Expected Number of Moves to Fall Off Table

There is a 20cm long table. You start 3 cm from the left. Each move to the left or to the right is 10 cm with equal probability. If you reach the edge of the table you fall. In how many moves do you expect to fall?

Let E represent the expected number of moves to fall.

$\frac{1}{2}$ chance we move to the left and fall the first time, in which case 1 move is used.

$\frac{1}{2}$ chance we move to the right, in which case there are 2 equally likely outcomes. We can move to the left, in which we are back to square one, or we can move to the right again in which case we fall.

3 distinct sequences - L(50%), RL(25%), RR(25%). L is 1 move and RR is 2. However, there are only 2 distinct ways to fall, as RL returns us to our original position, so RL entails 2 + E moves until we fall(recursive expectation).

The scenario is modeled by E = $\frac{1}{2}$(1) + $\frac{1}{4}$(E + 2) + $\frac{1}{4}$(2). Solving for E, we get 2. Is this correct, and if so, are there other approaches? If not, could someone please delineate what errors were made?

Solution 1:

The shortcut is to realize that if you don't fall off the table after your 1st move, you are (by symmetry) in the same position. That is, being $7$ cm from an edge is no different than being $3$ cm from an edge, given that the moves are $\pm 10$ cm.

Therefore $\displaystyle ~E = \frac{1}{2} + \frac{1}{2}(E + 1) \implies$

$\displaystyle 2E = 1 + (E + 1) = 2 + E \implies E = 2.$