A transfinite epistemic logic puzzle: what numbers did Cheryl give to Albert and Bernard?

We can map $n - 2^{-k} - 2^{-(k+r)}$ onto the ordinal $\omega^2(n-1) + \omega(k-1) + r$; this preserves order. So Cheryl effectively says "I have given you both distinct ordinals $a$ and $b$, both less than $\omega^3$. Which is bigger?"

Albert says "$a \geq 1$", Bernard says "$b \geq 2$", Albert says "$a \geq 3$", Bernard says "$b \geq 4$".

Cheryl says "Actually $a$ and $b$ are both $\geq \omega$".

Albert says "$a \geq \omega + 1$", Bernard says "$b \geq \omega + 2$", Albert says "$a \geq \omega + 3$", Bernard says "$b \geq \omega + 4$".

Eventually Cheryl says "Actually both are $\geq \omega^2 100 + 1$".

Albert says "$a \geq \omega^2 100 + 2$".

Bernard says "$b \geq \omega^2 100 + 3$".

Albert says "Aha! In that case I know the answer, which tells you that $a < \omega^2 100 + 4$".

From that, Bernard can work out Albert's number. How does he know whether Albert's number is $\omega^2 100 + 2$ or $\omega^2 100 + 3$? It can only be that $\omega^2 100 + 3$ is Bernards's number, so he knows Albert's must be $\omega^2 100 + 2$.

Translating back in to the language of the original question, this means that Albert's number is $101 - 2^{-1}- 2^{-3} = 100.375$ and Bernard's is $101 - 2^{-1}- 2^{-4} = 100.4375$.

Every time Albert and Bernard go back and forth they are eliminating a new ‘r’ , and every time she tells them that no matter how many times they go back and forth they won’t get it, they are eliminating a new ‘k.’

When Cheryl makes her big rant, she if effectively telling them that for all r and k when n = 1, they won’t find it, and furthermore that the first 100 times she makes that statement they won’t find it.

She also says that after saying it 100 times, neither will know which number is larger. This means that neither Albert nor Bernard has the number 100. (something a couple other solutions missed I think.) Albert says he does not know, ruling out him having 100 + 1/4, and then Bernard says he still does not know, ruling out Bernard having 100 + 3/8 and 100 + 1/4.

When Albert says that he suddenly knows, he can either have 100 + 3/8 or 100 + 7/16. The only way for Bernard to know both of their numbers is if he has 100 + 7/16, which gives us the answer:

Albert: 100 + 3/8 Bernard: 100 + 7/16

I have posted an extended solution on my blog:

Solution to my transfinite epistemic logic puzzle.

My answer agrees with that of Joe and Kellen and Paul on the other answers here, namely, that Albert has $100\frac38$ and Bernard has $100\frac7{16}$.