geometric.Rmd

# Geometric Distribution {#geometric}

## Motivating Example {-}

In the casino game craps, after the "point" has been set, two dice are rolled repeatedly until either the "point" or a 7 comes up, 
at which time the round ends. Suppose the point is 4. What is the probability that it takes more than 6 rolls for the round to end?

We can calculate the probability that the round ends on a given roll by directly counting the 36 possible outcomes of two dice:
\[ P(\text{roll a 4 or a 7}) = \frac{9}{36}. \]
But how do we use this to determine the probability that it takes more than 6 rolls for the round to end?

## Theory {-}

<iframe width="560" height="315" src="https://www.youtube.com/embed/-vvtrsS4rkA" frameborder="0" allow="accelerometer; autoplay; encrypted-media; gyroscope; picture-in-picture" allowfullscreen></iframe>

```{theorem geometric, name="Geometric Distribution"}
If a random variable can be described as the number of draws, _with replacement_, 
from the box 
\[ \overbrace{\underbrace{\fbox{0}\ \ldots \fbox{0}}_{N_0}\ \underbrace{\fbox{1}\ \ldots \fbox{1}}_{N_1}}^N \]
_until_ a $\fbox{1}$ is drawn, then its p.m.f. is given by 
\begin{equation}
f(x) = \dfrac{N_0^{x-1} \cdot N_1}{N^x}, x=1, 2, 3, \ldots
(\#eq:geometric1)
\end{equation}
where $N = N_1 + N_0$ is the number of tickets in the box.

We say that the random variable has a $\text{Geometric}(N_1, N_0)$ distribution, and $N_1$, 
$N_0$ are called **parameters** of the distribution.

Equivalently, \@ref(eq:geometric1) can be written as 
\begin{equation}
f(x) = (1-p)^{x-1} p, x=1, 2, 3, \ldots,
(\#eq:geometric2)
\end{equation}
where $p = N_1 / N$ is the proportion of $\fbox{1}$s in the box. So we can instead specify 
the distribution as $\text{Geometric}(p)$, where $p$ is the parameter.
```

Note that in contrast to the hypergeometric and binomial distributions, there is no upper 
bound on the possible values of the geometric distribution. We could draw an unlimited number of $\fbox{0}$s, so 
in theory, we might need to wait an arbitrarily long time for a $\fbox{1}$.

We will derive the formulas \@ref(eq:geometric1) and \@ref(eq:geometric2) later in this lesson. 
For now, let's see how these formulas can be used to solve problems.

```{example name="The Craps Problem"}
Let's model each roll as a draw from a box. There are $36$ equally likely outcomes when we 
roll two dice, so we place $N=36$ tickets into the box. Only 9 of these outcomes results in 
the end of the round, so we label $N_1=9$ of these tickets as $\fbox{1}$; the other $N_0=27$ tickets 
are labeled $\fbox{0}$.

\[ \overbrace{\underbrace{\fbox{1}\ \ldots \fbox{1}}_{N_1=9}\ \underbrace{\fbox{0}\ \ldots \fbox{0}}_{N_0=27}}^{N=36} \]

Now, the number of rolls until the end of the round is just the number of draws until a 
$\fbox{1}$ is drawn. So the number of rolls follows the $\text{Geometric}(N_1=9, N_0=27)$ distribution.
 
We can now write down its p.m.f. by plugging 
the values of the parameters $N_1$, $N_0$ into \@ref(eq:geometric1):
\[ f(x) = \frac{27^{x-1} \cdot 9}{36^x}, x=1, 2, \ldots. \]

Equivalently, we have $p = N_1/N = 9/36$, and plugging in this value into 
\@ref(eq:geometric2), we have:
\[ f(x) = \left(1 - \frac{9}{36}\right)^{x-1} \frac{9}{36}, x=1, 2, \ldots. \]

Finally, we can calculate probability that it takes more than 6 rolls for the round to end 
in a few different ways:

- Using the complement rule and the geometric distribution:

    \begin{align*}
    P(X > 6) &= 1 - P(X \leq 6)  \\
    &= 1 - (f(1) + f(2) + \ldots + f(6)) \\
    &= 1 - \left( \frac{27}{36} \right)^0 \frac{9}{36} - \left( \frac{27}{36} \right)^1 \frac{9}{36} - \ldots - \left( \frac{27}{36} \right)^5 \frac{9}{36}  \\
    &\approx .178.
    \end{align*}
  
- Directly, using the geometric distribution and the geometric series:

   \begin{align*}
    P(X > 6) &= f(7) + f(8) + f(9) + ...  \\
    &= \left( \frac{27}{36} \right)^6 \frac{9}{36} + \left( \frac{27}{36} \right)^7 \frac{9}{36} + \left( \frac{27}{36} \right)^8 \frac{9}{36} + \ldots  \\
    &= \frac{\left( \frac{27}{36} \right)^6 \frac{9}{36}}{ 1 - \frac{27}{36}} \\
    &\approx .178.
    \end{align*}
    In the second-to-last line, we used the formula for an [infinite geometric series](https://www.youtube.com/watch?v=-y1Ob0K63hc):
    \[ a + ar + ar^2 + ar^3 + \ldots = \frac{a}{1 - r}, |r| < 1. \]
    
- Without using the geometric distribution at all. In order for the round to end after more than 6 rolls, the first 6 rolls must all have 
failed to end the round. In other words, all 6 of these rolls resulted in one of the other 27 outcomes. The probability of this is 
\[ \frac{27^6}{36^6} \approx .178. \]
```

