Probability - three-headed dragon and three knights I wrote a test from probability today and totally screw it.
There was a exercise about a three-headed dragon and three possible knights. Chosen knight must cut off at least two dragon heads to save princess. When he cut off zero or one head he failed. Every knight also use different weapon with different probability to cut head off.
First knight uses a sword with constant probability $\frac{1}{2}$ to cut off one head in a single attempt. He has three attempts.
Second knight uses an axe and has constant probability $\frac{1}{2}$ to hit the dragon. Also when he hits the dragon, he has $\frac{1}{2}$ change to cut off one head or $\frac{1}{2}$ to cut off two heads. He has two attempts.
Third knight uses a bow. His first attempt has hit probability $\frac{1}{3}$, second attempt has probability $\frac{1}{2}$ and third attempt has probability $\frac{2}{3}$. Every successful attempt cuts off one head. He has three attempts.
Which knight should be chosen and what chance to save the princess does each knight have?
I would appreciate step by step solution from which I can learn. Thank you.
 A: You will need to calculate the probability of each knight cutting off at least two heads.
First Knight: Since he has three attempts and he cuts off either $1$ or $0$ heads per attempt, there are eight possible outcomes for the number of heads he cuts off in each of his three attempts. 
$000, 001, 010, 011, 100, 101, 110$, and $111$.
Since he has a $50$% chance in each attempt, all eight possibilities are equally likely, and in four of them he cuts off at least two heads. Therefore his probability of success is $\frac{4}{8}=50$%
Second Knight: There are three ways for the second knight to cut off at least two heads.
Either he cuts off two heads on the first try, which has probability $25$%
Or he cuts $1$ head on the first try and at least $1$ more on try two.  This would have probability $(\frac{1}{4})(\frac{1}{2})=12.5$%
Or he misses on try one and cuts off $2$ heads on try two, which has probability $(\frac{1}{2})(\frac{1}{4})=12.5$%
Overall, his probability of success is $50$%
Third Knight: The third knight also has three ways to cut off at least two heads.
Either he cuts off a head on each of his first two attempts, which has probability $(\frac{1}{3})(\frac{1}{2})=16.67$%
Or he hits, misses, and then hits, which has probability $(\frac{1}{3})(\frac{1}{2})(\frac{2}{3})=11.11$%
Or he misses, then hits twice, which has probability $(\frac{2}{3})(\frac{1}{2})(\frac{2}{3})=22.22$%
Overall, his probability of success is $50$%
So it looks like it doesn't really matter which knight you choose because they all have a $50$% chance of succeeding.  Although Knight 2 does have the best chance of getting all three heads.
