8 September 2014 - 19:10

# Calculating the Odds in Your Favor Cheat Sheet

RNG is a big factor in Hearthstone. But there's actual math behind it, so here's a cheat sheet and detailed breakdown…
Dot Esports

## Introduction

[cardinsert card="avenging-wrath" float=right]

In my last article I briefly went over what probability is and how to calculate it. I have received a lot of good feedback about creating a Cheat Sheet for common use cases.

In this article I will go over some common use cases for Arcane Missilesand Avenging Wrath, which seems to be a very popular topic. I will go over the math, but for those who don't want to read it, make sure to check the TL;DR section.

## Common Use Cases

Arcane Missiles

[cardinsert card="arcane-missiles" float=right]

1 minion:

• 1HP minion: 87.5% to kill
• 2HP minion: 50% to kill, 37.5% chance to hit leave at 1HP
• 3HP minion: 12.5% to kill, 37.5% chance to leave at 2HP, 37.5% to leave at 1HP (which means 12.5% to miss it)
• 4HP+ minion: Look at 3 HP minion

2 minions:

• 1HP (Minion A) and 1HP (Minion B): 61.1% Killing both A and B, 17.6 killing just A, 17.6 killing just B, 3.7% missing both. That means 96% killing anything
• 1HP (Minion A) and 2HP (Minion B): 19.4% killing both A and B, 72.222% killing A, 32.4% killing B. That means 85.2% chance of killing anything, and 14.8% nothing gets killed.
• 2HP (Minion A) and 2HP (Minion B): 25.9% killing A, 25.9% killing B. That means 51.8% chance of killing either minion. Also, a 25.9% chance nothing gets killed.
• 3HP (Minion A) and 1HP (Minion b): 3.7% chance killing A, 70.4% killing minion B, which gives 74.1% killing A or B, which means 25.9% chance nothing gets killed.
• 3HP (Minion A) and 2HP (Minion B): 3.7% chance killing A, 25.9% killing B. That means 29.6% to kill anything, and 70.4% chance of not killing anything.
• Are you seeing a pattern? Hint: 25.9% to kill a 2HP minion with another minion that is 2HP+, 3.7% chance killing a 3HP minion with any minion.

3 minions:

• 1HP (A), 1HP (B), 1HP (C) : 25% chance to kill A B and C, 18.05% to kill any 2, 6.4% to kill just one.  But it's 67.5% to kill A, 67.5% to kill B, 67.5% to kill C. Also 98.4% chance to kill anything, leaving 1.6% chance to not kill a thing.
• 2 / 1 / 1: 36.1% to kill both B and C, 6.9% chance to kill A and B, 6.9% chance to kill A and C. But it's 19.1% to kill A, 62.3% to kill B, 62.3% chance to kill C. Also 93.8% to kill anything, and 6.2% to not kill anything.
• 2 / 2 / 1: 6.9% chance to kill A and C, 6.9% chance to kill B and C. However it's 17.4% to kill A, 17.4% to kill B, and 58.9% chance to kill C.
• 2 / 2 / 2: 15.6% chance to kill either A, B, or C.
• 3 / 1 / 1: 36.1% chance to both B and C. 1.6% chance to kill A, 61.8% chance to kill B or C.
• 3 / 2 / 1: Seeing a pattern? 6.9% chance to kill B and C. 1.6% chance to kill A, 17.4% chance to kill B, 58.3% chance to kill C. Pretty much, there is a 1.6% chance to hit a 3+HP minion 3 times, 17.4% chance to kill a 2HP minion.

Anything over 4 minions is a bit ridiculous and you probably won't want to use Arcane Missiles. I will cover the common use case of there is at least 1 minion that has 1 life.

