Random Bernoulli n by n matrix is singular with probability (1/2+o(1))^n

You need to know: Matrix, determinant $\text{det}(M)$ of square matrix M, singular square matrix (matrix with determinant $0$ ), basic probability theory, independent and identically distributed (i.i.d.) random variables, $o(1)$ notation.

Background: Let $M_n$ be a random $n \times n$ matrix whose entries are i.i.d. Bernoulli random variables (each entry is equal to $+1$ or $-1$ with probabilities $1/2$ ). Let $P_n = P(\text{det}(M_n) = 0)$ be the probability that $M_n$ is singular.

The Theorem: On 21st December 2018, Konstantin Tikhomirov submitted to arxiv a paper in which he proved that $P_n = \left(\frac{1}{2}+o(1)\right)^n$ as $n\to\infty$ .

Short context: The question of estimating $P_n$ has attracted considerable attention in literature. There is an obvious lower bound $P_n \geq \left(\frac{1}{2}+o(1)\right)^n$ , and it has been conjectured by many authors that equality holds. For an upper bound, Komlós proved in 1967 that $\lim\limits_{n\to\infty} P_n = 0$ . In 1995, Kahn, Komlós, and Szemerédi proved that $P_n=O(0.999^n)$ . This was later improved by Tao and Vu to $P_n=O(0.958^n)$ and then to $P_n \leq \left(\frac{3}{4}+o(1)\right)^n$ , see here. In 2010, Bourgain, Vu and Wood improved it to $P_n \leq \left(\frac{1}{\sqrt{2}}+o(1)\right)^n$ . Finally, the Theorem establishes the upper bound $P_n \leq \left(\frac{1}{2}+o(1)\right)^n$ , which matches the lower bound up to $o(1)$ term and proves the conjecture.