Over in the National League Championship Series, the Washington Rationals and the St. Louis Ordinals (known as the “Ords” for short) are also evenly matched. Again, both teams are equally likely to win each game of the best-of-seven series.
You enter a competition in which you must predict the winner of each of the seven games before the series begins. If any or all of the fifth, sixth or seventh game are not played, you are not credited with predicting a winner.
You win the competition if you predict at least two games correctly. If you optimize your strategy for picking winners, what is the probability you will win the competition?
While it might seem like a counterintuitive real life strategy, you can win $65/70 = 92.86\%$ of the time by just picking a single team to win each games, say the Ords. As we saw in the Riddler Express problem this week, there are a total of $70$ equally likely outcomes for the seven game series. If you are betting the Ords in each game and only need for two of your guesses to be correct, then the only time you will fail to win the competition is if the Rationals (Rats?) win in $4$ or $5,$ which comprise $\binom{3}{3} + \binom{4}{3} = 5$ of the $70$ possible outcomes.
Again, it seems that while this may have made sense in there are only 70 equally likely outcome events, that we were talking about a different underlying probability space, whoopsy-doozles.
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