1
$\begingroup$

The professor asked us to imagine a scenario where we have a basketball player who isn't good at shooting free throws. He makes his first free throw with probability $0.2$ After the first free throw, he makes a free throw with probability $0.6$ if he made the preceding one, and he makes a free throw with probability $0.3$ if he missed the preceding one. During practice, he throws $10$ free throws.

What is the total number of outcomes? Let $A_i$ be the event that he makes the $i$th free throw. Find $P(A_i)$ for $i=2, 3, 4$.

I figured out the total number of possible outcomes using binomial coefficients ${n \choose r} = \frac{n!}{(n-r)!r!} $. The number of possible ways outcomes are as follows:

\begin{eqnarray} \text{Possible ways he will make 10 free throws} & = & {10 \choose 10} = 1 \\ \text{Possible ways he will make 9 free throws} & = & {10 \choose 9} = 10 \\ \text{Possible ways he will make 8 free throws} & = & {10 \choose 8} = 45 \\ \text{Possible ways he will make 7 free throws} & = & {10 \choose 7} = 120 \\ \text{Possible ways he will make 6 free throws} & = & {10 \choose 6} = 210 \\ \text{Possible ways he will make 5 free throws} & = & {10 \choose 5} = 252 \\ \text{Possible ways he will make 4 free throws} & = & {10 \choose 4} = 210 \\ \text{Possible ways he will make 3 free throws} & = & {10 \choose 3} = 120 \\ \text{Possible ways he will make 2 free throws} & = & {10 \choose 2} = 45 \\ \text{Possible ways he will make 1 free throws} & = & {10 \choose 1} = 10 \\ \text{Possible ways he will make 0 free throws} & = & {10 \choose 0} = 1 \\ \end{eqnarray}

These outcomes add up to $1024$. So that's done.

Now, if $B_j$ represents the event where this basketball player made $j$ number of free throws. Then $B_{10}$ has $1$ outcome, $B_{9}$ has $10$ outcomes, $B_{8}$ has $45$ outcomes, and so on until we get to $B_{0}$ which has $1$ outcome. From each of these I need to figure out how many outcomes in each $B_j$ made the $i$th free throw. These outcomes would then go into $A_i$.

Now, this is where I am a little overwhelmed. How would I find the number of outcomes in each $A_i$ without having to conduct it tediously. From there, how would I take into consideration the free throw probabilities $0.2, 0.6, 0.3$. I assume this has to do something with conditional probability, but since I am new to probability, I am not sure.


I am terribly sorry for the long post. I am truly grateful for all the help and advice you give me. Thank You for your time, take care, and have a wonderful day.

$\endgroup$
  • 3
    $\begingroup$ An easier way to do the first: Our player either makes or does not make any given shot, two possibilities. Hence the total number of outcomes is $2^{10}=1024$. $\endgroup$ – lulu Feb 1 '16 at 18:35
  • 2
    $\begingroup$ To clarify: the way you have phrased the question, I assume you are asking for the probability that our player makes the $i^{th}$ free throw, I mean, that's what it says. But in your discussion you appear to be asking for the probability that he makes exactly $i$ shots which is completely different. My posted solution addresses the first question, not the second. $\endgroup$ – lulu Feb 1 '16 at 18:58
2
$\begingroup$

For the second part, it is probably easiest to proceed recursively.

We are told $P(A_1)=.2$.

Now suppose we have computed $P(A_{j-1})$ and want to compute $P(A_j)$. Look at the first $j-1$ shots. We only care about the last one, which we know to be a hit with probability $P(A_{j-1})$ (and therefore a miss with probability $ 1-P(A_{j-1}))$ It follows that $$P(A_j)=.6\times P(A_{j-1})+ .3\times (1-P(A_{j-1}))=.3 \times P(A_{j-1})+.3 = .3\times (P(A_{j-1})+1)$$

In this way it is very easy to compute all the $P(A_j)$.

Note: it wasn't part of the question, but it is interesting to note that in the limit as the number of free throws gets large, the probability goes to $\frac 37$; this follows quickly from the recursion. Indeed, the recursion can be solved in closed form by standard methods. Doing so, we get: $$P(A_j)=\frac 37-\frac {16}{21}(.3)^j$$

$\endgroup$
1
$\begingroup$

Here are results from a simulation in R of 100,000 ten-throw sessions, baed on @lulu's analysis.

The simulation is based on a matrix MAT with $m = 100,000$ rows and $n = 10$ columns. At the end of the simulation, the matrix has a 1 in cell $(i,j)$ if the $j$th throw in the $i$th session was a success. The $m$-vector x contains the number of successes in each of the $m$ sessions. The table shows the simulated distribution of the number of successes per ten-throw session, and this is plotted in the figure.

 m = 10^5;  n = 10
 MAT = matrix(0, nrow=m, ncol=n)  # all 0's to start
 MAT[,1] = rbinom(m, 1, .2)       # fill first column
 for(i in 2:n) {
   MAT[,i] = rbinom(m, 1, (MAT[,i-1]+1)*.3) }
 x = rowSums(MAT)
 round(table(x)/m, 3)
 ##  x
 ##     0     1     2     3     4     5     6     7     8     9    10 
 ## 0.032 0.083 0.135 0.176 0.187 0.159 0.115 0.068 0.033 0.011 0.002 
 plot(table(x)/m, type="h", lwd=2, ylab="PDF", main="Distribution of Successes")
   abline(h=0, col="darkgreen")

enter image description here

As a reality check, we confirm that the key conditional structure is properly captured by the program:

(a) If throw #2 (in col 2) is a success, then throw #3 (in col 3) should be a success with probability 0.6.

 mean(MAT[,3][MAT[,2]==1])  # vec of results on 3rd given S on 2nd
 ## 0.6014073               # 'mean' is proportion of S's

(b) If throw #2 (in col 2) is a failure, then throw #3 (in col 3) should be a success with probability 0.3.

 mean(MAT[,3][MAT[,2]==0])
 ## 0.3002676

Finally, according to @lulu's analysis, by the tenth throw the proportion of successes ought to be approaching $3/7 = 0.4285714.$

 throw.10 = MAT[,10]
 mean(throw.10)
 ## 0.42746

Simulating 100-throw sessions, I got 3/7 to three place accuracy for the 100th throw.

I realize this is not a substitute for a formal mathematical solution to your problem, but it does a pretty good job of confirming @lulu's analysis, and may give you something against which to compare your analytical solution.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.