Probability matching is the strategy of betting on different events proportionally to the probability of them happening.
In standard decision theory, that's quite a bit worse than just betting all your money on the most likely event. (Assuming linear utility of money.)
However, when the utility of the good bet on is not linear, this can make sense. If there is enough food for a population, split over two locations, it makes sense for the population to split matching the ratios.
In Nassim Talebs Antifragility the Author writes:
The idea of comparative advantage has an analog in probability: if you sample from an urn (with replacement) and get a black ball 60 percent of the time, and a white one 40 percent, the optimal strategy, according to textbooks, is to bet 100 percent of the time on black. The strategy of 60 percent of the time on black and 40 percent on white is called "probability matching" and considered to be an error in the decision-science literature. (...) People's instinct to engage in probability matching appears to be sound, not a mistake. In nature, probabilities are unstable (or unknown), and probability matching is similar to redundancy, as a buffer. So if the probabilites change, in other words, if there is another layer of randomness, then the optimal strategy is probability matching.
I have tried some simulations using an urn-model as described, but with stochastically fluctuating $p$ (adding some uniform noise to $p$ after every draw), and in all my simulations, probability matching still performed worse.
So is the quote nonsense, or are there actual precise conditions making probability matching optimal in the urn model, or at least in a more sophisticated model (not invoking non-linear utilities)? If the quote is sound, but there are no such precise conditions (or they are not feasible to formulate for you), what is the intuition behind it?
Probability matching is equivalent to randomising between two options: it can be optimal only when the agent is indifferent between them.
Consider the example above. If you bet on black, you win with probability $p_B=0.6$ and $E(B)=0.6$; if you bet on white, you win with probability $p_W=0.4$ and $E(W)=0.4$. Then betting on black is (on average) better than betting on white.
If you do probability matching and choose Black with probability $p_B$ and White with probability $p_W$, you win when you choose Black and the ball is Black (probability $p_B^2$) or when you choose White and the ball is White (probability $p_W^2$). The overall win probability is $$p_B^2+p_W^2 = p_B^2+(1-p_B)^2 \le p_B$$
When the win probability for Black is random, with expected value $E(P_B)$, replace $p_B$ with $E(P_B)$ and $p_W$ with $1-E(P_B)$ to conclude again $$E(P_B^2+P_W^2)= E(P_B^2)+E(1-P_B)^2 \le \max \{ E(P_B), E(P_W)\}$$ Then all-out betting on the color with the highest $E(P)$ is (weakly) preferable to probability matching.
A different problem is to assume that $P_B$ is not known and the agent plays repeatedly, updating the estimate for $P_B$ after each round of play. If the estimate flips and White looks better than Black, then the agent may switch to the different color. But all-out betting remains optimal.