Monday, September 21, 2020

The Hunt in War of the Ring

First, let's review the rules for the Hunt roll in War of the Ring. When the Fellowship moves (before the Mordor track), the Shadow player rolls a number of dice equal to the number of dice showing an Eye on the Hunt Box, up to a maximum of five. Each roll of a $6+$ is a success. Every previous move of the Fellowship that placed a die in the Hunt Box previously that turn gives a $+1$ to each die roll (an unmodified 1 still misses). There are also three conditions that can grant up to three re-rolls. If there's at least one success, then a tile is drawn to determine damage. 

 Let's first find the probability of a successful hunt, given the following three parameters. 
\begin{align} e & \in \{0, 1, 2, 3, 4, 5\} \qquad &&\text{Number of eyes in the Hunt Box} \\ r &\in \{0, 1, 2, 3\} \qquad &&\text{Number of rerolls} \\ n & \in \{1, 2, 3, 4, 5\} \qquad &&\text{Number of Fellowship moves this turn} \end{align} 
The probability of succeeding on each die during the first move is $1/6$, as the shadow needs to roll a 6. We could look at the pmf of the number of successes, but instead, we can compute the probability of success as the complement of the probability of failure. To fail, all dice must have failed to roll a success. Each die here fails with probability $1-1/6=5/6$. For $e$ dice with no re-rolls, this is, 
\begin{align} P(\text{failed hunt}) = \left ( \frac{5}{6} \right) ^e . \end{align} 
In this framework, incorporating re-rolls is relatively simple also. To fail with re-rolls, first the main roll is failed, then the re-rolls must fail. This means a total of $e+r$ failures are rolled, as long as the number of dice is at least equal to the number of re-rolls. We could restrict $r \leq e$, or we could say that when $r> e$ we look for $2e$ failed dice instead. Ignoring this special case, equivalently, considering $r$ as the effective number of re-rolls, this updates the probability of a failed first hunt as follows. 
\begin{align} P(\text{failed hunt}) = \left ( \frac{5}{6} \right) ^{e+r} \end{align} 
To incorporate the effect of the number move involved, we just look at the probability of failing each roll. Instead of $5/6$, we can write this as, 
\begin{align} \frac{6-n}{6}. \end{align} 
This brings us a comprehensive equation, using the fact that $P(\text{successful hunt}) = 1 - P(\text{failed hunt})$. 
\begin{align} P(\text{successful hunt}) = 1 - \left ( \frac{6-n}{6}\right) ^{e+r} \end{align} 
The result of this equation is plotted in Figures 1–3. 

Figure 1: Probabilities of a successful first Hunt

Figure 2: Probabilities of a successful Hunt by number of moves

Figure 3: Probabilities of a successful Hunt by number of dice

Now that we have the probabilities, there are a number of questions about possible choices in the game that we can answer. First, is it preferable to have another eye in the Hunt Box, or move a Nazgul in order to get a re-roll. In terms of a successful hunt, there's no difference between the two. However, an aspect of the rules we haven't covered here is that if an Eye Tile is drawn, then the Hunt Damage is equal to the number of successes rolled. Thus, we prefer to have eyes to a re-roll. However, we have to consider the cost and how each are obtained. The shadow player can assign eyes to the Hunt Box before the action die roll (with some limits), but may roll more. Shadow players should consider the possibility of getting more eyes than desired and leaving few actions. There are also cases when the Shadow player rolls fewer eyes than expected. In such times, it makes sense to use action dice to obtain one or more re-rolls. This is especially attractive if Nazgul were going to be moved anyways, and one can incidentally be placed with the Fellowship. At worse, this costs one action die (the same), as well as the position of one Nazgul. Army movement costs vary more. Some cases will have an army nearby the Fellowship (possibly with a Nazgul as well). Each army die allows two armies to move, so in some sense the cost is half of that of assigning an eye. However, this likely leaves the army in an otherwise undesirable position (even if it is just one unit), so half of an additional army movement would be needed to put it back into position. 

Another question, which the Free Peoples player may be faced with is whether it is better to move again this turn, or wait until next turn. There are a couple of considerations here. First, the distribution of the number of eyes rolls on the next turn, which also depends on a choice that the Shadow player has yet to make. We'll set this aside here, and focus on what the Free People's player expects. Let's say there are usually two eyes in the Hunt Box with no re-rolls. However, this turn there are four eyes. Using Figure 3, we can compare the probability of a successful hunt on the first move this turn with a second move on the next turn with two dice. We can see that a first move against four dice actually has a lower probability of success (though not by much). Thus, while moving against four dice sounds much worse, because of the diminishing returns of additional dice, it's not as bad as we may think.


  1. Andrei Munteanu PrincePhocsJune 5, 2022 at 7:22 AM

    Your calculation is correct but in the FAQ you will see that 1 is considered always a miss for everything. Thus Frodo can move against 5 eyes with 5 moves before that and still have a slim chance of being unseen even with 5 rerolls :)

  2. This is a good point. I stopped the plots at 5 moves, but perhaps extending it to 6 would make this more clear, as rare as that would be to happen.