6  To Be Fair, It Must Be Biased

The return probability has the form 1 - 1/(Underoverscript[∑, n = 0, arg3] p(n, ϵ)) where p(n, ε) is the probability of return to the origin after 2n steps; this can be written in terms of binomial coefficients as p (n, ϵ) = 1/16^n (2  n) Underoverscript[∑, k = 0, arg3] (n)^2 (1 - 16&n ... sp;ϵ^2)^k n k. The sum can be estimated by using a partial sum or, for more digits, using an extrapolation technique to estimate the tail. Then standard root-finding techniques tell when the expression is 2. The answer is
0.06191395447399094284817521647321217699963877499836207606146725885993101029759615845907105645752087861.

There is also a closed form for this sum in terms of elliptic functions. Several people observed this. Here is how to use the elliptic function.

RowBox[{ϵ, /., RowBox[{FindRoot, [, RowBox[{( π   (2 + 16 ϵ^2 - 2 ( ... ^2)] 2, ,, RowBox[{{, RowBox[{ϵ, ,, 0.1}], }}], ,, , AccuracyGoal8}], ]}]}]

0.06191395447

And there is a well-known connection between this elliptic function and the very-fast-to-compute arithmetic-geometric mean. Thus, as observed by Bornemann, it is faster and more elegant to use the following, which gives the same result, and gives 500 digits in under a second.

RowBox[{RowBox[{ϵ, , /., , RowBox[{FindRoot, [, , RowBox[{2^(1/2) ArithmeticGe ... owBox[{{, RowBox[{ϵ, ,, 0.1}], }}], ,, , WorkingPrecision500}], ]}]}], //, Timing}]

RowBox[{{, RowBox[{RowBox[{0.766667, , Second}], ,, 0.061913954473990942848175216473212176999 ... 2665130410769275896429379457358898700313736928994949809378129095499896576724082600428539419}], }}]

Created by Mathematica  (June 27, 2004)