conditional probability: game with three different bags, balls in bags

We play a game with three bags of balls, where the game is best of five, and one wins when they select three winning balls. A bag is selected at the beginning of a game, and it does not change between selections within games. There are three bags described as follows, and selecting each is equally likely.

• The "lucky" bag, where selecting a winning ball has a 3/4 chance.

• The "impartial" bag, where selecting a winning ball has a 1/2 chance.

• The "unlucky" bag, where selecting a winning ball has a 1/4 chance.

Answer the following questions:

• What is the probability that the game ends in the first three consecutive draws?

This seemed to mean you select three winning balls in a row or three losing balls. By the law of total probability:$$P(\text{Game ends in 3 draws}) = \frac{1}{3}(\frac{3^3}{4^3} + \frac{1^3}{2^3} + \frac{1^3}{4^3}) + \frac{1}{3}(\frac{1^3}{4^3} + \frac{1^3}{2^3} + \frac{3^3}{4^3}) = \frac{3}{8}$$

• What is the probability that the you lose the game?

This seems to be the probability that you draw exactly three losing balls (regardless of the game ending before five balls are drawn, but this is where I'm confused). Again by the law of total probability:

$$P(\text{Lose the game}) = \frac{1}{3}(\frac{1^3}{4^3} + \frac{1^3}{2^3} + \frac{3^3}{4^3}) = \frac{3}{16}$$

• What is the probability that you selected one winning ball and three losing balls?

By the law of total probability:

$$P(\text{1 winning ball, 3 losing balls}) = \frac{1}{3}(\frac{1^3}{4^3}\cdot\frac{3}{4} + \frac{1^4}{2^4} + \frac{3^3}{4^3}\cdot\frac{1}{4})$$

• Given that you selected one winning ball and three losing balls, what is the probability that you were drawing from the "lucky" bag?

By the definition/formula for conditional probability, seems like this could be $$\frac{1}{4^3}\frac{3}{4}$$ divided by the answer to the part right above (i.e. the probability that we draw one winning ball and three losing ones).

• Not sure the set up is clear. I choose the bag (randomly) once, and that's it? When you say the outcomes of ball selections are independent...I take it you mean, independent conditioned on which bag you got, right? I mean...drawing a string of winners would be strong evidence that you had got the lucky bag, hence evidence that the next one would also be a winner. – lulu Sep 15 '18 at 22:52
• @lulu yes, I believe that is what is meant. We choose the bag at the beginning of the game and are stuck with the bag for that game. I hope that also answers your second question :) – 0k33 Sep 15 '18 at 22:53
• Ok, given that I'd say your first computation looks sensible. Second doesn't...if you imagine you draw all five times, regardless of whether the game is done or not, then you lose if you get $3,4$ or $5$ losers. Not "exactly $3$". – lulu Sep 15 '18 at 22:55
• As a useful (and fairly common) trick: Note that the situation is entirely symmetric. Thus, there is no bias between "winning" and "losing" so... – lulu Sep 15 '18 at 22:57
• Sure, but it is computationally useful to imagine that all five are drawn. Simplifies the algebra considerably...doesn't change the winner (since it is impossible to get both three winners and three losers out of five draws). – lulu Sep 15 '18 at 22:58