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Random Walk II


Alignments to Content Standards: F-IF.A.2

Task

Imagine Scott stood at zero on a life-sized number line. His friend flipped a coin 50 times. When the coin came up heads, he moved one unit to the right. When the coin came up tails, he moved one unit to the left. After each flip of the coin, Scott's friend recorded his position on the number line. Let $f$ assign to the whole number $n$, when $1 \leq n \leq 50$, Scott's position on the number line after the $n^{\rm th}$ coin flip.

  1. If $f(6) = 6$ what can you conclude about the outcomes of the first $6$ coin tosses? Explain. What if $f(6) = -4$?
  2. Is it possible that $f(7) = 2$? Explain.
  3. Find all integers $m$ so that the probability that $f(50) = m$ is zero.

IM Commentary

This task follows up on ''The Random Walk,'' looking in closer detail at what outcomes are possible. These problems form a bridge between work on functions and work on probability. In ''The Random Walk III'' we will return to similar scenarios and calculate probabilities. The task is better suited for instruction than for assessment as it provides students with a non standard setting in which to interpret the meaning of functions. Students should carry out the process of flipping a coin and modeling this Random Walk in order to develop a sense of the process before analyzing it mathematically.

Solution

  1. If $f(6) = 6$ this means that Scott has moved to the right on all six coin tosses so all six tosses must have come up heads.

    If $f(6) = -4$ then Scott must have moved five steps to the left and one to the right so the outcomes of the first six coin tosses were $5$ tails and $1$ heads.

  2. To find what values $f(7)$ might take, suppose that there are $t$ tails and $h$ heads in the first $7$ coin tosses. We have $h + t = 7$. Scott will have moved $h$ steps to the right and $t$ steps to the left so we will be standing on $h - t$. Substituting $t = 7 - h$ we find $$ h - t = h - (7-h) = 2h - 7. $$ Since $2h$ is an even number, $2h - 7$ is an odd number and so it is not possible that $f(7) = 2$.

    Alternatively, we can reason directly that landing on $2$ after $7$ tosses is not possible. We know that $h + t = 7$ after seven tosses. Landing on $2$ after the seven tosses would give us the additional equation $h - t = 2$. Solving these two equations in two unknowns gives $h = \frac{9}{2}$ and $t = \frac{5}{2}$ which is not possible since the number of heads (and also the number of tails tails) must be a whole number.

  3. Since Scott is taking a total of $50$ steps, he can not pass $+50$ to the right or $-50$ to the left. Between $+50$ and $-50$ the argument from part (b) will show that only even numbers are possible. To see this, if $h$ is the number of heads and $t$ the number of tails then Scott will be standing on $h - t$ after the $50$ tosses. We have $h + t = 50$ since there are $50$ tosses. Substituting $t = 50 - h$ shows that Scott will be standing on $h - (50-h) = 2h - 50$ after the $50$ coin tosses. Since $2h$ and $-50$ are even numbers this shows that Scott is on an even number.

    So far, we have shown that Scott must end on an even number between $-50$ and $50$. We have seen that with $h$ heads Scott ends up on $2h - 50$. When $h$ takes the values $0,1, \ldots, 50$, $2h-50$ takes the values $$ -50, -48, \ldots, 50 $$ or, in other words, all even numbers between $-50$ and $50$. So $P(m) = 0$ for any odd integer $m$ and also any even integer $m$ whose absolute value is larger than $50$.