What is the probability that some number of the form $10223\cdot 2^n+1$ is prime?

Solution 1:

First, regarding Didier's question about how to assign a probability to something which is either true or false, I highly recommend Towards a Philosophy of Real Mathematics by David Corfield.

In the present context, we can ask: If we had some way of determining whether there is at least one prime in this set, what would be the maximal price at which a rational self-interested actor would buy a bet to receive $€1$ in case it turns out that there is one?

This will depend in a complicated way on the actor's knowledge of the history of the Sierpinski problem: How far this set has been probed so far; how far the corresponding sets for other numbers that didn't turn out to be Sierpinski numbers had to be probed to find primes, etc.

But I'm guessing that your question isn't intended to take all this into account, and rather aims at a hypothetical rational self-interested actor who has no specific knowledge of the Sierpinski problem, but is aware of more general facts about primes, including density results related to the prime number theorem. Such an actor might calculate the "probability" of all numbers in this set being composite on the basis of an independence assumption roughly as

$$p=\prod_n\left(1-\frac1{\log\left(10223\cdot2^n+1\right)}\right)\approx\prod_n\left(1-\frac1{\log10223+n\log2}\right)$$

and thus

$$\log p\approx\sum_n\log\left(1-\frac1{\log\left(10223\cdot2^n+1\right)}\right)\approx-\sum_n\frac1{\log10223+n\log2}\;.$$

This logarithm diverges to $-\infty$, so our hypothetical self-interested actor would buy the bet at any price below $€1$.

Given that this model predicts that there shouldn't be any Sierpinski numbers at all but we know that there are, the independence assumption is unwarranted and it would be unwise to base any bets on it. To arrive at a more sound assessment of the value of such a bet would require an in-depth analysis of the history of the Sierpinski problem.

[Edit in response to the comments:]

It was rightly pointed out that I glossed over the possibility that there might be Sierpinski numbers despite the "probability" for this being $0$. However, my conclusion that a rational self-interested actor would abandon the naive model based on the independence assumption upon learning of the existence of Sierpinski numbers was correct.

Let's denote by $I$ the applicability of the independence assumption, by $S$ the existence of a Sierpinski number and by $A$ the existence of at least one prime in the set under consideration. Let's say the actor originally had degree $P(I)$ of belief in the applicability of the independence assumption and didn't know the first thing about Sierpinski numbers. Then the value of the bet to her would be $P(A)=P(I)P(A|I)+P(\neg I)P(A|\neg I)$, where $P(A|I)=1$ is the probability calculated above and $P(A|\neg I)$ is the probability she would ascribe to $A$ if the independence assumption were found not to apply. If her original degree of belief in the independence assumption is very high, that is, nearly $1$, the value of the bet is nearly $€1$.

Now upon learning that there are in fact Sierpinski numbers, assuming that this would influence her determination of $P(A)$ only via $P(I)$, she would perform a Bayesian update as follows:

$$ \begin{eqnarray} P(A|S) &=& P(I|S)P(A|I)+P(\neg I|S)P(A|\neg I) \\ &=&\frac{P(I\cap S)}{P(S)}P(A|I)+\frac{P(\neg I\cap S)}{P(S)}P(A|\neg I)\\ &=& P(A|\neg I)\; \end{eqnarray} $$

since $P(I\cap S)=0$, as the probability for $S$ under the independence assumption is zero. Thus $P(A|S) = P(A|\neg I)$, that is, learning of the existence of Sierpinski numbers has the effect of abandoning the independence assumption.