In a game, there are a series of attacks, each with a known probability of success.
As a specific example, let's say that there are two potential attacks that can be made against a single defender. Each has a probability of 0.59 to succeed. If the first attack succeeds, the defender is destroyed, and therefore no subsequent attack can be performed.
Here is a tree of how I think this can be modeled:
Based on the above tree, I think the following statements are correct:
- There is a $P(0.59)$ probability of winning in the first attack
- There is a $P(0.83)$ probability of destroying the defender (we don't care if it takes 1 or 2 attacks, just that it is destroyed in one of them). This can be calculated in two ways:
- $1 - (0.42^2) = 0.83$ (the probability that the bottom branch does not occur)
- $0.59 + 0.24 = 0.83$ (the probability that one of the two Win leaves is reached)
- There is a $0.41 * 0.59 = 0.24$ probability of losing the first attack, but winning the second
- It is not a coincidence that the leaf probabilities all add up to $1$; this simply indicates that we've fully captured the probability space
My questions are:
- Are the above statements correct?
- Is there an alternative way of deriving this? That is, for a given branch of the above tree, it seems that it could be represented instead as some variation of $P(B|A) = \frac{P(A|B) \ P(B)}{P(A)}$, as in this answer, but I have no idea how to get there
