I’m a few weeks into my statistics class and so far, it’s been a lot of “plug and chug.” But real mastery of statistics requires knowing when to use who’s test where…so for fun, I will attempt to analyze this year’s brackets for SLJ’s Battle of the Kids’ Books, especially now that the Battle schematics are up! (Aside: please correct me if I get the math wrong. Better now than on my exam.)
Suppose you didn’t read any of the battle books, what’s the probability of randomly guessing the correct winners?
Now, there are 14 brackets leading up to and including the semi-finals, and the probability of guessing right per bracket is 0.5, since there are two contestants, but only one winner each time. Each guess is independent of the others, since you’re no more likely to correctly guess the outcome, say, of Boxers&Saints vs. A Corner of White after successfully predicting the previous match-up.
To solve this problem, the gut instinct thing to do would be to multiply 0.5 fourteen times, which equals 0.000061, to get the probability of guessing correctly all the way through the semi-finals.
The textbook way to do it would be to use the binomial distribution method, where p=prob. of guessing right and q=prob. of guessing wrong and the probability of getting n trials correct k times equals kCnpkqn-k. Since p and q occur 50% of the time, and we want to get all fourteen out of fourteen matchups correct, then the probability equals (14!/14!0!) (0.5)14(0.5)0, although why you’d do it this way is beyond me. (Note: because you want to know the probability of getting k out of 14 match-ups correctly, and so forth.)
Either way, you have a 0.0061% chance of getting all the brackets through the semi-finals. But then, The Undead complicates matters, as usual, because for the Big Kahuna Round, the probability of guessing right is 1 out of 3, or 0.33, and the probability of guessing wrong is 0.67. I’m just going to find the union of the two events–getting everything right all through the semi-finals, and then guessing correctly in the Big Kahuna round–and because they’re independent, I can multiple p1=0.0.000061 by p2=0.33 to get an overall probability of 0.00002. In other words, we’ve a 0.002% chance of randomly guessing correctly the results for every match of every round.
Luckily, we don’t need to guess correctly every round. Next time, if I haven’t bored you, I will attempt to calculate the probability of guessing the fewest possible match-ups correctly and still make it all the way through to the end. But first, I have eight more books to read and a problem set due Tuesday!