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 |

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.

That is, how to tell if the game box is lying by looking at the rules (it almost always is, but still...).

My basic framework here is that for many games, the length is simply the product of the number of turns and the amount of time taken per turn. This obviously doesn't work for games without turns, such as real-time games like Space Cadets Dice Duel. There also may be overhead that doesn't scale the same way, e.g. once per day activities in Robinson Crusoe, or once per round activities in Agricola.

Many games are the same length to first order, independent of player count: Battlestar Galactica, Carcassonne, Pandemic, Kingdomino. These games have a roughly fixed number of total turns. Some of these may have strong second-order terms to affect game length, for example Pandemic, which tends to last more turns with more players as it's harder to get cards in a single hand. Other games could take more time per turn with more players due to group discussion, which may be play group dependent. These games can be identified by the fact that the there's some shared resources (e.g. cards or tiles) that advances the players towards the end (Battlestar Galactica, Pandemic), or ends the game when it runs out (Carcassonne, Kingdomino*).

Let me expand my thoughts on Kingdomino. First, for 2-player I'm focused on the "Mighty Duel" 7x7 variant (which I've almost always played). This alone makes the 2 player and 4 player games take the same amount of time, at least to first order. If instead we looked at 2 player 5x5 games, we'd expect them to be roughly half the length of a 4 player game. Second, we should probably break up the two aspects of the game: selecting tiles and placing them. In Mighty Duel and 3-4 player games, all tiles/dominos are available to choose at some point. Also, each round three tiles are chosen, and the fourth is forced to whoever is left. Thus, this part of the game is the same whether there are 2, 3, or 4 players. The question is whether placing takes a different amount of time. Perhaps it's slightly different in a 3 player game, but I often think of it as incidental to your choice of tile in the next round (as it happens at the same time). I can see how 4 player games we may expect to take more time, but I don't think it'd be linear.

Many games are proportional to the number of players to first order: Ticket to Ride, Splendor, Agricola. These games have a roughly fixed number of turns per player. Some of these games (e.g. Agricola) are honest and state that the play time is a certain number of minutes per player. Others, like Ticket to Ride, do not. These games can be identified by an explicit number of rounds (Agricola), the end of the game triggered by a single player's resources depleting (Ticket to Ride), or games which are a race to a victory condition (Splendor).

Sometimes, though, a game doesn't fit neatly into one of the above categories, as some elements are adjusted by player count. First, let's consider Bohnanza. There's a shared deck of cards and the game ends as soon as the players exhaust it three times (see more below). At first, we'd think that this would be a fixed length game, however things get more complicated when we consider the variant rules below, which allow the full 7 players.

Bohnanza, 45 minutes

104 normal cards

expansion: +50 cards: 154

3 player: -4 cards: 150 cards

4-5 players: -24 cards: 126 cards

6-7 players: -4 -6 cards: 144 cards

3 players: only two times through the deck. 3 bean fields

4-7 players: 3 times, 2 bean fields

new rules: flip 2 cards, draw 1 card per player per turn (original: draw three cards)

https://www.riograndegames.com/wp-content/uploads/2013/02/Bohnanza-Rules.pdf

There are three things coming into play here. First, the number of cards in the deck varies with player count. Second, with 3 players you only go through the deck twice and each player has three bean fields instead of two. Third, the number of cards drawn per turn is a function of the number of players.

Let's calculate the number of turns it takes to get through the deck the first time.

3: 150 / (2+3) = 30

4: 126 / (2+4) = 21

5: 126 / (2+5) = 18

6: 144 / (2+6) = 18

7: 144 / (2+7) = 16

If subsequent times through deck has no card removal (almost certainly not true), then we can calculate the following number of total turns.

3 players: 60 turns

4 players: 63 turns

5 players: 54 turns

6 players: 54 turns

7 players: 48 turns

Practically, each time through the deck actually takes much less time than the first. Some of this may just be as players get into the flow of the game, but there are two structural reasons as well. First, cards are essentially removed from play as they are sold and converted to coins. Second, a number of bean cards are still in players' fields as they go through the deck the second and third time. Let's look at the total number of bean fields in each player count.

3 players: 9 fields

4 players: 8 fields

5 players: 10 fields

6 players: 12 fields

7 players: 14 fields

This effect would further skew decrease the number of turns in large player count games relative to smaller player count games. However, with more players there are more possible negotiation considerations. It's more likely that there are multiple other players who have something you want, and similarly more likely that there are multiple players who want something you have. Three player games can go very quickly, as players get committed to what they are going for and you have more need to manage the large number of plantings you'll do. But in a 7 player game, your turn is much more rare, so trading is more important. While I do not have data to quantify it, I suspect that this at least equalizes the length of the game and probably even pushes larger player length games to be longer. If there's any data on this, it'd be interesting to see.

As another example, let's look at Century: Spice Road. Here what's interesting is that the game ends when one player gets a certain number of point cards, but that number depends on the number of players in the game.

Century: Spice Road, 30-45 minutes

2-3 players: 6 cards

4-5 players: 5 cards

Thus, if each card takes "x" turns to obtain, then the relative game length is as follows.

2 players: 12x

3 players: 18x

4 players: 20x

5 players: 25x

As we can see, we'd expect a 5 player game to take more than twice as long as a 2 player game. This ignores the fact that obtaining the 6th point card likely takes fewer turns than the first, as you've already built up an engine. This clearly indicates that the 30-45 minute ranges is incorrect, as the max range should be closer to twice the minimum.

Some games get shorter with more players: Sushi Go Party, at least to first order, given negligible overhead for scoring more players, actually has fewer cards per player in higher player count games.

Sushi Go Party!

2-3: 10 cards each

4-5: 9 cards each

6-7: 8 cards

8: 7 cards

There are a host of other considerations in game length that I haven't touched on here, many of which have clear implications as to whether they increase or decrease game length, but how to quantify that effect is less clear.

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