Now, let's derive the p.m.f. of the geometric distribution.

```{proof, name="Theorem \\@ref(thm:geometric)"}
To calculate the p.m.f. at $x$, we need to determine the probability that it takes exactly $x$ draws to get 
the first $\fbox{1}$. 

First, there are $N^x$ equally likely ways to draw $x$ tickets from $N$ with replacement, taking order into account. 
(We must count ordered outcomes 
because the unordered outcomes are not all equally likely. See Lesson \@ref(replacement).)

Next, we count the outcomes where the first $\fbox{1}$ is on the $x$th draw.
In order for this to happen, the first 
$x-1$ draws must all be $\fbox{0}$s and the $x$th draw a $\fbox{1}$:
\begin{equation}
\underbrace{\fbox{0}, \ldots, \fbox{0}}_{x-1}, \fbox{1}. 
(\#eq:geometric-outcome)
\end{equation}
The number of ways to do this is:
\begin{equation}
\underbrace{N_0 \cdot \ldots \cdot N_0}_{x-1} \cdot N_1, 
(\#eq:geometric-count)
\end{equation}
since we are drawing with replacement (so the composition of the box does not change from one draw to the next).
 
Dividing this by the total number of outcomes, we obtain the p.m.f.
\[ f(x) = P(X = x) = \frac{N_0^{x-1} \cdot N_1}{N^x}. \]

To see that this formula is the same as \@ref(eq:geometric2), we write $N^x = N^{x-1} \cdot N$. 
Then, we have:
\begin{align*}
f(x) &= \frac{N_0^{x-1} \cdot N_1}{N^{x-1} \cdot N} \\
&= \left( \frac{N_0}{N} \right)^{x-1} \frac{N_1}{N} \\
&= (1 - p)^{x-1} p,
\end{align*}
where in the last line, we used the fact that $\frac{N_0}{N} = \frac{N - N_1}{N} = 1 - \frac{N_1}{N} = 1 - p$.
```

### Visualizing the Distribution {-}

Let's graph the geometric distribution for different values of $N_1$ and $N_0$.

```{r geometric-pmf-1, echo=FALSE, fig.show = "hold", fig.align = "default", fig.asp=0.5}
library(latex2exp)

par(mfrow=c(1, 3))
n <- 10
N1 <- 5
N0 <- 15
plot(0:n, dgeom(0:n, 0.25), pch=19, xlab="x", ylab="f(x)", ylim=c(0, 0.75),
     main=TeX(paste("Geometric$(N_1=", N1, ", N_0=", N0, ")$")))
for(x in 0:n) {
  lines(c(x, x), c(0, dgeom(x, 0.25)), lwd=2)
} 

N1 <- 10
N0 <- 10
plot(0:n, dgeom(0:n, 0.5), pch=19, xlab="x", ylab="f(x)", ylim=c(0, 0.75),
     main=TeX(paste("Geometric$(N_1=", N1, ", N_0=", N0, ")$")))
for(x in 0:n) {
  lines(c(x, x), c(0, dgeom(x, 0.5)), lwd=2)
}

N1 <- 15
N0 <- 5
plot(0:n, dgeom(0:n, 0.75), pch=19, xlab="x", ylab="f(x)", ylim=c(0, 0.75),
     main=TeX(paste("Geometric$(N_1=", N1, ", N_0=", N0, ")$")))
for(x in 0:n) {
  lines(c(x, x), c(0, dgeom(x, 0.75)), lwd=2)
}
```

As the number of $\fbox{1}$s in the box increases, we tend to draw a $\fbox{1}$ sooner, so 
not surprisingly, smaller values of the geometric random variable become more probable.

## Essential Practice {-}

1. Carl is a bad driver, so each time he takes the driving test, he only has a $30\%$ chance 
of passing, independently of all previous attempts. If he takes the driving test over and over 
until he passes, what is the probability that he passes within his first 5 attempts?

2. Samuel Pepys was interested in the probability of getting at least 1 ⚅ in 
6 throws of a die. Previously, we saw how this probability could be calculated 
using the complement rule or the binomial distribution. Show how this probability 
can also be calculated by defining an appropriate geometric random variable.