Stone, Paper, Scissors Game Winning Probability between two players in 1 match I am required to find winning probability and algorithm of winning a game between two players in the above mentioned game.  The catch is to find the winning stone, paper, scissor pattern so that 

The first player wins every time i.e. at least 2 times in a row in a given match. 

 A: Technically it doesn't matter what pattern you choose, because you don't know what the opponent will choose. The probability to win a single game is 0.5 therefore the match is symmetric and the probability to 0.5.
However, when playing in a real world against a real human there are psychological trics to manipulate your opponent to make then choose what you want. You can read more about that on Wikipedia.
A: Since for any choice you make your opponent has a winning choice, a losing choice and a draw and visa versa then assuming your opponent chooses randomly there is no optimum strategy. I did once write a computer program once though that would beat most humans more often than not in extended games by looking for patterns. People are not random.
The sequence can't be calculated in advance: If I know what move you are going to play I can always pick a move that will beat you.
What I did was record every move by the player and computer and try to find the longest matching sequence of past moves that matched the moves recently played to see what my opponent was likely to play next.  I then chose the winning move.  The program in effect learnt the player. At the start of the game the player and computer were winning roughly the same number of games but after a minute or so of playing the computer was able to guess the humans next move more often than not.  
When I tested it using against a random opponent me throwing a die (singular of dice) and selecting "Rock" for 1 or 2, "Paper" for 3 or 4 and "Scissors" for 5 or 6.  It again returned to the winner being more or less random.  
A: This is a matrix game with payoff matrix
$A = \left( \begin{array}{ccc}0 & -1 & 1\\1 & 0 & -1\\-1 & 1 & 0\end{array}\right)$. It's easy to see (using an equilibrium-finding program, etc.) that the pair $(x^*, y^*)$ is a Nash-equilibrium for the game, where $x^* = y^* = (1/3, 1/3, 1/3)$. $x^*$ is an optimal strategy for the row player, whilst $y$ is an optimal strategy for the column player. I'm not sure any other Nash equilibria exist. Also, the value of the game is $\langle x^*, Ay^*\rangle = 0$.
Thus the game is totally balanced, and so there is 50% probability for any one player to win a single match, assuming both players play optimally. In real life, an opponent can be psychologically confounded, rendering them thoroughly exploitable.
