Probabilities at House on the Hill
31st July 2016

"This...this usually doesn’t happen so quickly." I say sheepishly.
I look up and can see in their eyes both pity and derision.
"This has never happened to me before." I mumble and avert my gaze.

Maybe I made a mistake? I recheck the dice hurriedly: I needed to roll a total of 3 or higher and there are 4 blanks and 2 1s looking back at me. I have failed a Haunt roll after only 2 and a half rounds.

"Betrayal at House on the Hill" is a board game in which heroes adventure around a spooky house, building it as they go along, until "the Haunt" occurs and one of them turns traitor. The traitor and remaining heroes now have separate objectives; the heroes must generally kill the traitor and/or flee the house while the traitor must generally slaughter all of their former comrades. Good family fun! (Watch the Betrayal episode on Tabletop for full rules.)

The traitor dynamic is one of the more interesting that I have seen in board games. As rooms are discovered and placed in the house players will sometimes have to draw an Omen card. These are beneficial items which can help the player but also requires them to make a Haunt roll. Using 6 dice with 2 blank faces, 2 1s and 2 2s on each the player must roll a number greater than or equal to the total number of Omen cards that have been revealed by all players. Success and the game continues, failure and 1 of 50 scenarios activates. Before the Haunt players are free to wander the house, discovering rooms and items that boost their stats and while they can lose stat points they can not be killed in this phase. After the Haunt monsters which can kill the heroes are often released in the house and there is now a set objective which often hampers further exploration. It is therefore usually desirable to hold off the Haunt for as long as possible so the players can get stronger and enjoy exploring the house.

Playing with some friends recently I took my third turn and revealed a room that required me to draw an Omen card. This was my second Omen card and there was 1 other in play for a total of 3, so I would need to roll a 3 or higher. These early rolls are a mere formality because come on, what’s the probability of failing to roll a 3?

So after failing to roll a 3 I asked myself that very question. So did the guy chosen by the rulebook as the traitor and before the next round started we basically plucked numbers out of the sky as to how likely this happening was. After daemons from hell stormed the house and killed everyone, leaving me as the last to die so I could see what I had wrought upon my friends, I crunched some numbers in python. I had been doing some of the probability questions on hackerrank in the days beforehand and was in a probability crunching mood.

Each dice is in effect a 3 sided dice as there is a 2/6 chance of rolling a blank, 2/6 of rolling a 1 or 2/6 of rolling a 2. 6 dice are rolled for the Haunt roll giving us (3 outcomes per dice ˆ 6 dice) 729 different combinations. Totalling up each combination and getting a count of each total gives the below histogram. We can see there are 141 ways of rolling a total of 6 ( [1,1,1,1,1,1], [2,2,2,0,0,0] …) and only 21 ways out of 729 (7/243, 2.8%) of rolling a 2, which is what I got.

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Because I wanted to know what the chance was of failing a Haunt roll with only 3 Omen cards in play I added together all the combinations of dice rolls that totaled a number less than each possible number of Omen cards (as the highest possible roll with the 6 dice is 12 and rolling this will allow the game to continue if there are 12 Omen cards in play, 13 cards is the upper limit). So for 3 Omen cards this was all combinations that totalled 0, 1 and 2 (1 + 6 + 21) which equalled 28/729 or 3.84%. With 3 Omens in play there was a 96.2% chance of rolling a number that would allow the game to continue…

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Knowing the probabilities doesn’t really help in the game but it can be fun to work out what the likelyhood of something happening was. In a different game I stepped into a Mystic Elevator and pressed the button (rolled 2 dice). A total of 0 (1/9, 11.11% chance) is the only roll that would result in something negative happening and that is exactly what I got. The Mystic Elevator crashed to the basement, causing me to roll a dice and take that number as damage. I coaxed my friend whose turn it now was into the elevator, he was stuck in the basement with no way out, and told him it could get both of us out of there. Because come on, what’s the probability he’d roll double blanks right after me? Which is exactly what he did.

Before I stepped into the Mystic Elevator the chances of 2 rolls of double blanks in a row was 1/9 * 1/9 for 1/81 or 1.23%. However, after I rolled a double blank the probability that the next roll would be a double blank was back up to 11.11%. I have often been blind to the probability of a second unlikely event happening, both in games and in life in general, and guilty of assigning probabilities dependent on earlier events to independent events as above. If nothing else Betrayal at House on the Hill has taught me the value of learning a bit more probability.



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