Probability of winning after adding coin condition Assume that we have two games $X, Y$ with probability of winning $P(X)$ and $P(Y)$ respectively. Now I create a new game $Z$: I throw a coin and if I get heads I play $X$. Otherwise I play $Y$. What is the probability of winning in $Z$? Do we need more info on $X, Y$ in order to solve this? This might be a newbie question, I'm just so bad at probability.

Let $P(Z)$ be the probability of winning game $Z$. Then we have that $P(Z) = \frac{1}{2}P(X) + \frac{1}{2}P(Y)$. If you draw a probability tree you can visualise why this is true. If you flip a coin - you either get heads or tails, each with probability $\frac{1}{2}$. Then if you get heads, the probability of winning is $P(X)$, and if you get tails the probability of winning is $P(Y)$. Since the coin flip, and $P(X), P(Y)$ are independent events, we can therefore find the probability using the product of the branches in the probability tree. 
