I came across this probability puzzle:
William is playing archery blindfolded. The first arrow he shoots unfortunately misses the bull's-eye. The second arrow misses the bull's-eye even further. William still shoots a third arrow. How big is the chance that his third shot is also worse than his first shot?
Now the given solution is: We have first two marks as [A B] (here, A is left to 'B' means A is closer to Bull's eye than B). So C has following $3$ possibilities to take on, viz.:
[A B C][A C B][C A B]
So, the probability that C is worse than A is $\frac 2 3$ i.e. the cases [A B C] and [A C B].
My doubt here is that this probability of $\frac 2 3$ would be true only when above three possibilities are equally likely. But they don't seem to be equally likely because if we look a bull's eye:
Now, farther we move from bull's eye, we have more area to hit on and so probability to perform worse than a shot should be more than the probability to perform better than it, for e.g. Suppose I first hit at the Red region, then to perform better than my first hit I need to hit in Yellow Region and to perform worse than my first hit I need to hit in Black or White or even out of the board and so probability of performing worse is more than that of performing better.
So, what is going wrong here? Or there is some other solution?
Best Answer
There is a certain distribution of the distance $X\geq0$ of a hit from the aim. Given three independent shots we obtain three distances $x_i\geq0$ $(1\leq i\leq3)$ whose $<$-order is such that each permutation of $\{1,2,3\}$ has equal probability ${1\over6}$. We now are told to compute the conditional probability $$P\bigl[x_3>x_1\bigm| x_2>x_1\bigr]\ ,$$ and your case analysis shows that this probability is ${2\over3}$.