Question
I have two blue dice, with which I play a game. If I throw a double six (i.e. if I get two six on both the dices) then I win the game. I separately throw a red die. If I get a one, then I tell truth about whether I win/loose in the previous game, otherwise I lie. I just rolled the three dice. I turn around to you and say, "I won!". What is the probability that I actually won the game?
My Approach
For red die, if it is truth then probability of winning the game is
$\frac{1}{6} \times \frac{1}{36}$
If I lie, then the probability of winning the game is
$\frac{5}{6} \times \frac{1}{36}$
So required probability =$\frac{1}{6} \times \frac{1}{36}+\frac{5}{6} \times \frac{1}{36}=\frac{1}{36}$
Am I right?
You got this wrong, since you will say that you've won if either you threw a $1$ with the red die and won, or didn't throw a $1$ with the red die and lost. Using Bayes' theorem, we find, with $W$ the event in which you win and $T$ the event in which you tell that you've won:
$$P(W | T) = \frac{P(W \cap T)}{P(T)} = \frac{P(W \cap T)}{P(W \cap T) + P(\lnot W \cap T)} = \frac{\frac{1}{36} \frac{1}{6}}{\frac{1}{36} \frac{1}{6} + \frac{35}{36} \frac{5}{6}} = \frac{1}{176}$$