A Simple Application of Probability: Deck Building in a Digital TCG

April 4, 2022

When asked a question with a binary-esque answer - such as “will it rain today or not?”.  Your instinctual reply would “50%.  Either it rains or it doesn’t”.  This half-joke comes from the fact that, without prior information, we often cannot predict the outcome of certain events.  Even when we do have such information, ignorance can leave us in a state of guessing.  However, the study of probability does allow us to peer into the possibilities of certain effects happening or not – if we do have prior information.  


One of my favourite ways to study and teach probability is via Tree Diagrams.  Tree diagrams are very illustrative ways of showing the outcomes of an event.  Outcomes may be dependent if the result of one outcome affects the one after it (such as taking a coloured ball out of a bag of balls and not replacing it before you take the next one) or independent if the outcomes do not affect the outcome of each other – such as rolling a dice several times.  Below is an example.  The “event” is flipping a coin to get heads or tails.  The “outcome” is whatever combination of heads/tails was obtained:  

unnamed (7)

But we still need to keep this relatively relatable, so let’s have a closer look at Tree diagrams by considering a game I sometimes play: Hearthstone.


Hearthstone is a digital card game developed by Blizzard TM and is similar to Yi Gi Oh, Magic the Gathering and Pokémon in some of its mechanics.  It has a graphical interface and art style that is pretty simple to understand. In such a game, the main objective is to make a deck of cards (with different mechanics and abilities) and play said deck of cards on a board versus an opponent. Decisions made whilst playing or even making the deck are a matter of understanding the probability of certain events occurring.


A deck I’m trying to build is centred around a card called “Floating Watcher” as its set-piece (its one of my favourite cards in the game).  

unnamed (8)

The Floating Watcher is seen as “too expensive” and non-impactful in today’s meta.  It is understatted on the turn its played and it comes out at turn 5 when games can easily be decided by then.  So the basic premise is to cheat out the card either via a mana discount or summoning and then ramp up its stats using self-damage until the point where its out of control and the enemy cannot check it.  So, the question remains “how do I cheat it out”?  I start to make some decisions. I can use a card called “Free Admission” which draws 2 minions and discounts their mana costs by 2 if they are BOTH Demon type:

unnamed (9)

Lets have a closer look at the deck.

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It’s a bad deck. I have 14 minions in my deck and 8 of them are Demons whilst 6 of them are neutral minions.  A basic understanding of probability would have us confirm that I have an \( \frac{8}{14} \) chance to draw a demon out of 14 total minions. In Tree diagrams, we describe the probability of events A and B occurring simultaneously by using the keyword AND.  The individual probabilities are multiplied.  If however we have two events A and B where it isn’t possible for them to occur at the same time and we want to know if A or B will occur, then we use the keyword OR.  The individual probabilities are added.  On Tree diagrams, its as simple as multiplying along a branch for AND, and adding across adjacent branches for OR:


unnamed (12)

Let’s use this Tree diagram to calculate the probability of success (Free Admission draws 2 demons and discounts them) or failure (Free Admission doesn’t draw exactly 2 demons, in whatever combination).  If we simplify things by assuming the match begins with all minions in the deck by the time we use Free Admission, we can see that drawing 1 minion (whatever it is) would affect the probability of drawing the next.  So, this is a Tree diagram describing dependent events. To draw a Demon AND a Demon, the total probability is \( \frac{8}{14} \times \frac{7}{13} = \frac{56}{182} \)


This is approximately 31%.  To draw a Demon AND a Neutral OR a Neutral AND a Demon OR a Neutral AND a Neutral, the total probability is mixture of multiplication (for the AND’s) and addition (for the OR’s).  We should already expect the total probability to around 70%, as 100% of the time, something must occur and therefore the probabilities of individual outcomes must all add up to 100%.

For the math of getting a Demon AND a Neutral, OR a Neutral AND a Demon, OR a Neutral AND a Neutral:

$$ \left( \frac{8}{14} \times \frac{6}{13} \right) + \left( \frac{6}{14} \times \frac{8}{13} \right) + \left( \frac{6}{14} \times \frac{5}{13} \right) = \frac{126}{182}$$

This amounts to approximately 70%.


What is the conclusion here?  If I played this deck in a match and expected to cheat out Demons with Free Admission, I’d be disappointed around 70% of the time.  That’s bad.  However, its good in that I can now make a more informed decision of how best to modify the deck so that Free Admission can either do its job by:


  • Increasing the ratio of Demons to Neutral minions
  • Keeping more neutral minions in my opening hand at the mulligan stage so that the pool of minions in the deck is smaller with Demons being the majority population
  • Remove Free Admission entirely and figure out a more consistent way to cheat it out my Floating Watcher


Given the climate of Hearthstone right now and how prevalent mana cheating and tempo is, I can’t afford to play fair.  This Tree Diagram helped me at least to make better and more informed deck building decisions.  


I hope learning about Tree Diagrams can help you make better decisions too!

Author: Dr. Adetokunbo Ayilaran

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When asked a question with a binary-esque answer - such as "will it rain today or ...
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