• 4 minions, 1/1/1/1: 99.2% chance you will kill at least 1, 87% chance to kill at least 2, 40% chance to kill 3. 56.7% chance for each specific minion to die.
• 4 minions: 1/X/X/X: 48.8% chance you will kill the 1HP minion.
• 5 minions, 1/1/1/1/1: 99.5% chance you will kill at least 1, 91.1% to kill at least 2, 50% to kill 3. 48.1% chance for each specific minion.
• 5 minions, 1/X/X/X/X: 42.1% chance you will kill the 1HP minion
• 6 minions, 1/1/1/1/1/1: 99.7% chance you will kill at least 1, 93.5% you will kill at least 2, 57.1% to kill 3. 41.7% chance for each specific minion.
• 6 minions, 1/X/X/X/X/X: 37% chance to kill the 1HP minion
• 7 minions,  1/1/1/1/1/1/1: 99.8% chance to kill at least 1, 95.1% chance to kill at least 2, 62.5% to kill 3. Each minion has 36.8% of dying.
• 7 minions, 1/X/X/X/X/X/X: 33% chance of killing the 1HP minion

Avenging Wrath

1 minion. I'm going to skip 1, 2, and 3HP minions, because you should play against those.

• 4HP minion: 63.7% chance to kill it
• 5HP minion: 36.3% chance to kill it
• 6HP minion: 14.5% chance to kill it
• 7HP minion: 3.56% chance to kill it
• 8HP minion: 0.39% chance to kill it

2 minions:

• 1 / 1: 98.5% chance to kill both
• 2 / 1: 92.6% chance to kill both
• 2 / 2: 81.3% chance to kill both, 90.1% that you kill A or B
• 3 / 1: 76% chance to kill both, 77.7% to kill A, 97.8% to kill B
• 3 / 2: 58.3% chance to kill both, 68.2% to kill A, 85.9% to kill B
• 3 / 3: 33.3% chance to kill both, 59.2% to kill A or B
• 4 / 1: 49% chance to kill both, 50.5% to kill A, 97% to kill B
• 4 / 2: 29.9% chance to kill both, 38% to kill A, 82.7% to kill B
• 4 / 3: 11.9% chance to kill both, 29.6% to kill A, 55.1% to kill B
• 4 /4: 2.2% chance to kill both, 26.4% to kill A or B.
• 5 / 1: 22.1% chance to kill both, 23.2% to kill A, 96.4% chance to kill B
• 5 / 2: 9.4% chance to kill both, 13.8% chance to kill A, 81% chance to kill B
• 5 / 3: 1.9% chance to kill both, 9.6% chance to kill A, 53.4% chance to kill B
• 5 / 4: 8.8% chance to kill A, 25.9% chance to kill B
• 5 / 5: 8.8% chance to kill either.
• 6 / 1: 6.1% to kill both, 6.6% chance to kill A, 96.2% chance to kill B
• 6 / 2: 1.3% to kill both, 2.8% chance to kill A, 80.6% chance to kill B
• 6 / 3: 2% to kill A, 53.2% chance to kill B
• 6 / 4: 2% to kill A, 25.9% chance to kill B
• 6 / 5: 2% to kill A, 8.8% chance to kill B. Seeing a pattern?
• 6 / 6: 2% to kill A, 2% to kill B... If the sum of the health of the two minions is greater than 8, then the probability of killing it is lower than 2%. Let's stop here.

3 minions:

• 1 / 1 / 1: 95.7% chance to kill all 3 minions
• 2 / 1 / 1: 85.2% chance to kill all 3 minions
• 2 / 2 / 1:  67.6% chance to kill all 3 minions
• 2 / 2 / 2: 44.3% chance to kill all 3 minions, 88% chance to kill at least 2.
• 3 / 1 / 1: 61.8% chance to kill all 3 minions
• 3 / 2 / 1: 39.7% chance to kill all 3 minions, with individual chances of killing A at 55.8%, B at 78.9%, and C at 94.1%
• 3 / 2 / 2: 18.6% chance to kill all 3, with individual chances of killing A at 44.9%, and B and C at 71.9%.
• 4 / 1 / 1: 32% chance to kill all 3, with individual chances of killing A at 36.1%, and B and C at 95.1%.
• Note: Anything above 4HP is going to be worse than 36.1%.
• 4 / 2 / 2: 3.5% chance to kill all 3, with 16.7% chance to kill A, 69.5% chance to kill B, 69.5% chance to kill C.
• 5 / 2 / 2: 3.7% chance to kill A, 68.8% to kill either B or C.
• 10 / 2 / 2: 68.8% chance to kill B or C.
• 10 / 10 / 2: 63.3% chance to kill C.
• 10 / 10 / 3: 32.1% chance to kill C.

