This balls in the urn problem is relatively straightforward with Bayesian Updating. We have a uniform prior for the number of red balls that were placed in the

This balls in the urn problem is relatively straightforward with Bayesian Updating. We have a uniform prior for the number of red balls that were placed in the Urn, P(U) = 1/101 where U ∈ [0, 100]. We then have the probability the first ball was red given a value of U, P(B1=Red|U) = U/100. Lastly we have the P(B1=Red) = ∑ P(B1=Red|U)P(U), where the sum is over all values of U. But we don't actually need to compute this sum because from symmetry arguments we can see that P(B1=Red) = 1/2. This follows because there are the same number of configurations with U red balls as U green balls and each configuration is equally likely. So now we can compute P(U|B1=Red) using Bayes Rule. P(U|B1=Red) = P(B1=Red|U)P(U)/P(B1=Red) = (U/100)*(2/101) But how does that help us? Well, after we remove a red ball the probability the next ball will be red given the first was red and there were U red balls initially in the urn is: P(B2=red|B1=red,U) = (U-1)/99. From this we see that if U was in the range of 51 to 100, then P(B2=red|B1=red,U) > 1/2, i.e., the second ball is more likely to be Red. So let's compute the following where we are summing over U ∈ [51,100]: P(U ∈ [51,100]|B1=Red) = ∑ P(B1=Red|U)P(U)/P(B1=Red) = ∑ (U/100)*(2/101) = (2/101) ∑ U/100 = (2/101) (50/100) * 75.5 P(U ∈ [51,100]|B1=Red) = 75.5/101 So we find that P(U ∈ [51,100]|B1=Red) = 75.5/101 ≈ 3/4 which means that given the first ball we picked was red, then roughly 3/4 of the time we would expect the next ball being red to be more likely. This seems like a much more straightforward way to think about it for me.