“How rational are you?”
Are there degrees of rationality?
Y is the random variable associated with you picking the winning color. It is equal to 1 with probability p and 0 with probability 1-p (bernoulli distribution). If p itself is a random variable X as in the second case, then the conditional expectation and conditional variance of Y are as follows:
E[Y|X]=X, Var[Y|X]=X(1-X)
By the law of total expectation, we have:
E[Y]=E[E[Y|X]]=E[X]
By the law of total variance, we have:
Var[Y]=E[Var[Y|X]]+Var[E[Y|X]]=E[X(1-X)]+Var[X]=E[X]-E[X^2]+E[X^2]-E[X]^2=E[X]*(1-E[X])
For a risk-averse player, use some sort of Sharpe-like ratio to measure which choice is more attractive:
ratio = E[Y]/Var[Y]=E[X]/{E[X]*(1-E[X])=1/(1-E[X])
so as long as the you don’t have any knowledge that certain color will be favored, i.e. X is a symmetrical distribution, then E[X]=1/2, and your ratio is 1/(1-1/2)=2, regardless of what the variance of X is. You should be indifferent between case 1 (X=1/2=constant, VarX=0) and case 2 (say X is uniform, or normal, or any distribution centered around 1/2 with positive variance Var[X]>0).