A series of \(2014\) games is being played between two players \(A\) and \(B\). There are no ties-- exactly one of the players wins in a game and the other loses. The first game is won by \(A\), and \(B\) wins the second game. In the consequent games, the probability of \(A\) winning is equal to \(\dfrac{P}{P+Q}\), where \(P\) and \(Q\) denote the number of games won by \(A\) and \(B\) respectively so far (for example, the probability of \(A\) winning the third game is \(\dfrac{1}{2},\) if \(A\) wins the third game, the probability of \(A\) winning the fourth game is \(\dfrac{2}{3},\) etc). The probability that the series is tied, i.e. \(A\) and \(B\) both win exactly \(\dfrac{2014}{2}\) games., can be expressed as \(\dfrac{a}{b},\) where \(a\) and \(b\) are coprime positive integers. Find \(a+b+1\).

**Details and assumptions**

- This problem is not original.

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