Time for a little lesson on probability
To answer this question we need to set a few guidelines
- The probability that Fire Fang flinches(A) is 0.1(10% in decimal form) which can be restated as P(A)=0.1
- The probability that it burns(B) is also 0.1, which can then be restated as P(B)=0.1
- The probability that it flinches AND burns is 0.01, which can be restated as P(A and B)
Now the question we are trying to answer is what is the chance that it flinches OR burns, which can be restated as P(A orB)
This is a very simple question when one knows the right formulas, and for this case, it would be the one below
P(A or B)= P(A)+P(B)-P(A and B)
so basically the probability that fire fang flinches or burns IS the chance of it flinching plus the chance of it burning minus the chance of both of them happening, and when plugging in the numbers, this is what you get
P(A or B)= 0.1+0.1-0.01=0.19
Therefore the chance that it both flinches and burns is 19%, your initial intuition was correct! Though I hope this helped you understand why
As for your question on the probability of crunch getting a defense drop 2 times over 3 uses: it is a different type of probability involving a binomial distribution.
To answer the question, we need to set a few guidelines
- There is a chance for "success" or "failure." Let's say that crunch getting the defense drop is "success," that probability would be 20%
- There are a set number of "trials," three in this instance that you stated
- Each trial is "independent" of another, meaning that getting a defense drop does not affect the chances of getting another defense drop
The Binomial Formula is a bit complicated and I'm sure you are already tired of my math and such but here it is anyways
You will probably ignore that and I won't blame you!
But if I were to plug it into the formula(which I won't since I can just use calculator functions ;) then I would get....
This is the chance that you land two defense drops in three crunches, a surprisingly high probability actually
hoped I helped