www.lesswrong.com/posts/MsjWPWjAerDtiQ3Do/on-the-independence-axiom

In the standard Ellsberg setup, the known-probability bet is the bet on red, and its win probability is 30/90 = 1/3. If you put a uniform prior over the 61 possible numbers of black balls among the 60 non-red balls, the expected number of black balls is 30, so the expected probability of drawing black from the full urn is 30/90 = 1/3 as well.

The post's 30/60 = 1/2 calculation is the expected share of black among the 60 non-red balls, not the probability of drawing a black ball from the 90-ball urn. So it does not “match the known-probability case” as stated.

Put differently:

• expected black count among the ambiguous 60 balls = 30;
• expected black fraction among those 60 balls = 30/60 = 1/2;
• expected probability of drawing black from the urn = 30/90 = 1/3.

The article's sentence conflates the middle quantity with the last one.

Using the formula stated immediately above this claim,

g(p) = p·log(1+f) + (1-p)·log(1-f),

we can simplify:

g(p) = log(1-f) + p·(log(1+f) - log(1-f)).

That is an affine (linear) function of p, not a concave one. Its second derivative with respect to p is 0. For affine functions, Jensen's inequality gives equality, not a strict inequality:

E[g(p)] = g(E[p]).

So the post's mathematical step is incorrect as written. The strict inequality claimed here does not follow from the Kelly/log-growth formula it gives, and the subsequent claim that averaging over urn compositions makes the average growth rate strictly lower than the known-probability case is unsupported by this argument.

This sentence compounds two errors.

First, for the log-growth formula the post itself gives,

g(p) = p·log(1+f) + (1-p)·log(1-f),

g is affine in p, so averaging over urn compositions yields

E[g(p)] = g(E[p]),

not a strictly smaller value.

Second, in the classic Ellsberg urn the known-probability benchmark is the bet on red, whose probability is 1/3 (30 red balls out of 90 total), not 1/2. A uniform prior over the number of black balls gives an expected black count of 30, which corresponds to an overall draw probability of 30/90 = 1/3.

So the statement is incorrect both about the inequality and about the benchmark probability.