Anything over 4 minions is also ridiculous so let's take a look at common use scenarios instead:

• 4 minions, 1/1/1/1: 89.6% chance killing all minions, 97.2% chance to kill A, B, C, or D
• 4 minions, 2/2/2/2: 6.1% chance of killing all minions, 47.8% chance to kill at least 3, 88% chance to kill at least 2, 99.1% chance of killing at least 1 minion. Each minion has a 60.3% chance of dying.
• 5 minions, 1/1/1/1/1: 77.9% chance to kill all minions, 97.4% chance to kill at least 4, 99.8% chance to kill at least 3, 99.995% chance to kill at least 2, and 99.999% chance to kill at least 1. Each minion has a 95% chance of dying.
• 5 minions, 2/2/2/2/2: 4.3% chance of killing at least 4, 42.8% chance of killing at least 3, 85.6% chance of killing at least 2, 98.9% chance of killing at least 1. Each minion has a 46.3% chance of dying.
• 6 minions, 1/1/1/1/1/1: 58.8% chance of killing all minions, 91.6% chance of killing at least 5, 99.1% chance of killing at least 4, 99.9% chance of killing at least 3, 99.998% chance of killing at least 2, and practically 100% chance of killing at least 1. Each minion has a 91.6% chance of dying.
• 6 minions, 2/X/X/X/X/X: 32% chance of killing A. After 6 minions there really isn't a point to play against high HP minions. Each minion past 2HP will be less than 32% chance of killing.
• 7 minions, 1/1/1/1/1/1/1: 34% of killing all 7 minions, 77.5% chance to kill at least 6, 96% to kill at least 5, 99.6% chance to kill at least 4, 99.98% chance to kill at least 3, 99.999% chance to kill at least 2, and pretty much 100% chance of killing at least 1. Each individual minion has an 86% chance of dying.
• 7 minions, 1/X/X/X/X/X/X: 65.6% chance of killing the 1HP minion.
• 7 minions, 2/X/X/X/X/X/X: 26.4% chance of killing the 2HP minion. Any minion higher than 2HP will take less than 26.4% chance to kill.

## Summary

Arcane Missiles

If there are less than 3 minions on the opponent's board and at least one of them only has 1 HP, FIRE THE MISSILES (Overall better than 50% chance to killing the 1HP minion).

If there is a 2HP minion on the board, the BEST you can do is 50%, which is the situation where it is the only minion on the board. Adding any additional minions drops it the chances significantly (just adding a 1HP minion drops it to 32.4%). Try to decrease it to 1HP first like using . Doing such immediately increases your chances by at least 37.5%.

LightsOutAce brought a very good point that ing before playing Arcane Missiles isn't a good idea. I've tried very hard to refute him, but the more I looked into the matter, the more concrete his idea is.

I was trying to argue that decreasing a minion's HP will help your odds of killing it. While this statement is in fact mathematically true, strategically this is a BAD idea because you have the potential of getting your desired outcome WITHOUT spending 2 mana for Fireblast. Here's why:

Let's say you have a 3HP minion on the board and you play Fireblast on it, it now has 2HP, giving it a 50% chance to kill it with Arcane Missiles.

Supposed you played Arcane Missiles first. There is a 12.5% chance you will kill it, which is worse than 50%. However there's also a 37.5% chance you will hit it twice, thus leaving it at 1 life, and now you can Fireblast it for the kill. Total these two outcomes you will get 50%, but of that 50%, 12.5% of it was you saving 2 mana without using Fireblast.

Here's another scenario:

There is a 2HP minion and a 1HP minion.

Play Fireblast on 2HP minion, now it's 1 and 1. Playing Arcane Missiles now will yield 61.1% chance to kill both. There's a 78.7% chance of killing minion A (the originally 2 HP minion) in total, so that leaves a 21.3% chance of not killing it.

