This year, Daniel Litt, a mathematician at the University of Toronto, has distracted and delighted social media users with simple-sounding but deceptively difficult probability problems. In a conversation with Quanta, Litt discusses what makes a great puzzle, and why questions about chance can throw people off. https://lnkd.in/e3K7nm-v
Tim Caber
2mo
The answer follows from Laplace’s rule of succession, which can be derived as the expected value of the posterior distribution for Bernoulli trials, using a Beta(1,1) prior on the probability of red versus green balls (i.e., it’s just like the coin flip problem).
Researcher (Quantum Information, Condensed Matter, Data Science)
2moCool article! I believe this is yet another example of generic counterintuitive nature of conditional probabilities. But it is simple if we think through standard terminology and math. The standard way of thinking about the problem would be to consider the conditional probability of finding the second ball red, given that the first one was red, without any knowledge of the number of red or green balls. Therefore, the standard brute-force solution involves finding the joint probability of both balls being red and dividing it by the probability of first being red. Given N balls, the instances where we have m red balls for different values of m=0,1,…,N are equally likely. For any m, the probability P_m that the first ball is red is m/N, and the probability of both balls in a row being red, P_m (both balls are red) is m(m−1)/[N(N−1)]. To obtain the overall probabilities, we need to average over m for both of these probabilities. This gives P(first ball is red)=(1/2)(N+1)/N and P(both balls are red)=(1/3)(N+1)/N, which finally results in the conditional probability P(second ball is red|first ball is red)=2/3.