1 Leftward Jump

Question

A frog began its journey on the first stone in a sequence of seven stones, numbered 1 through 7 from left to right. With each move, the frog hopped to a stone it hadn't already landed on. What is the probability that after touching all the stones, the frog made precisely one jump to the left during its hops?

QuantSoc · Accepted Answer

We begin by counting the total number of possible sequences the frog can follow:

$$
6! = 720
$$

To count the sequences with exactly one leftwards jump, consider splitting the set

$$
[2,3,4,5,6,7]
$$

into two subsets, $A$ and $B$.

The frog first visits the stones in $A$ in increasing order, and then visits the stones in $B$ in increasing order. This construction guarantees exactly one leftwards jump when transitioning from the last element of $A$ to the first element of $B$.

However, if $B$ is empty or if $B$ forms a suffix of

$$
[2,3,4,5,6,7]
$$

then no leftwards jump occurs. These cases must be excluded.

Since each element may independently belong to $B$, the total number of choices for $B$ is:

$$
2^6 = 64
$$

There are 7 invalid cases that produce no leftwards jump, so the number of valid sequences is:

$$
2^6 - 7 = 57
$$

Therefore, the required probability is:

$$
\frac{57}{720}
$$

Final Answer:

$$
\boxed{\frac{57}{720}}
$$