Suppose I am betting on binary outcomes, lets say randomly drawing a red ball from an urn with only red and yellow balls That is, I say either "red" or "yellow", then a ball is drawn from the urn, and I win the bet if I am correct.
If I choose to say "red" or "yellow" on the basis of flipping a fair coin will I be correct half the time?
Assume the coin is independent of the distribution of balls in the urn, as well as of the process of choosing a ball from the urn (because random and Its weird to have a coin flip related to this)
My attempt at this is as follows. (In what follows suppose I guess red when the coin is heads)
Suppose there are $N$ balls in the urn, $K$ of which are red. Then the probability of a red ball being drawn is $\frac{K}{N}$.
Define a random variable $X$ such that $X=1$ if I win the bet (I guess correctly) and $X=0$ if I guess incorrectly. Then the question reduces to finding the expected value of $X$ (when the probability that $X=1$ is calculated according to the betting strategy).
The expected value of $X$ is $$1\cdot P(X=1) = P(X=1) = \\.5P(Red\vert Heads) + .5P(Yellow\vert Tails) = .5P(RED)+.5(P(Tails) = .5$$
So I would be correct half the time.
Could someone tell me how I could formally derive the probability of being correct, i.e. $P(X=1)$. That is, how can I mathematically show that $$P(X=1) = .5P(Red\vert Heads) + .5P(Yellow\vert Tails)$$
I guess I would need to define a random variable for the outcome of the urn, and a random variable for the outcome of the coin flip, find the joint distribution of these random variables, and then $P(Correct)=P(Red,Heads) + P(Yellow, tails)$?
By independence though I guess $P(Red,Heads) =P(RED)\cdot P(Heads)) and similar for tails, which gives the formula.
You will win with probability $\dfrac KN$ half of the time and win with probability $\dfrac{N-K}N$ the other half. On average, $\dfrac12$.
Said differently, whatever the outcome, you will choose this outcome half of the time.