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 ** _{k}C_{n}p^{k}q^{n-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!

on February 23, 2014 at 12:43 pm |Karen MaurerThis is such a fun may to look at Bob.

on February 23, 2014 at 9:16 pm |SondyYour math is great, Jen! You’re actually not finding the union of the events at the end, but the intersection. In other words, you’re finding the probability that you guess all matches right up to the semi-finals AND the probability you guess the last match correctly. They’re independent, so you multiply, as you did.

The other problem you suggested is intriguing. What is the probability that you just guess the final winner correctly? I’m assuming you mean that you guessed that book to win each round to get into the finals and then win it all OR that it loses some time and then comes back from the dead and wins it all.

So let’s figure that out. First, the traditional way of winning. What is the probability that you randomly guess the ultimately winning book to win at each level? In other words, that it wins the first round, second round, semifinals, and finals? That would be (1/2)x(1/2)x(1/2)x(1/3) = 1/24

However, there’s also the nontraditional path, and that way it could have lost in the first round OR the second round OR the third round and then come back from the dead.

The probability you guess the ultimately winning book (let’s call it W) to lose in the first round is 1/2.

The probability you guess W to lose in the 2nd round is the probability it wins in the first round and loses in the 2nd round, (1/2)x(1/2) = 1/4.

The probability you guess W to lose in the 3rd round is the probability it wins in the first and second rounds and loses in the 3rd round, (1/2)x(1/2)x(1/2) = 1/8.

Thus, the probability that you guessed W to lose in the 1st OR 2nd OR 3rd round is 1/2 + 1/4 + 1/8 = 7/8.

We’re saying all this is random, which gives W a 1/16 chance of being your pick for undead winner.

So the probability you randomly have W in the finals by being the undead winner is the probability it loses one of the earlier rounds AND is the undead winner. So that’s (7/8) x (1/16) = 7/128 Then that it wins that final match is (7/128) x (1/3) = 7/384

So finally, the chance of randomly choosing the correct book to win the Battle of the Books is the probability that you guess it to win all its rounds and the finals OR that it advances to the finals by the undead route.

So that’s (1/24) + (7/384) = 16/384 + 7/384 = 23/384 = 0.060

So if I did all that math right, you’ve got a 6% chance of just guessing the final winner correctly, if you guess all the matches randomly. (Assuming your guesses follow the rules of the tournament. So you don’t guess a second round winner that didn’t win the first round.)

This was fun! Thanks for doing this, Jen!

on February 23, 2014 at 9:33 pm |SondyHuh. Now I’m having doubts. It seems like EVERY book should have the same chance of being picked as the final winner, if it’s all random. So the final probability should be 1/16 = 24/384. Can you figure out where I went wrong? I think I’m off by 1/384.

on February 23, 2014 at 9:39 pm |SondyI’ve got it! My error was in the probability W is picked as the Undead winner. The two books that advanced to the finals are NOT eligible, so the probability W is picked is only 1/14, not 1/16.

That gives us (7/8) x (1/14) = 7/112

Then the probability it wins the final round that way is (7/112) x (1/3) = 7/336

Finally, the last probability should be (1/24) + (7/336) = (14/336) + (7/336)

= 21/336 = 1/16

!

What do you know? We could have just said, right at the beginning, that there’s a 1/16 chance of picking the final book correctly. 🙂

on February 23, 2014 at 9:51 pm |JenSondy–thanks for your thorough analysis, and the anti-climactic resolution to the second question posed. Too funny! (Also, great catch. I must learn the difference between union and intersection before my midterm!)

I wonder what else we can do statistics-wise…any thoughts?

on February 24, 2014 at 11:56 am |SondyI’ve always liked that with probabilities, you can usually look at the problem more than one way — and you *should* get the same answer. After I did it the first way, I thought why would it be different for the book that happens to win than any other? That actually helped me catch my mistake. Anyway, the first way was a lot more fun, so it’s all good! 🙂

on February 24, 2014 at 11:57 am |SondyNow, we could figure out the probability that we guess the correct three books that are in the Finals…. Would it just be choosing 3 out of 16, or more complicated? (Probably we could again do it *both* ways.)

on February 24, 2014 at 2:49 pm |SondyI *think* it’s more complicated than just choosing 3 out of 16. Because given that we know the brackets, it’s not even possible for two books from the same half of the tournament to go to the finals, unless one is the undead winner.

