## Saturday, October 6, 2012

### Jeremy Lin vs The Ghost (Finding Patterns)

This is a follow up post to Jeremy Lin vs The Ghost.

Note: For this post, I decided to stick to the basic rules of probability. There are formulas and models that could be used for this problem, but we'll focus on doing some exploration.

Before answering our question, let's tackle the same question with smaller numbers to see if we can find any pattern that we can use to help us arrive at the solution. Instead of finding the probability of making 21 shots and missing at most 6, for now, let's use making 4 shots and missing at most 3. We will still assume his probability of making a shot is 71%.

There are several arrangements possible for 4 shots made and up to 3 missed shots. Let's write out the arrangements and see if we can get some insight. To keep things simple, we will be using S for success (made shot) and F for failure (missed shot). Here are the possible arrangements for wins (4 made shots in 4 to 7 attempts):
 Shots Attempted Outcome Summary Possible arrangements (disregarding rules) 4 5 6 7 4S & 0F 4S & 1F 4S & 2F 4S & 3F SSSS SSSSF SSSFS SSFSS SFSSS FSSSS SSSSFF SSSFSF SSFSSF SFSSSF FSSSSF SSSFFS SSFSFS SFSSFS FSSSFS SSFFSS SFSFSS FSSFSS SFFSSS FSFSSS FFSSSS SSSSFFF SSSFSFF SSFSSFF SFSSSFF FSSSSFF SSSFFSF SSFSFSF SFSSFSF FSSSFSF SSFFSSF SFSFSSF FSSFSSF SFFSSSF FSFSSSF FFSSSSF SSSFFFS SSFSFFS SFSSFFS FSSSFFS SSFFSFS SFSFSFS FSSFSFS SFFSSFS FSFSSFS FFSSSFS SSFFFSS SFSFFSS FSSFFSS SFFSFSS FSFSFSS FFSSFSS SFFFSSS FSFFSSS FFSFSSS FFFSSSS
One thing to note is that certain arrangements are just not possible. For example:
SSSSF
is not a possible arrangement since 4 shots were already made (S), there wouldn't be a 5th attempt that results in a miss (F).

In fact, every arrangement that has 4 made shots (S) and where the last shot is a miss (F) is not possible, the game would end before the last shot was even attempted. In the above table, the arrangements which are not possible are bolded. A summary of the arrangements are in the table below.
Shots Attempted Outcome Summary Number of arrangements Number of impossible arrangements (disregarding rules) (last attempt is F) 4 5 6 7 4S & 0F 4S & 1F 4S & 2F 4S & 3F 1 5 15 35 0 1 5 15 1-0=1 5-1=4 15-5=10 35-15=20

The question is essentially "What is the probability that he makes 4 shots out of 4 attempts, 4 shots out of 5 attempts, out of 6 attempts, or out of 7." We will multiply the probability of those events occurring by the number of ways (different arrangements) it can occur and add them together.

We assumed earlier that the probability of making a shot is .71, so the probability of missing a shot (its complement) is 1-.71=.29.

For 4 successes in 4 attempts, we see there's only 1 way it can happen:
\begin{align*} & 1 * (.71)(.71)(.71)(.71) \\ & 1 * (.71)^4 \\ & 1 * .25411681 \\ & .25411681 \end{align*}For 4 successes in 5 attempts, we see there are 4 ways it can happen:
\begin{align*} & 4 * (.71)(.71)(.71)(.71)(.29) \\ & 4 * (.71)^4(.29) \\ & 4 * .25411681 * .29 \\ & 4 * .0736938749 \\ & .2947754996 \end{align*}For 4 successes in 6 attempts, we see there are 10 ways it can happen:
\begin{align*} & 10 * (.71)(.71)(.71)(.71)(.29)(.29) \\ & 10 * (.71)^4(.29)^2 \\ & 10 * .25411681 * .0841 \\ & 10 * .021371223721 \\ & .21371223721 \end{align*}For 4 successes in 7 attempts, we see there are 20 ways it can happen:
\begin{align*} & 20 * (.71)(.71)(.71)(.71)(.29)(.29)(.29) \\ & 20 * (.71)^4(.29)^3 \\ & 20 * .25411681 * .024389 \\ & 20 * 0.00619765487909 \\ & .1239530975818 \end{align*}Adding all together we get:
\begin{align*} &= .25411681 + .2947754996 + .21371223721 + .1239530975818 \\ &= .8865576443918 \end{align*}
So assuming Jeremy Lin's probability of making a 3-point shot is 71% and the shots are indepedent, then the probability of him making 4 shots before he misses 4 (up to 3 missed shots) is about 88.66% (assuming I made the right assumptions and did the calculations correctly).

Can we use this pattern to answer the question in our original post?

Question 1: Assuming each shot is independent of the others and his probability of making a shot is 71%, what is the probability that Jeremy Lin beats the ghost (essentially he makes 21 shots before he misses 7)?

I'll put up a simulator in the next post so we can check our answer when we arrive at it.

Jeremy Lin vs The Ghost
Jeremy Lin vs The Ghost (Finding Patterns)
Jeremy Lin vs The Ghost (Simulation)