What is the probability that I actually won the game

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?

• The arithmetic in the last line should yield $1/36$, and indeed this is the probability of winning the game. (You didn't need to consider the red die to compute this.) However, the problem is asking for the probability of winning the game given that you say that you won which is a different quantity. Try writing down the definition of this conditional probability and think about what quantities you need to compute. Bayes's rule may be helpful. – angryavian Feb 18 at 20:58

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:
$$P(W | T) = \frac{P(W \cap T)}{P(T)} = \frac{P(W \cap T)}{P(W \cap T) + P(\lnot W \cap T)} = \frac{\frac{1}{36} \frac{1}{6}}{\frac{1}{36} \frac{1}{6} + \frac{35}{36} \frac{5}{6}} = \frac{1}{176}$$