Since the sum of all tile numbers does not change in a merge, the only way for the sum to increase is by the random tile generation. Therefore, after any turn, the sum increases by 2 with a 90% chance or by 4 with a 10% chance.
Let P(n) be the probability that at some point in the game the total sum is exactly N = 2n and Q(n) = 1 − P(n) be the probability that the sum will never be equal N.
At the start of the game the sum is zero with a probability of 100%, i.e. P(0) = 1 and Q(0) = 0.
The only possibility that an even number N cannot be the sum of all tiles is that N−2 occured as a sum (with probability 1−Q(n−1) ) AND a 4 is generated (with probability ¹⁄₁₀). The probability is therefore Q(n) = (1−Q(n−1))/(10) which is the linear recurrence relation 10Q(n) + Q(n−1) = 1.
The general solution of this inhomogeneous linear recurrence relation (with a 1 on the right-hand side) is equal the general solution of the homogeneous version (with a 0 on the right-hand side) plus a particular solution of the inhomogeneous version.
The general solution of the homogeneous version 10Q(n) + Q(n−1) = 0 is Q(n) = a(−¹⁄₁₀)ⁿ with a being an arbitrary number.
A particular solution of the inhomogeneous version is Q(n) = ¹/₁₁ for all n because 10(¹/₁₁) + ¹/₁₁ = 1.
The constant a is determined from the case n=0: 0 = Q(0) = a(−¹⁄₁₀)⁰ + ¹/₁₁ = a + ¹/₁₁ and therefore a = −¹/₁₁,
Substituting a into the formula for Q gives Q(n) = ¹/₁₁ − ¹/₁₁(−¹⁄_{1⁄_{1⁄₁₀)ⁿ and therefore}}
P(n) = 1 − Q(n) = ( 10 + (−¹⁄₁₀)ⁿ )/(11). ■

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