I'm not an expert in statistics, but I gather there is disagreement whether a "frequentist" or "Bayesian" interpretation of probability is the "right" one. From Wagenmakers et. al p. 183:

Consider a uniform distribution with mean μ and width 1. Draw two values randomly from this distribution, label the smallest one $s$ and the largest one $l$, and check whether the mean μ lies in between $s$ and $l$. If this procedure is repeated very many times, the mean μ will lie in between s and l in half of the cases. Thus, $(s, l)$ gives a 50% frequentist confidence interval for μ. But suppose that for a particular draw, $s = 9.8$ and $l = 10.7$. The difference between these values is 0.9, and this covers 9/10th of the range of the distribution. Hence, for these particular values of $s$ and $l$ we can be 100% confident that $s < μ < l$, even though the frequentist confidence interval would have you believe you should only be 50% confident.

Are there really people who believe that there is only 50% confidence in this case or is it a straw man?

I guess more generally, the book seems to be saying that frequentists can't express a conditional claim like "Given s = 9.8 and l = 10.7, $s<\mu<l$ with probability 1". Is it true that conditioning implies Bayesian reasoning?

I'm not an expert in statistics, but I gather there is disagreement whether a "frequentist" or "Bayesian" interpretation of probability is the "right" one. From Wagenmakers et. al p. 183:

Consider a uniform distribution with mean $\mu$ and width $1$. Draw two values randomly from this distribution, label the smallest one $s$ and the largest one $l$, and check whether the mean $\mu$ lies in between $s$ and $l$. If this procedure is repeated very many times, the mean $\mu$ will lie in between $s$ and $l$ in half of the cases. Thus, $(s, l)$ gives a 50% frequentist confidence interval for $\mu$. But suppose that for a particular draw, $s = 9.8$ and $l = 10.7$. The difference between these values is $0.9$, and this covers 9/10th of the range of the distribution. Hence, for these particular values of $s$ and $l$ we can be 100% confident that $s < \mu < l$, even though the frequentist confidence interval would have you believe you should only be 50% confident.

Are there really people who believe that there is only 50% confidence in this case or is it a straw man?

I guess more generally, the book seems to be saying that frequentists can't express a conditional claim like "Given $s = 9.8$ and $l = 10.7$,   $s<\mu<l$ with probability 1". Is it true that conditioning implies Bayesian reasoning?

I'm not an expert in statistics, but I gather there is disagreement whether a "frequentist" or "Bayesian" interpretation of probability is the "right" one. From Wagonmakers et. al p. 183:

Consider a uniform distribution with mean μ and width 1. Draw two values randomly from this distribution, label the smallest one $s$ and the largest one $l$, and check whether the mean μ lies in between $s$ and $l$. If this procedure is repeated very many times, the mean μ will lie in between s and l in half of the cases. Thus, $(s, l)$ gives a 50% frequentist confidence interval for μ. But suppose that for a particular draw, $s = 9.8$ and $l = 10.7$. The difference between these values is 0.9, and this covers 9/10th of the range of the distribution. Hence, for these particular values of $s$ and $l$ we can be 100% confident that $s < μ < l$, even though the frequentist confidence interval would have you believe you should only be 50% confident.

Are there really people who believe that there is only 50% confidence in this case or is it a straw man?

I guess more generally, the book seems to be saying that frequentists can't express a conditional claim like "Given s = 9.8 and l = 10.7, $s<\mu<l$ with probability 1". Is it true that conditioning implies Bayesian reasoning?

I'm not an expert in statistics, but I gather there is disagreement whether a "frequentist" or "Bayesian" interpretation of probability is the "right" one. From Wagenmakers et. al p. 183:

Consider a uniform distribution with mean μ and width 1. Draw two values randomly from this distribution, label the smallest one $s$ and the largest one $l$, and check whether the mean μ lies in between $s$ and $l$. If this procedure is repeated very many times, the mean μ will lie in between s and l in half of the cases. Thus, $(s, l)$ gives a 50% frequentist confidence interval for μ. But suppose that for a particular draw, $s = 9.8$ and $l = 10.7$. The difference between these values is 0.9, and this covers 9/10th of the range of the distribution. Hence, for these particular values of $s$ and $l$ we can be 100% confident that $s < μ < l$, even though the frequentist confidence interval would have you believe you should only be 50% confident.

Are there really people who believe that there is only 50% confidence in this case or is it a straw man?

I guess more generally, the book seems to be saying that frequentists can't express a conditional claim like "Given s = 9.8 and l = 10.7, $s<\mu<l$ with probability 1". Is it true that conditioning implies Bayesian reasoning?