A large, regular hexagon is drawn on the ground, and a man stands at one of the vertices. The man flips a coin. If the coin lands heads, he

Thanks Alan, I did it exactly the same way BUT

can you see any way of getting the total number of paths possible?

I suppose you could do it backwards.

Any individual path has the probability of   $${\frac{{\mathtt{1}}}{{{\mathtt{2}}}^{{\mathtt{6}}}}}$$      

So

$${\frac{{\mathtt{1}}}{{{\mathtt{2}}}^{{\mathtt{6}}}}}{\mathtt{\,\times\,}}{\mathtt{x}} = {\mathtt{1}} \Rightarrow {\mathtt{x}} = {\mathtt{64}}$$

So there must be 64 possible paths taken, 22 of these are favourable outcomes

so   P(ending at startpoint) =      $${\frac{{\mathtt{22}}}{{\mathtt{64}}}} = {\frac{{\mathtt{11}}}{{\mathtt{32}}}} = {\mathtt{0.343\: \!75}}$$

This is interesting -

I cannot remember ever using probability backwards like this to work out the number of possible outcomes.  

There are 22 possible ways of ending up where he started after 6 steps.  11 of them where his first step is in a clockwise direction and another 11 where his first step is in an anti-clockwise direction.  The direction of each step is chosen with probability 1/2, so overall probability = 22*(1/2)⁶ ≈ 0.34375

If we number the vertices as 0 1 2 3 4 5, where 0 is the initial vertex, then the following list shows the 11 routes from 0 back to 0 in which the first step goes to vertex 1 (the other 11 are obtained by interchanging the 1s and 5s and interchanging the 2s and 4s).

.