Solution #35

The point (2,2) can be reached in either 4 or 6 steps. If it is reached in 4 steps, those steps must be R, R, U, U, in some order. There are 4C2 = 6 ways of placing the two "R" moves, and hence 6 ways of producing the required four-step sequence. There are 4⁴ possible four-step sequences, so there is a 6/(4⁴) probability that the bug reaches (2,2) after four steps.

To reach (2,2) in six steps, the steps must include R, R, U, U, and then must also include a "cancelling pair" of the form R, L or U, D. We consider the first case; the second will be equivalent.
There are (6C3)(3C2) ways of arranging R, R, R, L, U, U, for a total of 60 ways. However, some of these ways will have the bug reach (2,2) in four steps, and must be eliminated. As we saw above, there are 6 ways to arrange R, R, U, U, and each such arrangement can then be completed either R, L, or L, R. Thus 12 arrangements must be eliminated, leaving 48.

Arrangements of R, R, U, U, U, D will produce another 48 ways, for a total of 96. There are 4⁶ arrangements of six steps, so there is a 96/(4⁶) probability that the bug reaches (2,2) in exactly six steps.

The final probability is thus 6/(4⁴) + 96/(4⁶) = 3/64.