Tim throws two darts at a dartboard, aiming for the bullseye (center of the board). The second dart lands farther from the center than the first dart. If Tim is aiming for the bullseye, what is the probability that the third dart lands farther from the center than the first? Assume that Tim's throwing accuracy is constant.

Write out all possible cases in which you rank the throws. Tim throws three darts, so there are 3*2*1 = six different scenarios. For this problem, use the conditional probability equation. \begin{equation} P(A|B)=\frac{P(A \cap B)}{P(B)} \end{equation} P(B) is the probability that the second dart lands farther from the center than the first dart. Writing out all possible scenarios, from close to farther away (123, 132, 213, 231, 312, 321), show that 3 outcomes satisfy P(B), being the outcomes (123, 132, 312). Now we know that P(B) = 1/2, because 3 out of the 6 possible outcomes are satisfying event B. Furthermore, probability P(A) is the probability that the third throw is farther from the center than the first. The only outcomes that satisfy both event A and event B are (123, 132). Because 2 out of the 6 possible outcomes are satisfying, this gives you $P(A \cap B)=1/3$. So, we have \begin{equation} P(A|B)=\frac{P(A \cap B)}{P(B)}=\frac{1/3}{1/2}=\frac{2}{3} \end{equation}

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