There are 2014 houses in a circle. Let $A$ be one of these houses. Santa Claus enters house $A$ and leaves a gift. Then with probability $1/2$ he visits $A$'s left neighbor and with probability $1/2$ he visits $A$'s right neighbor. He leaves a gift also in that second house, and then repeats the procedure (visits with probability $1/2$ either of the neighbors, leaves a gift, etc). Santa finishes as soon as every house has received at least one gift. Prove that any house $B$ different from $A$ has a probability of $1/2013$ of being the last house receiving a gift.