What are the odds? Crackthorning around

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Cracking it out

In match two of the Grand Finals in Monthly Cup XI, Maihem plays what seems likely to be a game-winning Crackthorn Beast. But just how likely was it?

We need to go deeper.

First, let’s take a look at the clip here:

In the final stages of the game, Maihem is left in an interesting position.

Here is the board state right before the Crackthorn was played:

You can see that Maihem has two Axegrinders totalling 12 damage ready to strike at Kumpel’s 15 remaining life. He also has a Horsemaster sitting out of range. KumpelKefer has only a 5 health Mystic Beast on board.

##Enter the Beast

Suddnely, this guy gets plopped on to the field.

How do we know how likely it is for lethal to be achieved here?

First of all, let’s remember that Crackthorn Beast has a chance to buff itself. This means that each of Crackthorn’s +1/+1’s has four potential targets. Two of those targets are in range of the enemy orb and ready to attack.

On the other hand, each of Crackthorn’s 1 damage rolls has only two potential targets: Mystic Beast or face.

All that needs to happen is for three extra damage to make its way onto the opponent orb to pull the win here - whether by buffing either Axegrinder, or by hitting the face directly. That’s 8 total chances, right?

Well, not quite.

We need to consult our friend: Mr. Probability Tree.

#Probability Trees

Probability trees are useful tools to determine multiple successive outcomes involving chance. It’s most easily illustrated by the example of flipping a coin twice and predicting the results:

(Credit to mathisfun.com for the above image)

At their simplest level, you need only continue branching out each potential outcome until you’re left with a series of probabilities. By following a branch of a completed tree to its end, you can multiply the chance of each branch to figure out the probability that outcome will present itself.

In the case of flipping a coin twice, you have an equal 25% chance of any of the four potential outcomes:

  • Heads Heads
  • Heads Tails
  • Tails Heads
  • Tails Tails

Crackthorn, of course, is much more complicated than that. I’ve made a series of terrible images to demonstrate this.

##Crackthorn Christmas

In this case, I have drawn out a probability tree for this specific scenario.

Here I have abbreviated each possible outcome for both buffs and damage. We’ll consider each “wave” of the Crackthorn ability to produce one +1/+1 buff, and one 1 damage ping.

With four potential buff targets and two potential damage targets, we end up with 8 possibilities for the first wave of damage. That means each potential outcome has a 1/8 chance of occurring.

Neat! However, this would make for an incredibly large probability tree. Tracing out 8 potential outcomes four times leaves us with 8^4 potential outcomes in the end. That is 4096 potential final outcomes. Pfft, I’m not drawing that out.

Let’s find a way to simplify it.

###Simplifying for sanity

First, we can say that since there are two potential “hits” for buff targets (the two Axegrinders), and two potential “misses” for buff targets (Horsemaster and the Crackthorn itself). We can therefore say that each buff has an equal chance of targetting either an Axegrinder or nothing.

This would shrink our probability tree to only four potential outcomes each wave:

  • Buff Axegrinder, Hit Face
  • Buff Axegrinder, Hit Mystic Beast
  • Buff Miss, Hit Face
  • Buff Miss, Hit Mystic Beast

Tracing this out after four waves, we’d then only be left with 256 (4^4) potential outcomes.

We’re getting somewhere! But it’s still too complicated…

What if we consider each possibility to represent a certain amount of damage to the orb?

  • Buff Axegrinder, Hit Face = 2 damage
  • Buff Axegrinder, Hit Mystic Beast = 1 damage
  • Buff Miss, Hit Face = 1 damage
  • Buff Miss, Hit Mystic Beast = 0 damage

Interesting… now we have only 3 potential outcomes each wave: 2, 1, or 0 damage.

  • 2 damage: 25% chance
  • 1 damage: 50% chance
  • 0 damage: 25% chance

This is looking like something we can work with. Four waves of three results is only 81 (3^4) possibilities! This is made a lot easier by the fact that Mystic Beast can never possibly be destroyed.

The only thing left for us to do is to draw it all out while watching day two of E3 on Twitch.


According to my calculations, in this instance Maihem had a roughly 85.5% of achieving lethal orb damage after playing Crackthorn Beast.

In only 14.5% of instances would this not have been the case.

But what actually happened?

  • Two buffs land on Crackthorn
  • Two buffs land on Horsemaster
  • Two damage is dealt to Mystic Beast
  • Two damage is dealt to face

This left him one damage short of lethal. KumpelKefer surrendered the following turn, making this entire series of calculations irrelevant.

###The end.


This is great. :smiley: Love to see more position/game analysis like this and puzzles.

I didn’t expect the chance to be so high! Thought it would be below 50%.

God, its definitly all i had calculate in 18 sec during the game.
Wait … No, i simply drop it with a big : “YOLOOOOOOO” .
Ty for this post man :stuck_out_tongue:

1 Like

Or you could just use the formula which gives you the chance of x hits of y overall hits being successful when the individual chance is z. Takes probably 5 mins to find the correct formula :smiley:
Plug in 3 out of 8 times with 50% chance - profit!
When you put it like that it is even intuitively clear that the chance is >50%, since 4+ out of 8 events with 50% chance would give you 50%.

I’d be more than willing to learn the 5 minute formula, but the fact that the individual chances of each thing happening are not equal makes it much more difficult. Feel free to post it.

Anyway, the point of the post is that it was intuitively clear that playing the Crackthorn was a likely win, but just how likely, exactly? You can look at the board and figure out you’re over 50% pretty easily.

It’s reasonable to assume that at least half of the 8 instances of damage will end up on the orb - that’s all you really need to know in this scenario while actually playing the game. The correct answer in this case, as Maihem said, is YOLO.

Actually the chances are always 50%, that is why it is rather simple. You need some basic knowledge about probability mathematics for that (i learned it in school), namely you need faculties, binominalcoefficients and the bernoulli formula (or you are lazy and skip to the end where i put a link to a program giving the result).
In short, faculity is written as !, e.g. 5!=12345
A binominal coefficient “(n choose k)” is the amount of ways you can choose k elements out of n elements and can be calculated with n!/(k!(n-k)!).
The bernoulli formula is usually written as P(i)= (n choose i)p^i(1-p)^(n-i) with P(i) being the probability of i successes in n events with p being the individual probability.
Now if we break down our case we have 8 events from which 3 or more have to be successful, while the successrate is 50%, since the buffs can land on 2 of the 4 creatures to be considered a success and the dmg pings on 1 of the 2 enemies.
For the 50% chance the formula is actually easier since 1-p is the same as p, so “p^i*(1-p)^(n-i)” becomes “p^n” or in our case “0.5^8”.
So in our case the probability to have exactly 3 buffs out the 8 waves which all have the probability 0.5 is "(8 choose 3)0.5^8". Not a complicated formula is it? :wink:
Since we want the chance for 3 or more successful events we add the other possibilities: (8 choose 3)+(8 choose 4)+(8choose 5) and so on…
(0.5^8) and calculate the result.
Or since we are lazy we put it all in some calculation program which even understands some kind of english:

scroll down to “3 or more successes” and the program tells us it is “0.8555” which confirms OPs results! :slight_smile: