# Will the Warriors go 82-0?

My favorite basketball team the Golden State Warriors is on an historic run. After winning the NBA championship last year, the Warriors have rattled off 19 consecutive wins to open the season. This is the first time that a team has won that many games to open a season, and some might be starting to wonder whether the Warriors can complete an undefeated regular season. The cruel realities of probability will step in to show us why that won’t happen.

First, I’ll admit that that this analysis will omit several complicating considerations, most notably injuries. Chances are that several of the key players for the Warriors will get hurt this year, which will reduce their chances to win the games where those players are out. There are a number of other variables that cannot be accounted for, but as you’ll see the math is pretty overwhelming and is unlikely to be overcome by anything else.

Sports bettors are very interested in how likely a team is to win each game, and as a result there is quite a bit of analysis put into this idea. As a result we can say that in their toughest games the Warriors have about a 52% chance to win (on the road in Cleveland or San Antonio). In their easiest games (at home against the Lakers or the 76ers) the Warriors have about a 95% chance to win.

So what are the chances that the Warriors win the next 63 games? Well, let’s establish the range. What if the Warriors were 52% favorites in all of their remaining games? We’d have 68 independent events all of which we’d need to happen. To calculate independent probabilities, multiply them out and the product is your total probability. So, in this case we get: 0.52*0.52*0.52*… 0.52 sixty-three times! In other words, we’d get 0.52^63 = 0.000000000000000001.28295271. So that’s the low end of our range. Not very optimistic is it?

Let’s look at the high end. Let’s assume that every opponent is just as bad as the Lakers or 76ers at home. We calculate that probability the same way, only this time it’s 0.95^68. This number should be a lot bigger, right? I mean, the Warriors are huge favorites in every single game. Right?

0.95^63= 0.0395

Right around a 4% chance. That seems downright possible! But, of course all the teams the Warriors will play aren’t that bad. So how do we best estimate the rest of their schedule? We could go game-by-gam0e and look at estimates for each one to develop a number, but instead let me give you a choice that will test your estimating skills. If you were the Warriors would you rather play 32 opponents that you had a 95% chance to beat and 31 you have a 52% chance to win or all your games in the middle where you have a 75% chance to win?

0.75^63= .0000000134542531

0.52^31*0.95^32= .00000000030426992e-10

The Warriors are much better off facing a lot of pretty good teams, because of the big impact that the games against the equally matched teams have.

But as you can see in either case… don’t hold your breath for an undefeated season.

# Is Warren Buffett’s \$1 Billion Safe?

March is my favorite part of the year, and the NCAA March Madness is no small part of that. For those of you that are unfamiliar, after months of regular season basketball, weeks of conference tournaments, and two days of preliminary matches the field is narrowed to 64 teams. Those teams are put into a single-elimination bracket with four regions, and after 63 games a champion is crowned.

This year’s tournament has a little extra excitement as businessman Warren Buffett has offered \$1,000,000,000 to anyone who can complete a perfect bracket. But Buffett didn’t luck into his fortune. Let’s use basic probability to determine just how likely it is that someone will be collecting that billion dollar check!

First of all, we need to determine whether we’re working with independent or dependent events. In this case, our outcomes are independent. The result of one game does not alter the odds in the next game. Dayton beating Ohio State doesn’t change the likelihood of Harvard beating Cincinnati. In contrast, a dependent event would be something like pulling marbles out of a bag without replacement. If we pull a blue marble out of the bag there is one less blue marble to pull out the next time, so the odds will change.

With independent events, we can compute the probability of all n events in a series happening by the following formula:

P(1)* P(2)* P(3)*…*P(n) = Probability of all events occurring

In this case, we need to know the probability of predicting all 63 events correctly. In our first case, let’s assume we know nothing about college basketball. In that case, we’ll randomly guess a winner from each matchup. Since we know nothing about the teams, we will estimate our likelihood of getting each guess correct at 1/2 or 0.5. In that case our likelihood of predicting a perfect bracket will be 1 in 2^63 or 1 in 9,223,372,036,854,775,808. That’s more than 9 billion billions! So without any knowledge it’s safe to say that your chances are not good.

But what if you know a little bit more? What if you know that a 1 seed has never lost to a 16 seed? That makes four games slam dunks and just leaves 59 left to figure out. Maybe you’re a college basketball expert and you can predict 60% of outcomes correctly. Your chances of getting a perfect bracket are now 1/(0.6)^59 or 1 in 12,275,963,663,147. We’re down to one in twelve million billion, but Mr. Buffett’s money is still looking pretty safe.

Let’s try one other example. Let’s say you’re a college basketball expert with a time machine that allows you to look two days into the future. That will allow you to know the first 32 results with absolute certainty, and just leave the final 31 games that you can predict correctly 60% of the time. Now your chances are 1/(0.6)^31 or 1 in 7,538,956. Even with expertise and a time machine your chances are just one in seven and a half million!

Three quick takeaways:

1. When events are independent, multiply the probabilities to find the total probability of all events happening

2. Exponential growth makes numbers get really big (or really small) really fast

3. Warren Buffett’s money is probably safe… until we build a better time machine!

It happens every March. You lean forward to the edge of your chair and watch in horror, with a crumpled bracket sheet clutched in your hand as that desperation shot by that unheralded kid from that underdog school arcs slowly through the air. As it drops through the net and knocks out the team you picked to win it all you mutter “What are the odds?”

I understand the exasperation that missed predictions in the NCAA Tournament can bring, but I’m afraid that I don’t understand the exasperation of students who don’t understand basic probability. Basketball brings quite a few opportunities to illustrate the basics of probability, so let’s take a look at a few of those situations, so that you can get probability under control (even if your bracket stinks).

One of the areas of basketball most ripe for probability analysis is the free throw. Without the variables of defense or different areas of the floor, we can make a pretty good prediction about a player’s likelihood of making a free throw based on the results of his previous free throw attempts. For instance, if Player A has made 700 out of the 1000 free throws he has attempted, we would say that he is a 70% free throw shooter and that the probability of him making his next free throw is 0.70. It’s important to note that a probability is not a prediction. We don’t know whether Player A will make his next free throw, nor do we know that Player A would make 7 out of his next 10 free throws. All the probability does is reflect the likelihood of an individual event taking place.

As you might have noticed, probabilities are expressed as a decimal value between 0 and 1, where 0 is an event that will not occur and 1 is an event that is certain to occur. When we have multiple independent events and we want to know the probability of both occurring, we multiply the probabilities together.

Here’s an example. Player A gets fouled attempting a shot and is awarded two free throws. Seeing this, Chad says to Jimmy, “He’s a good free throw shooter, he’ll probably make both.”

“Probably not,” Jimmy replies.

Let’s take a look and see who is correct. First, we’ll take the probability of Player A making the first free throw, which we know is 0.7. Then, since the events are independent (the result of the first free throw doesn’t affect the likelihood of making the second free throw), we’ll multiply that by the probability of Player A making the second free throw.

$P(Making Both)=P(Making First)*P(Making Second)$

$P(Making Both)=0.70*0.70$

$P(Making Both)=0.49$

So, Jimmy is correct. Player A will make both free throws less than half of the time, even though he is likely to make any single free throw attempt. Just as Jimmy finishes bragging and Chad goes back to his drink, Player A is fouled shooting a 3-pointer and is awarded three free throw attempts.

“I’ve got it this time,” Chad says. “I know he’s going to make at least two of these free throws almost all time time.”

“Don’t count on it,” Jimmy says.

To figure out how often Player A will make at least two free throws, we first need to figure out how many different ways that could happen. Here is a table that summarizes that information:

 Attempt 1 Attempt 2 Attempt 3 Outcome 1 Make Make Make Outcome 2 Make Make Miss Outcome 3 Make Miss Make Outcome 4 Make Miss Miss Outcome 5 Miss Make Make Outcome 6 Miss Make Miss Outcome 7 Miss Miss Make Outcome 8 Miss Miss Miss

So, in 4 of 8 of the sequences we could have Player A makes at least two shots. However, we’re not done yet because not all of these outcomes are equally likely. Player A is a pretty good free throw shooter, so he’s more likely to make all three than to miss all three. So, we need to find the probability of each of the four different scenarios we want and add them together. We add in this “or” situation because either Outcome 1 or Outcome 2 or Outcome 3 or Outcome 5 will satisfy the condition we’ve set.

$P(O1)=0.7*0.7*0.7=0.343$

$P(O2)=0.7*0.7*0.3=0.147$

$P(O3)=0.7*0.3*0.7=0.147$

$P(O5)=0.3*0.7*0.7=0.147$

$P(Making At Least Two)=P(O1)+P(O2)+P(O3)+P(O5)$

$P(Making At Least Two)=0.343+0.147+0.147+0.147=0.784$

Is that the outcome you expected? Whether it’s intuitive or not, working with probabilities will help you get more comfortable with them on your test… even if it doesn’t help your bracket.

Good luck with probabilities and with your picks!

# What Are the Hardest Questions On the GRE?

As you undoubtedly know very well, the GRE contains some very tricky brainteasers. In this post I am going to discuss some (but definitely not all!) of the question types that give students the most trouble.

Multiple-Blank Text Completion Questions

Text completion questions are not necessarily inherently difficult, but in their advanced stages they can cause quite the headache. This is because they have the potential to test two disparate yet challenging skill sets: logical relationships and difficult vocabulary. Take a look at the following question:

Q. From the beginning of her career, Donna Sutherland was mostly perceived by her peers as a __________; only through her repeated publications was she inevitably considered a true _________ of story and structure.

 (1) (2) dilettante cognoscente macrocosm neophyte caprice opprobrium

This question contains zero contextual clues about the meanings of the words in the blanks – only relationships. This makes for a pretty tough question, especially considering the difficulty of the words offered for the blanks!

As written, the sentence makes it clear that the two blanks must contain opposite meanings – as such, the only possible words that fulfill this relationship are “dilettante” (which means beginner) and “cognoscente” (which means master).

Argument Analysis Essay

The Argument Analysis Essay is not for the faint of heart. To gain full credit for this task, a student must have a fairly comprehensive grasp of the fundamental parts of an argument (premises, assumptions, and conclusions), as well as the ability to identify some of the most common logical flaws that appear on the GRE (causal arguments, argument by analogy, and so on). And don’t forget – you’ve got to write  well too!

Probability and Statistics

Although these questions don’t appear as frequently as, say geometry questions, students who intend to score highly on the GRE must be familiar with this subject matter.

Q. A card is drawn randomly from a regular deck of cards (52 cards in total, 13 in each of four suits). What is the probability that the card is either a 10 or a spade?

This question requires students to understand both basic probability and the ability to calculate probabilities of events connected by the word “or”. To answer this question, we may either calculate the probability of each event independently and take their sum, or try to find a way to combine them.

There are 13 cards of each suit in the deck, so that means that there are 13 spades in this deck. Also, there are four tens in each deck. But, before we move directly to calculation, we must consider their potential overlap: there is one ten that is also a spade, thus we cannot simply add the numbers together. We must add them together and then subtract their overlap to find the total number of chosen elements in this set:

4 tens + 13 spades – 1 ten of spades = 16 cards

Probability of choosing 16 cards from a total of 52 = 16/52 .

These question types are not meant in any way to be exhaustive; it is possible to turn even the simplest operation, such as addition, into a daunting problem on the GRE. Having a solid understanding of the underlying fundamental concepts is really the only way to prepare for complex question types, so study up!