Play Arcane first, and there's a 19.4% chance you'll kill both, but the chances of killing just minion A is 32.4%, and the chances of leaving it to 1HP is 46.3%. That means 46.3% of the time you will have to play Fireblast on it anyway, but if you add the chance of killing it + chance of leaving it at 1HP, it is 78.7%; the same as if you played Fireblast first then tried to kill it. What does that mean? It means 32.4% of the time you will kill minion A with JUST Arcane Missiles and of that 32.4%, 19.4% of that chance will be an added bonus of killing the 1HP minion sitting next to it. So now you have 2 extra mana to do whatever you want.

So now I will say ALWAYS play Arcane Missiles first before playing Fireblast.

If there is at least one 1HP minion on the board, your chances of killing one are at least 33% (which is the worst case scenario: 7 minions 1/X/X/X/X/X/X).

Avenging Wrath

If there is any 1 or 2HP minions on the board and up to 3 minionsUNLEASH THE KRAKEN (because the Kraken has 8 legs, and Avenging Wrath has 8 hits, and they're both scary...never mind). The chances of killing the 2HP minion is greater than 63.3%.

If there is a 4HP minion on the board, your best chances of killing it is 63.7%, which is if it's the only minion on the board. 5HP minions are 36.3% chance and that is not very good. Adding minions onto the board decreases these odds significantly.

If there is a 1HP minion on the board, your worst chances of killing it is at least 65.6% (which is the worst case scenario of 1 minion with another 6 buff minions). For a 2HP minion, your worst chances of killing it is at least 26.4% (also where the other 6 minions are buff minions). That's not very high.

Anything above a 4+ / 1 +/ 1+ situation is going to be worse than 36.1% to kill the 4HP+ minion.

Like with the "Arcane missiles before Fireblast" argument, ALWAYS play Avenging Wrath before doing anything (except maybe Consecrationbecause you are eliminating minions with less than 2HP "sinkholes" from the pool. What use is hitting these minions if they are gonna die with consecration anyway?)

Here's a scenario: there's a 7/5 Illidan Stormrageand a X/3 minion and you have a 3 attack minion on the board.

If you use your 3 minion to attack the X/3 to get rid of that minion first, the probability of killing Illidan is 36.3%.

If you use your 3 minion to attack Illidan first, it's now a 2 / 3 situation, and the probability to kill Illidan is 85.9%, which is good

If you use Avenging Wrath first, the probability to get Illidan down to Dead, 1, 2, or 3 HP for you to finish him off with your 3 minion is -- wait for it -- also 85.9%. However there's a 28% added bonus you will kill the 3HP minion on the way.

## Back To School

So how did we come up with all these calculations? In my last article we talked about writing down all the possibilities and then counting the winners. Well how many possibilities are there if there are 8 attacks like in Avenging Wrath? The answer is way too many, but thankfully some mathematicians in history have figured a shortcut for this. For this lesson you will need a probability calculator of some sort (you can Google for one or just use this one).

Let's imagine we have Michael Jordan and he is going to make free throws. His free throw average is roughly .83, so let's just say 80%. He has 3 free throws to shoot. What is the probability he will make exactly two of them?

We can list out all the scenarios, which would look like: (H) denotes Hoop and (M) denotes Miss. H has a 80% chance of happening and M has a 20% chance of happening.

• H, H, H
• H, H, M
• H, M, H
• H, M, M
• M, H, H
• M, H, M
• M, M, H
• M, M, M

So we have to take each winning scenario and calculate the outcomes: (H, H, M), (H, M, H), (M, H, H). That's (80% * 80% * 20%) + (80% * 20% * 80%) + (20% * 80% * 80%) = 38.4%. But wait, are you seeing a pattern? That's just (The number of outcomes for 2 H's) * (The probability of H happening twice * The probability of M happening once) = (# of 2 H scenarios) * (80%)^2 * (20%).

That can be rewritten as (# of X H scenarios) * (80%)^X * (20%)^(Total # of shots - X).

In our case, we wanted 2 H's so X = 2, thus (# of 2 H Scenarios) was 3, 80% was raised to the 2nd power, and 20% was raised to the Total # of shots - 2, which is 3 total shots minus 2, = 1, hence 20% was raised to the 1st power. What a mouthful!

All you need to remember is c * (Pa)^x * (1-Pa)^y where:

• c = the number of combinations
• Pa = Probability of A occurring
• x = # of times you want x to occur
• 1-Pa = The probability where A won't occur = 100% minus the probability of A will occur
• y = # of times x won't occur

So what if I asked What is the probability he will get all 3?

