The chance of winning is 1/13 and of losing is 12/13.
If they play three games, the chances of winning two out of the three are 36/2197, or about 1 in 61.
This comes from 3 x (12/13) x (1/13) x (1/13), ie 3 x P(Lose) x P (Win) x P (Win).
The factor of 3 is because the losing game could be first, second or third in the sequence.
If, however, you discount the case where the first two games were won, so the third game would never be played, then the chance is only 2 x (12/13) x (1/13) x (1/13), or about 1 in 91.5.
I think.
Someone will correct me if I am wrong.