Season 2 Episode "Mind Games"Statistics / Probability

NUMB3RS - Season 2 - "Mind Games" - Walkabout

Workbook Guide · Step 1 of 5

Mapping Random Walks to Pascal's Triangle

Imagine starting at position $0$ on a vertical number line. At each step, you flip a coin: heads you move up 1 unit ( $U$ ), tails you move down 1 unit ( $D$ ).

Let's construct a table of final positions for short walks.

For a 0-step walk, there is only $1$ way to end at $0$ .
For a 1-step walk, there is $1$ way to end at $U1$ and $1$ way to end at $D1$ .
For a 2-step walk, the paths are $UU$ (ends at $U2$ ), $UD$ (ends at $0$ ), $DU$ (ends at $0$ ), and $DD$ (ends at $D2$ ). This gives us $1$ way to end at $D2$ , $2$ ways to end at $0$ , and $1$ way to end at $U2$ .

Notice that these counts ( $1, 2, 1$ ) match the rows of Pascal's Triangle! Continuing this pattern, let's look at a 4-step walk.

QFor a 4-step random walk, how many different paths will end exactly back at the starting position $0$ ?

Real-Time Stochastic Simulator

1D Lattice Path

Number of steps (N): 10

Probability of Up step (U): 0.50

Pi Approximation Diagnostics

Total Runs0

Avg distance E(d)0.000

Estimated Pi0.000

Why does this approximate

\pi

As $N \to \infty$ , the binomial distribution is approximated by the continuous Gaussian normal distribution. The integration of its absolute expected distance returns the constant $\sqrt{\frac{2}{\pi}}$ , hence establishing $\pi \approx \frac{2N}{[E(d)]^2}$ !