1. Check the combinations, there is only 1 (H, H, H) so c = 1.
2. We know what Pa is, it's the probability of H which is 80%.
3. We want H to occur 3 times so x = 3
4. We know what 1-Pa is, it's 20% because that's how often he misses, or 100 - 80.
5. We don't want missing to occur at all, so y = 0.

So now you have 1 * (80%)^3 * (20%)^0 = 51.2%.

But what if MJ had to take 5 free throws, and we want to know what the probability is of him making exactly 3? Are we really going to list out all the scenarios?

Luckily there is another way to think about this. Of the 5 free throws, we want to choose 3 that make it. This is known as "combinations" in probability and that's what the calculator I linked to earlier does for you. What Combinations does is it calculates the c in the above formula; that is, it finds out the number of combinations for your desired outcome without having to list out all the possible outcomes and picking the winners!

The simple way to think of this is "I have 5 open boxes in a line, and I have 3 H cubes and 2 M cubes to put into these 5 boxes. How many different ways can I put these into the boxes?" Or even simpler, "I have 5 open boxes and I have to choose 3 of these boxes to put my 3 H cubes in." So in the Combination Calculator, put n = 5 for total boxes, and r = 3 for H cubes. The short form of this term is "n Combination r" or sometimes nicknamed as "n Choose r."

In our case, we have 5 total free throws, and we want 3 hoops. So "5 Choose 3" yields 10 possible outcomes. Don't believe me? Here they are:

1. H, H, H, M, M
2. H, H, M, H, M
3. H, M, H, H, M
4. M, H, H, H, M
5. H, H, M, M, H
6. H, M, H, M, H
7. M, H, H, M, H
8. H, M, M, H, H
9. M, H, M, H, H
10. M, M, H, H, H

So now the formula is (nCr) * (Pa)^r * (1-Pa)^(n-r). In this previous example, n = 5, r = 3, Pa = 80%, 1-Pa = 20%, so it's 10 * (80%)^3 * (20%)^2 = 20.48%.

Let's go back to Arcane Missiles with just 3 shots. Let's assume there is only 1 minion, hence two targets, the Hero (H) and the Minion (M). Let's assume the Minion has 3 HP. So if the question is "What are the chances of hitting the minion EXACTLY twice?"

n = 3, r = 2, Pa = 50%, 1-Pa = 50%, so (3C2) * (50%)^2 * (50%)^1 = 37.5%.

So now we're ready for Avenging Wrath because nobody wants to write all the combinations of 8 missile outcomes! We have a minion at 5HP. What is the probability we kill it with Avenging Wrath? Easy right? n = 8, r = 5

n = 8, r = 5, Pa = 50%, 1-Pa = 50%, so we have (8C5) * (50%)^5 * (50%)^3 = 56 * (.390625%) = 21.875%.

But that's not the answer I gave above. What gives? If you read my previous article, you will remember that we forgot the scenarios where maybe Avenging Wrath would have hit it 6 times, or 7, or 8 times! So we have to add r = 6, 7, and 8 as well. Without going into all the formulas, 8C6 = 28, 8C7 = 8, 8C8 = 1, so P(r=5) + P(r=6) + P(r=7) + P(r=8) = 36.3%.

Let's step up the notch! What about having two minions, A has 5HP and B has 4HP? What is the probability we kill B?

n = 8, r = 4, 5, 6, 7, and 8. Pb = 33.3%, 1-Pb = 66.7%. so:

r = 4 => 8C4 * (33.3%)^4 * (66.7%)^4 = 17.03%

r = 5 => 8C5 * (33.3%)^5 * (66.7%)^3 = 6.8%

r = 6 => 8C6 * (33.3%)^6 * (66.7%)^2 = 1.7%

r = 7 => 8C7 * (33.3%)^7 * (66.7%)^1 = .7%

r = 8 => 8C8 * (33.3%)^8 * (66.7%)^0 = .015%

Add them together you get: 26.24% (this value is a little higher than the one mentioned above because I rounded to 33.3% instead of using 33.333...)