# The Probability of Drawing a Betrayer in the Dead of Winter game

games boardgame probability math

A friend proposed a new way of distributing secret role cards to the players. Does his scheme increase the chance of having a betrayer in your team?

I recently introduced some of my colleagues into the world of tabletop games. The particular game we are playing right now is Dead of Winter. It is a game where a group of humans in a colony try to survive the post-apocalyptic world inhabited by zombies. The survivors need to cooperate to survive. The twist is that some of the survivors might be betrayers, people who have their own agenda that are antithesis to the colony’s goals.

The official rules for distributing the secret role cards involve mixing betrayer and non-betrayer cards for each player. A friend of mine suggested an alternative way to distribute cards that is easier and faster to do. The catch is that there can only be one betrayer at most. At that time, we were trying to go for a fast, after work game. So I adopted his alternative distribution scheme.

The game was a blast! But it got me thinking how much the probability of having betrayers in the group was affected by the alternative. In order to satisfy this curiousity, I decided to study basic probability and publish my findings in this blog post.

## The Setups

### The official distribution scheme for secret role cards

1. Separate the secret role cards into a betrayer and non-betrayer pile
2. For each player, do the following
1. Get two cards from the non-betrayer pile randomly face-down
2. Get one card from the betrayer pile face-down
3. Shuffle the cards into one pile
4. Let the play pick one randomly

The official setup allows for scenarios where there are 0, one or more betrayers.

### The alternative distribution scheme for secret role cards

1. Separate the secre role cards into a betrayer and non-betrayer pile
2. Pick 5 cards from the non-betrayer pile and 1 card from the betrayer pile randomly
3. Deal 1 card randomly to each player

In this setup, there can only be one betrayer at most.

## The probabilities of the distribution scheme

### The probability of having a betrayer in the group using the official distribution scheme

We deal 2 non-betrayer and 1 betrayer cards to each player.

That means that each player has a 13 chance of getting a betrayer card.

Alternatively, we can also say that each player has a 23 chance of getting a non-betrayer card.

To calculate the chance that all 5 players don’t get betrayer cards

$$$$\begin{split} \bigg(\frac{2}{3} \cdot \frac{2}{3} \cdot \frac{2}{3} \cdot \frac{2}{3} \cdot \frac{2}{3}\bigg) \cdot 100 &= \bigg(\frac{2}{3}\bigg)^5 \cdot 100 \\ &= \bigg(\frac{32}{243}\bigg) \cdot 100 \\ &= \text{13.17%} \end{split}$$$$

There is a 13.17% chance of having no betrayers in the group

What are the chances of having at least one betrayer in the group?

$$$$\begin{split} 100 - 13.17 = \text{86.63%} \end{split}$$$$

There is a 86.63% chance of having at least one betrayer in the group

### The probability of having a betrayer in the group using the alternative distribution scheme

To get the probability of a 5 man play group having at least one betrayer, we must first calculate the probability of having no betrayers in the team

The chance that the first player gets a non-betrayer role card is 56 The chance that the second player gets a non-betrayer role card(assuming that the previous player doesn’t get the card) is 45 The chance that the third player gets a non-betrayer role card(assuming that the previous player doesn’t get the card) is 34

We can express the probability as the product of the individual probabilities

$$$$\begin{split} \bigg(\frac{5}{6} \cdot \frac{4}{5} \cdot \frac{3}{4} \cdot \frac{2}{3} \cdot \frac{1}{2} \bigg) \cdot 100 &= \bigg(\frac{5!}{6!}\bigg) \cdot 100 \\ &= \bigg(\frac{120}{720}\bigg) \cdot 100 \\ &= \text{16.67% chance of having no betrayer in the group} \end{split}$$$$

What are the chances of having at least one betrayer in the group?

$$$$\begin{split} 100 - 16.67 = \text{83.33% chance of having a betrayer in a group} \end{split}$$$$

## Simulation

In order to verify the probabilities empirically, I created a jupyter notebook to simulate these distributions.

## Conclusion

Comparing the results, it looks like the 2nd process has a lesser chance of producing a group with a betrayer(83.33%) compared to the 1st process (86.83%).