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Running an exploratory R code like
o=rep(0,3) for(n in 1:1e6){ b=c(0,0,1) for(t in 1:N){ i=sample((1:3)[!b],1);b=0*b;b[i]=1} o=o+b}
shows that the difference in probability is between the rightmost position and both others, starting at zero, and evolving as p⁺=(1-p⁻)/2, with the successive values 0,1/2,1/4,3/8,5/15,11/32,… Very quickly converging to 1/3.
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