Calculate the probability for an another fibonacci...

Below is one way to arrive at a very rough “back‐of‐the‐envelope” answer. (Any answer is highly speculative because both the infinitude of Fibonacci primes and the “randomness” of their digits are unproved assumptions.)

Let’s recall:

• A “cyclops number” is an odd number with an odd number of digits in which the only 0 appears in the very middle.
• One known Fibonacci prime, namely $$F(83)=99\,194\,853\,094\,755\,497,$$ has 17 digits and is cyclops (its 9th digit is 0, and no other digit is 0).

Now, if we assume that for large numbers the digits behave “randomly” (an assumption that seems to hold approximately for Fibonacci numbers) then for a number with $d=2k+1$ digits the chance that the middle digit is 0 is about $1/10$ while the chance that each of the other $2k$ digits is nonzero is about $9/10$ (one may ignore the fact that the first digit can’t be 0 since we are only looking at numbers of exactly $d$ digits). In other words, the probability that a “random” $d$–digit odd number is cyclops is roughly

$$P_{\rm cyclops}(d)\approx \frac{1}{10}\Bigl(\frac{9}{10}\Bigr)^{d-1}\,.$$

For example, when $d=17$ (as for $F(83)$) this is about

$$\frac{1}{10}\Bigl(\frac{9}{10}\Bigr)^{16}\approx \frac{1}{10}(0.9^{16})\approx \frac{1}{10}(0.185)\approx 0.0185\quad (1.85\%).$$

However, the next Fibonacci primes (if any exist) will be much larger. In fact, the $n^{\rm th}$ Fibonacci number is roughly

$$F(n)\approx \frac{\varphi^n}{\sqrt5}\quad\text{with}\quad\varphi\approx1.618\,,$$

so its number of digits is about

$$d(n)\approx n\log_{10}\varphi-\log_{10}\sqrt5\,.$$

Thus as $n$ increases, the number of digits (and hence the exponent in the factor $(9/10)^{d-1}$) increases linearly so that the cyclops “chance” decays exponentially. For instance, a Fibonacci prime having 28 digits (roughly what one might expect for the next candidate) would have

$$P_{\rm cyclops}(28)\approx \frac{1}{10}\Bigl(\frac{9}{10}\Bigr)^{27}\approx 0.0058\quad (0.58\%)\,.$$

On the other hand, it is known (or at least conjectured) that the “probability” for a “random” number of size $N$ to be prime is roughly $1/\ln N$. For $F(n)$ one has $\ln F(n)\sim n\ln\varphi$, so very roughly the chance that $F(n)$ is prime is about

$$P_{\rm prime}(n)\sim \frac{1}{n\ln\varphi}\,.$$

(There is a long‐standing conjecture that there are infinitely many Fibonacci primes but they are extremely sparse.)

Now, assuming that the events “$F(n)$ is prime” and “$F(n)$ is cyclops” are (roughly) independent, the probability that a given Fibonacci number (with index $n$ large enough so that it has an odd number of digits) is both prime and cyclops is roughly

$$P(n)\sim \frac{1}{n\ln\varphi}\cdot \frac{1}{10}\Bigl(\frac{9}{10}\Bigr)^{d(n)-1}\,.$$

Since $d(n)\sim n\log_{10}\varphi$, we see that for larger $n$ the extra cyclops condition gives an extra factor that decays roughly exponentially in $n$.

If we now “sum” over the (very sparse) Fibonacci indices beyond $n=83$ one obtains an expected total number of additional cyclops Fibonacci primes that is very small (on the order of a hundredth, say 0.01 or so). In a Poisson‐model this expected count corresponds to a probability

$$P(\hbox{at least one more})\sim 1-\exp(-0.01)\approx 0.01,$$

i.e. about a 1% chance.

In summary, under these rough heuristic assumptions the probability that there is another Fibonacci prime that is also a cyclops number (other than $99194853094755497$) is on the order of one percent or even less.

Caveats

1. Digit randomness: Although many “random‐looking” sequences (including Fibonacci numbers) appear to have uniformly distributed digits in many respects, this is not proven in any rigorous sense.

2. Fibonacci primes: It is still an open question whether there are infinitely many Fibonacci primes. Our “probability” here is conditional on there being more Fibonacci primes and that they behave like random numbers in terms of their digit distribution.

3. Independence assumption: We assumed that the event “$F(n)$ is prime” is independent of the event “$F(n)$ is cyclops.” This is a very rough assumption.

Thus, while one might “calculate” an answer of roughly 1% chance, it should be taken as a very speculative, back‐of‐the‐envelope estimate rather than a rigorous probability.

Final Answer

Under heuristic assumptions the chance that another Fibonacci prime is cyclops is on the order of 1% (or even less).