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The critical hits problem of Fire Emblem

Introduction

I came across this image on Discord the other day:

Some commented that the probability of both hits being critical is 25%. Another shared a screenshot from a forum arguing that the probability is 33%. Intrigued, I decided to crunch the numbers.

Demonstration

Using Bayes' theorem:

Where:

$P(A|B)$ is the probability of event $A$ occurring given that event $B$ has occurred.

$P(B|A)$ is the probability of event $B$ occurring given that event $A$ has occurred.

$P(A)$ is the probability of event $A$ occurring.

$P(B)$ is the probability of event $B$ occurring.

Let $A$ be the event where both hits are critical.

Let $B$ be the event where at least one hit is critical.

Since the critical hit probability is always 50% (assuming each hit is an independent event that neither influences nor is influenced by the outcomes of other hits), the probability of both hits being critical is $0.50 \times 0.50 = 0.25$, meaning:

If both hits are critical, then at least one of them is critical. Basic logic, hence:

The probability that at least one of the hits is critical is more interesting to calculate.

Let ⭕ represent a critical hit and ❌ a non-critical hit. The possible outcomes for two hits are:

1. ❌❌

2. ❌⭕

3. ⭕❌

4. ⭕⭕

This assumes a 50% chance of landing a critical hit and a 50% chance of missing it.

Here lies the crux of the problem: What is the value of $P(B)$?

If we consider all possible outcomes, in three of them there is at least one ⭕. This means that in 3 out of 4 cases, at least one critical hit occurs, therefore

However, Robin states that at least one of the hits is critical, so the first scenario is excluded from all possible cases, leaving:

1. ❌⭕

2. ⭕❌

3. ⭕⭕

Now, in all these cases, there is at least one ⭕. Does this mean $P(B) = 1$?

Bayes' theorem only applies to mutually exclusive events, meaning each hit is an independent event that neither influences nor is influenced by the outcomes of the other hits.

We must not fall into Robin's trap and set $P(B) = 1$. We must stand firm and declare that $P(B) = 0.75$.

Substituting $(2)$, $(3)$, and $(4)$ into $(1)$:

Therefore, the probability that both hits are critical is 33%. You're welcome, Robin.