So I think it can be done like this. P(Guessing correctly the three books in the final round) = P(Guessing first half winner correctly AND guessing second half winner correctly AND guessing Undead winner correctly.)

P(Guessing first half winner correctly) = 1/2 x 1/2 x 1/2 = 1/8

P(Guessing second half winner correctly) = 1/2 x 1/2 x 1/2 = 1/8

See above for this one:

P(Guessing Undead winner correctly) = 7/112 = 1/16

So P(Guessing final round participants exactly correctly) = 1/8 x 1/8 x 1/16 = 1/1024

However, what if you actually guessed the Undead winner to win its bracket, and the actual winner of that bracket as the undead book? So you flipped the two books, but ended up with the right books in the finals. By the same reasoning, there’d be a 1/1024 chance of doing that, and still having the right books in the final round.

So the total probability of having the correct three books in the final round should be 1/1024 + 1/1024 = 2/1024 = 1/512.

Clear as mud? You really shouldn’t get me going, Jen! 🙂

on February 25, 2014 at 9:27 pm |JenWait a sec–would the probability of guessing the undead correctly be 1/14, since two out of sixteen books are out of the running, being semi-finalists?

At some point, I’m going to collates all your awesome comments for a follow-up point, but phew this week is full…

on February 25, 2014 at 10:21 pmSondyYeah, I think you’re right. This is after I was writing out a whole big comment explaining why I was right!

So I *think* it should be (1/8) x (1/8) x (1/14) + (1/8) x (1/8) x (1/14)

= 1/896 + 1/896 = 2/896 = 1/448

Let’s test it by figuring it another way. (Because that’s the way I roll.)

Let’s call T the actual top bracket winner.

Let’s call B the actual bottom bracket winner.

Let’s call U the actual undead winner.

The probability of guessing that T wins the top bracket is 1/8.

The probability of guessing that B wins the bottom bracket is 1/8.

So then, yes, there are only 14 books left to choose from, so now the probability of guessing that U is the undead winner is 1/14.

So P(Guessing T wins top bracket AND B wins bottom bracket AND U wins undead poll) = 1/8 x 1/8 x 1/14 = 1/896

But suppose you actually guess that U wins its bracket, and whichever of T or B it knocked out, you guess to be the undead winner. You would still have the right three books in the finals, you just didn’t guess their path to get there correctly, but that doesn’t matter.

Let’s break it into two cases and add up the two probabilities.

Suppose you guess that T is the undead winner. Since you’re starting there, that has a 1/16 probability.

Now, the probability that U is in the top bracket is 1/2.

Then the probability that you guess U wins the top bracket is 1/7, since we already know that T is not the winner.

The probability that you guess B wins the bottom bracket is 1/8.

So, P(guessing T is the undead winner AND U is in the top bracket AND U wins top bracket AND B wins bottom bracket) = 1/16 x 1/2 x 1/7 x 1/8 = 1/1792

Now suppose you guess that B is the undead winner. Probability is 1/16.

The probability that U is in the bottom bracket is 1/2.

The probability that you guess U wins the bottom bracket is 1/7.

The probability that you guess T wins the top bracket is 1/8.

So, P(guessing B is the undead winner AND U is in the bottom bracket and U wins bottom bracket and T wins top bracket) = 1/16 x 1/2 x 1/7 x 1/8 = 1/1792.

Adding up those two probabilities gives us

P(guessing U to win its bracket and the winner it knocked out to win undead poll) = 1/1792 + 1/1792 = 1/896

So, yes, we end up with the same numbers! We add up those two probabilities for our final probability, which *does* match what we got above, *with Jen’s correction*!

P(guessing the right three books in the finals regardless of how they get there) = 1/896 + 1/896 = 1/448

Isn’t this fun?! (It’s at this point that all the students in my Statistics 101 classes would totally HATE me!)

(Aren’t you glad I switched careers and became a librarian?)