Question:

Let the unknown probability that a basketball player makes a shot successfully be Θ. Suppose your prior on Θ is uniform on [0,1] and that she then makes two shots in a row. Assume that the outcomes of the two shots are independent.

What is the posterior density of Θ?

What would you estimate the probability that she makes a third shot to be?

My attempt:

If I let X be the event that she made two shots then I have the posterior as

f(Θ|X) = f(X,Θ)/f(X)

Given that making a shot is binomial,

f(X,Θ) = (n C x)*Θ^x*(1-Θ)^n-x

where (n C x) is the binomial coefficient, n choose x.

Then f(Θ|X) = f(X,Θ)/f(X)

If I integrate out the marginal distribution from f(X|Θ) over the interval [0,1], I get

f(X) = 1/(n+1)

Hence,

f(Θ|X) = [(n+1)*n!/x!(n-x)!]*Θ^x*(1-Θ)^n-x

= [Γ(n+2)/Γ(x+1)*Γ(n-x+1)]*Θ^x*(1-Θ)^n-x

which is beta(x-1,n-x+1)

Does this look correct?

Then for the second question, would I just take the expected value of the beta with these parameters? I.e. if y ~ beta(1,1), E[y] = 0.5?

Here, I got the parameters from the fact that we now need 1 success in 1 try.

Thanks in advance for any input you may have.