I'm going to ask the question and then, just for fun, answer it. Does this make the thread pointless? Probably so.
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OK, here's how I decided to think about this question.
Ignoring the play-in game, there are 63 games in the tourney.
16 of these games are usually pretty easy to pick: the first-round games that involve the #1, #2, #3, and #4 seeds. Yes sometimes they lose but I would estimate they win roughly than 90% of the time (combined). Let's say, therefore, that you have a 90% chance for each of those games of picking them correctly. Odds of picking them ALL correctly therefore is
(9/10) ^ 16 ==> about 18.5%.
So just those 16 games, you have about an 18.5% chance of picking correctly.
For the rest of the games, there tends to be some parity. For any given game, team X may be better than team Y, but not so good that they will beat them every time, or even 90% of the time.
Let's assume that you are a tremendous basketball scout. You're so good that you know which team is better for each and every matchup. But let's also assume that the better team is only going to win 3/4 of the time. Then the odds of picking the rest of the games (all 47 of them) correctly is (3/4) ^ 47. This works out to be 1 time in about 745,000!
So the odds of picking all the games correctly would be about 1 in 745,000 * 18.5 percent, or roughly 1 in 4 million.
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But let's say 75% is too high an estimate for the 47 "close" games. Say you can only predict it 2 times out of 3, or 67%. How much of a difference does that make? It's huge. At that point, your odds of choosing all 63 games correctly drops to 1 in about 1 billion.
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If you can only predict the 47 "close" games with 60% accuracy, the odds fall to 1 in 144 billion. You could give every person in the USA 500 chances to pick and still not get 144 billion entries.
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On the other hand, if you can predict the 47 games with 80 percent accuracy, your chances of getting all 63 games correct go way up, all the way to abut 1 in 200,000.
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OK, nobody cares, but I had fun.
