# 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.

# Where Do I Know You From?

I’m at a fairly popular pizza place in my hometown just a few miles from the house where I grew up. I came to get some lunch and get some work done, but a few minutes ago someone walked in and I’ve been distracted ever since.

I’m pretty sure we went to high school together, and if the faintest of memories serves correct we may have even run track together for a year. I’m fairly confident about his first name, but the last name I’m associating with him would make his name the same as a popular action movie actor, so I’m pretty sure I’ve got that wrong.

It’s an uncomfortable feeling to be sitting across the room from someone you recognize incompletely. It’s slightly awkward enough to interrupt my lunch, and not familiar enough to start a conversation and reminisce for a bit.

This uncomfortable feeling is something many test-takers are familiar with. You see a problem and recognize that it’s somewhat similar to one that you’ve seen before. But what was the trick to that one? What was that one crucial step that you figured out that allowed you to unlock the problem? If you don’t remember, you’re left frustrated and second-guessing yourself.

Here’s how to avoid that feeling: treat review as an integral part of your prep process. Many students spend a great deal of time doing problems, and limit their review to checking to see whether or not they got the problems right and perhaps what the correct answer was. Sometimes someone else’s solution gets read for a particularly difficult problem, but that is rare. As they say in the infomercials, but wait, there’s more!

The crucial step in review comes after everything I’ve already listed. The last thing you have to do is work back through the problem with your fresh knowledge to make sure that you make sure you can do a similar problem in the future. Ideally, you’ll do this a couple days after you first encounter the problem. This cements the idea of the problem into your mind and ensures that the next time you see something similar you’ve got the understanding of what it will take to get it done.

Now if you’ll excuse me I’ve got to take a circuitous route to the soda machine to avoid an awkward encounter.

I recently had a very encouraging session. A girl whose progress had stalled out around 650 on the GMAT wanted to come in and do one session before her test. She already had the test date scheduled and wanted to at least get a sense of direction so that we could plan out her next attempt. As we worked together however, I found that her skills were excellent. She knew all of the concepts and formulas and had a strong enough math background to do very well on the test. The problem was that she had mad scientist disease.

Now mad scientist disease is a phrase I coined for someone who upon seeing a math problem erupts in a flurry of activity. Notes are scribbled, equations laid out and many important things are figured out. The problem is that many of those things are irrelevant! What the mad scientist does is largely a waste of time. She mixes together lots of chemicals and hopes that one of the compounds she finds is useful and that nothing blows up in her face!

In the one session we had before this student’s test we talked about how to approach problems. We focused on patiently organizing thinking and planning on what needed to be solved before solving anything. I got the student to slow down and relax.

Now I can’t claim that this is a typical result, but after that one session this student’s score went up to 730! An 80-point gain in a week! And all it took was a simple cure for mad scientist disease. Slow down and plan what you’re going to do before you do it.

# Parabolas

I recently came across a problem that required knowledge of parabola formulas. That surprised me. Although parabola problems show up relatively frequently, they usually require little more than logic. However, I thought that provided a nice opportunity to refresh on all things parabola.

A parabola is defined mathematically by this formula: y= ax^2 + bx + c. We see parabolas in nature most often when we look at projectiles, like a cannonball shot out of a cannon or a jump shot out of the hand of Steph Curry. Parabolas are generally u-shaped and are symmetrical about the vertex, which is either the highest or lowest point of the parabola, depending on the orientation.

Whether our parabola is cupped upward or downward is determined by the sign of the “a” term in the formula we saw above. When a is positive, the parabola will have a vertex at the bottom and open upward. When a is negative, the parabola will have a vertex at the top and open downward.

However, there is more than one way to define a parabola mathematically. We can also solve a parabola if we have the vertex and another point on the parabola. We do that by using the similar formula y = a(x-h)^2 + k. The coordinates of the vertex are (h,k).

So using that information, find the equation of a parabola with vertex (-2,1) containing the point (1, 19). The first thing we need to do is solve for a by inserting our points into the formula. We get:

19 = a(1-(-2)^2 + 1

19= a(3^2) + 1

19= 9a + 1

Now, we put our a into the formula with our vertex (h,k), but instead of using the x and y from a specific

point we’re going to solve for the generic x and y.

y= 2(x-(-2)^2 + 1

y= 2(x+2)^2 +1

Now expand:

y=2(x+2)(x+2) +1

y=2x^2 + 8x + 9

Now we’ve solved for the equation of this parabola and we could mathematically figure out all of the

points on this curve.

I hope that’s been a good refresher on parabolas!