Pascals first method So pascals first method was to first solve a simple problem,this was before the pascal triangle. 
This is in relation to De Meres problem: 
Each player stakes $32$ pistoles. One player has 1 round the other player has none. $3$ rounds are needed to win. How do you divide if the game is interrupted before completion? 
This is how my professor solved it, im sort of confused where he got some of the values from. 
$$(2,0) \implies 1/2 (64) + 1/2 (48) =56 $$
Player $1$ gets $56$ and player $2$ gets $8$ so the ratio is $7:1$.
The $(2,0)$ stands for how many games each player won. Im trying to figure out how he got the $64$ and $48$. Any help is appreciated.
Thank you.
 A: Your professor uses the law of total expectation, where he conditions on the result of one more round. 
In order to do this, you first need to have solved the case where the first player has won two rounds and the second player one round. If the players were to play one more round, then either player one would win the entire stake of $64$ by winning the third round, or the players would be tied if player one lost the third round. In the last case it would be fair to share the stake evenly, with $32$ pistoles to each. Computing the expected value for player one, we get
$$ (2,1) \quad \Longrightarrow \quad {1\over 2}\cdot 64 + {1\over 2}\cdot 32 = 48$$
We can now consider the case where the first player has won two rounds and the second player none. Imagine that they play one more round. Either player one wins, and gets the entire stake of $64$, or he loses and we are in the situation above where his fair share is $48$. So
$$ (2,0) \quad \Longrightarrow \quad {1\over 2}\cdot 64 + {1\over 2}\cdot 48 = 56$$
