In order for "Sn" to be equal to a number "k" it is necessary and sufficient that the number of +1 in the walk exceeds those of −1 by "k". It follows +1 must appear ("n" + "k")/2 times among "n" steps of a walk, hence the number of walks which satisfy formula_20 equals the number of ways of choosing ("n" + "k")/2 elements from an "n" element set, denoted formula_21. For this to have meaning, it is necessary that "n" + "k" be an even number, which implies "n" and "k" are either both even or both odd. Therefore, the probability that formula_20 is equal to formula_23. By representing entries of Pascal's triangle in terms of factorials and using Stirling's formula, one can obtain good estimates for these probabilities for large values of formula_24.