Second, Paiva and Gil suggested that rounding errors in K_eq can be avoided by giving students 10 boxes of 40 matches, instead of just 40 matches. A much simpler solution to the problem of round-off errors would be to simply start with the "correct" number of matches. For example, as stated by Wilson, if the R-to-P transfer fraction (k_fwd) is ³/₄ and the P-to-R transfer fraction (k_rev) is ¹/₈, then K_eq should be 6. If the total number of matches between the two teams is 40, then solving for (40 - x)/x = 6, at equilibrium R (= x) must have 5.7 matches and P, 34.3. Clearly, without breaking matches into pieces, an accurate value for K_eq is not possible when starting with 40 matches. Starting with 42 matches, however (x= 6), will yield an accurate value of K_eq = 36/6 = 6.

The trick is to choose the total number of matches (#m_tot) such that each team, R and P, ends with an integral number of matches at equilibrium and the quotient P/R yields an accurate value for K_eq. Solving the algebraic expression for the equilibrium constant, (#m_tot - x)/x = K_eq, we get #m_tot = x(1+K_eq). In this expression, x is the number of matches held by team R at equilibrium. If x is chosen so that the product x(1+K_eq) = #m_tot is an integer, then round-off errors will cancel out, and an accurate value of K_eq will be obtained from the exercise.

For example, for Wilson's initial transfer rates of k_fwd = ¹/₂ and k_rev = ¹/₄, K_eq = 2; one would have to start with either 39 or 42 matches (x = 13 or 14, respectively) in order to end with an accurate value of K_eq. Similarly, for transfer rates of k_fwd = 0.88 and k_rev = 0.11, K_eq = 8.0; starting with either 36 or 45 matches (x = 4 or 5) will yield an accurate value of K_eq. Starting with 40 matches would yield K_eq = P/R = ³⁵/₅ = 7 instead of 8. Hence using the algebraic expression #m_tot = x(1 + K_eq) allows one to judiciously choose the initial number of matches so that for any values of k_fwd and k_rev, the equilibrium situation will yield an accurate value of K_eq.