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 walks counterclockwise along the edge of the hexagon until reaching the next nearest vertex. If the coin lands tails, he walks clockwise around the hexagon until reaching another vertex. Once there, he repeats the process. The man flips the coin a total of six times. What is the probability that the man is standing where he started when he is finished?

Guest Apr 18, 2015

#2**+10 **

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.

Melody
Apr 19, 2015

#1**+10 **

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).

.

Alan
Apr 18, 2015

#2**+10 **

Best Answer

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.

Melody
Apr 19, 2015