Suppose you test Fermat’s Last Theorem for n up to 10^10, and don’t find a counterexample. How much evidence does that give you for FLT being true? In other words, how do you compute P(a counterexample exists with n<=10^10 | FLT is false), since that’s what’s needed to do a Bayesian update with this inductive evidence? (Assume this is before the proof was found.)
I don’t dispute that mathematicians do seem to reason in ways that are similar to using probabilities, but I’d like to know where these “probabilities” are coming from and whether the reasoning process really is isomorphic to probability theory. What you call “heuristic” and “intuition” are the results of computations being done by the brains of mathematicians, and it would be nice to know what the algorithms are (or should be), but we don’t have them even in an idealized form.
The most influential books on the topic I know of are Mathematics and Plausible Reasoning Volume I: Induction and Analogy in Mathematics, and Mathematics and Plausible Reasoning Volume II: Patterns of Plausible Reasoning (amazon link) by George Pólya.
There’s a difficulty here involving the fact that every finite set of possible counterexamples has measure zero in the set {(a, b, c, n) | a, b, c, n in N} equipped with the probability measure that assigns each possible counterexample an equal probability.
So all the usual biases and cognitive gotchas involving probability and infinity come into play (even when the people doing the thinking are mathematicians!), and all bets are off.
My hypothesis is that the commonly used prior for counterexample distributions is exponential. As the lower bound K >= a, b, c, n on possible counterexamples increases, the exponential is updated into something close to uniform on the rest of {(a, b, c, n) | a, b, c, n in N}.
This does somewhat dodge the question, but it does make a difference that an infinite set of counterexamples can be associated with each counterexample. That is, if (a,b,c,n) is not a solution to the Fermat equation, then (ka,kb,kc,n) isn’t either for any positive integer k.
Suppose you test Fermat’s Last Theorem for n up to 10^10, and don’t find a counterexample. How much evidence does that give you for FLT being true? In other words, how do you compute P(a counterexample exists with n<=10^10 | FLT is false), since that’s what’s needed to do a Bayesian update with this inductive evidence? (Assume this is before the proof was found.)
I don’t dispute that mathematicians do seem to reason in ways that are similar to using probabilities, but I’d like to know where these “probabilities” are coming from and whether the reasoning process really is isomorphic to probability theory. What you call “heuristic” and “intuition” are the results of computations being done by the brains of mathematicians, and it would be nice to know what the algorithms are (or should be), but we don’t have them even in an idealized form.
The most influential books on the topic I know of are Mathematics and Plausible Reasoning Volume I: Induction and Analogy in Mathematics, and Mathematics and Plausible Reasoning Volume II: Patterns of Plausible Reasoning (amazon link) by George Pólya.
There’s a difficulty here involving the fact that every finite set of possible counterexamples has measure zero in the set {(a, b, c, n) | a, b, c, n in N} equipped with the probability measure that assigns each possible counterexample an equal probability.
So all the usual biases and cognitive gotchas involving probability and infinity come into play (even when the people doing the thinking are mathematicians!), and all bets are off.
My hypothesis is that the commonly used prior for counterexample distributions is exponential. As the lower bound K >= a, b, c, n on possible counterexamples increases, the exponential is updated into something close to uniform on the rest of {(a, b, c, n) | a, b, c, n in N}.
This does somewhat dodge the question, but it does make a difference that an infinite set of counterexamples can be associated with each counterexample. That is, if (a,b,c,n) is not a solution to the Fermat equation, then (ka,kb,kc,n) isn’t either for any positive integer k.