The puzzle: You, your brother and your father all play chess. Experience has taught you that when you play your brother, you have a decent chance of winning - say 50/50 - but when you play your father, you have a very slim chance - say 1 in 20. They issue you a challenge: you play 3 games with them, and to win the challange you must win some two-in-a-row (either the first and second game, or the second and third). You get to choose the format of the three games: EITHER you play your father first, then your brother, and then your father again, or you play your brother first, then your father, then your brother again.
Which of the two formats should you choose (FBF or BFB) to maximize your chance of winning two-in-a-row? (And, of course, your answer should come with a clear justification.)
A solution: We need a probability model for each of the two possible tournaments. For the first, father first, it makes sense to let S consist of the 8 sample points WWW (1), WWL (2), WLW (3), WLL (4), LWW (5), LWL (6), LLW (7) and LLL (8) (with obvious meanings). The event that you win at least two-in-a-row consists of sample points (1), (2), and (5).
Next, we need to assign probabilities to the three interesting sample points. For WWW, you beat your father (probability .05), then your brother (.5), then your father (.05). A sensible probability to assign to that sample point is .05*.5*.05=.00125 (5% of the time you beat your father, so say 50,000 times out of 1,000,000. Of those 50,000 times, you also beat your brother 25,000 times, and of those 25,000, you beat your father again 1,250 times - so you expect to beat father, then brother, thn father 1,250 times out of 1,000,000). Similar reasoning suggests a probability of .05*.5*.95=.02375 for the point WWL, and .95*.5*.05=.02375 for the point LWW. Adding these three up, we get probability .04875 of winning the father-brother-father tournament.
The same analysis suggests a probability of .5*.05*.5 + .5*.05*.5 + .5*.05*.5 = .0375 of winning the brother-father-brother tournament.
So it is better to pick father-brother-father. This seems counterintuitive, since it is the choice that pits you twice against your father, by far the better player. The key observation to explain this paradox is that by the rules of the tournament, you *must* win the second game (at least). By choosing the option that maximizises your chances of winning that pivotal game, and gives you two shots at the Herculanean task of beating your father, you do yourself a favor.
If the rule was changed to simply ``win at least two out of three'', presumably you would instantly make the opposite choice!