rolling at least one ten from the regular dice, we have an additional 1/10 chance of a ten on the hunger die. That means that your probability of a messy crit is just 0.1 * [probability of at least one ten from the regular dice].

This does, indeed, increase the chances of a messy crit as dice pools get larger.

*Narration always matters

Sometimes in TTRPGs mechanics alone make for odd-seeming results in narrative terms. I'm not familiar with V5 rules overall, but I suggest that, narratively, a messy crit of "ripping the door of its hinges" during a lockpicking check really is unreasonable. It's hard for me to believe that anything done with a lockpick and torque wrench, by anyone or anything, would lead to the door being torn apart.

I don't have a good understanding of what a critical success in picking a lock would be... it's a binary event, you succeed or you don't, unless we factor in elements like how long it takes.

Instead, a messy crit (in the spirit of a botched roll from Revised or V20, which may not be appropriate) might more plausibly lead to something like jamming a lock pin permanently out of place in the lock cylinder. Maybe it makes a loud noise and attracts attention. Certainly the lock is obviously broken, and no key will ever work in it again.

Conversely, if the hunger die represents the character going out of control (similar to a frenzy in earlier editions), it might represent the character losing control and becoming so impatient with picking the lock that they lash out and tear the door off its hinges.

If there is a mechanism for things like automatic successes, a character with a large dice pool for a check might be a good candidate for one due to this messy crit mechanic. If you don't need, or can't use, a critical success, then there's little point risking a messy crit.

That said, Vampire has never been known for being really technically airtight and well-balanced. This may be another example of that, carried forward into the latest edition!

  • \$\begingroup\$ Does the fact that you can use willpower to reroll non-Hunger dice make a difference here? In such cases, you could reroll a 10 that gave you a messy critical by pairing with a hunger 10. \$\endgroup\$
    – Jadasc
    Commented Feb 12, 2021 at 17:48
  • \$\begingroup\$ "I'm assuming that a hunger die is a specific die marked out before the roll" correct. The hunger die is just marked differently. \$\endgroup\$
    – VLAZ
    Commented Feb 12, 2021 at 17:48
  • 1
    \$\begingroup\$ @Jadasc I'm not actually counting that. I just want to find how likely it is to get it from the first roll. \$\endgroup\$
    – VLAZ
    Commented Feb 12, 2021 at 17:49
  • \$\begingroup\$ @Jadasc Having a mechanism to override a roll doesn't change the underlying odds of events, but a gives the player a chance to "shift" to another result combination by replacing a known, rare result (1/10) with a more likely outcome (9/10). I'm not so familiar with 5e, but if willpower is a valuable, depletable resource it is hard to factor in how much of it a character might have to draw on. \$\endgroup\$
    – Upper_Case
    Commented Feb 12, 2021 at 17:53
  • 1
    \$\begingroup\$ "They are very good at picking locks, but have a chance of their hunger complicating the effort." OK, this still bothers me. The issue I have with this is that a non-critical success still works.So, if you try to pick with a dicepool of 4 you have a good chance of simply succeeding but low chance of criticals and messy criticals. If you roll a dicepool of 8, then you have a higher chance to succeed, a higher chance to succeed spectacularly, and a higher chance for a messy crit. It seems...like the an odd result from the mechanics. \$\endgroup\$
    – VLAZ
    Commented Feb 12, 2021 at 18:17

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