This link: https://www.youtube.com/watch?v=5UQU1oBpAic, points to a youtube video that presents the following problem:
suppose you have a chess match against an inferior opponent where 3/4 of the games end in a draw, and for the non-drawn games, you win 2/3 of the time and lose 1/3 of the time (i.e. the chance of a draw/win/loss for you in each game is (3/4, 1/6, 1/12).
The winner of the match is the first player to win two games in a row. What are your chances of winning the match?
The youtube video presents what seems to be a sensible approach that gives your chances of winning as 26/33. However, my intuition suggests that the answer is 4/5, based on the following reasoning.
The moment the match ends, either you or your opponent has just won the last two games. All game results prior to the last two games are irrelevant. Your chances of having won the last two games, given that neither game was drawn is (2/3 x 2/3) = 4/9, while the chances that your opponent won these games is (1/3 x 1/3) = 1/9. Therefore, you are 4 times as likely to have won these games as your opponent. Therefore, your chances of winning are 4/5.
Based on the youtube video, my above reasoning must be wrong. What is my mistake?
Possible explanation
I am speculating here: perhaps my mistake is in assuming that all previous results before the last two games are irrelevant.
Your argument would be valid if you were 4 times more likely to win after any number of games.
You are 4 times as likely as your opponent to win after the second game, but less than 4 times as likely to win after the third game because to win in 3 requires that you don't win the first game, and that probability is different for the two players Y and O
$$ \frac{P(Y_3) }{P(O_3)}= \frac{\frac 56 \times \frac 16 \times \frac 16 }{ \frac{11}{12} \times \frac{1}{12} \times \frac{1}{12} } =\frac{40}{11}$$