Let's roll a die four times and let $X_{k}$ $\left(k=1,2,3,4\right)$ denotes the rolled number during the $k$th rolling. How can we calculate
$$\mathbb{E}\left(\frac{X_{k}}{\min\left\{ X_{1},X_{2},X_{3},X_{4}\right\} }\right)=?$$
I know the CDF of the denominator can be calculated as $F_{Y}\left(i\right)=\left(F_{X_{k}}\left(i\right)\right)^{4}$, where $Y\overset{\circ}{=}\min\left\{ X_{1},X_{2},X_{3},X_{4}\right\}$, so
$$\mathbb{P}\left(Y=i\right)=\left(\frac{i}{6}\right)^{4}-\left(\frac{i-1}{6}\right)^{4},\;\;\;i=1,2,\ldots,6.$$
(Here I define the CDF as a right continuous function.) I wanted to calculate $\mathbb{E}\left(\frac{X_{k}}{Y}\right)$ with the law of total expectation:
$$\mathbb{E}\left(\frac{X_{k}}{Y}\right)=\mathbb{E}\left(\mathbb{E}\left(\frac{X_{k}}{Y}\mid Y\right)\right)=\mathbb{E}\left(\frac{1}{Y}\mathbb{E}\left(X_{k}\mid Y\right)\right).$$
From this expression:
$$\mathbb{E}\left(X_{k}\mid Y=i\right)=\frac{1}{7-i}\sum_{i=j}^{6}j,\;\;\;i=1,2,\ldots,6.$$
So using everything above:
$$\mathbb{E}\left(\frac{X_{k}}{Y}\right)=\mathbb{E}\left(\frac{1}{Y}\mathbb{E}\left(X_{k}\mid Y\right)\right)=\sum_{i=1}^{6}\left[\left(\frac{1}{i}\cdot\frac{1}{7-i}\sum_{i=j}^{6}j\right)\left(\left(\frac{i}{6}\right)^{4}-\left(\frac{i-1}{6}\right)^{4}\right)\right],$$
which is approximately $1.008$. But it doesn't feel good. Anyway, I ran MC simulations where $\mathbb{E}\left(\frac{X_{k}}{Y}\right)\widetilde{=}2.35$. Where do I go wrong with my calculation?
Your expression for $\ \mathbb{E}\big(X_k\,|\,Y=y\big)\ $ is incorrect. You appear to have assumed that given $\ Y=y\ $, $\ X_k\ $ is equally likely to take any of the values from $\ y\ $ to $\ 6\ $. Unfortunately, that isn't the case (except for $\ y=6\ $, when $\ X_k\ $ could only have the value $6$). If $\ y<6\ $, then given $\ Y=y\ $, $\ X_k\ $ is equally likely to take any of the values from $\ y\color{red}{+1}\ $ to $6$, but more likely than that to have the value $\ y\ $ In fact, for $\ x>y\ $, $$ \frac{P\big(X_k=x\,\big|\,Y=y\big)}{P\big(X_k=y\,\big|\,Y=y\big)}=1-\left(\frac{6-y}{7-y}\right)^3\ . $$
The equations $\ X_k=Y=y\ $ hold if and only if $\ X_k=y\ $ and $\ \min_\limits{j\ne k}(X_j)\ge y\ $. Since $\ X_k\ $ and $\ \min_\limits{j\ne k}(X_j)\ $ are independent, this has probability \begin{align} P\big(X_k=y,Y=y\big)&=P\left(X_k= y,\min_\limits{j\ne k}(X_j)\ge y\right)\\ &=P\big(X_k=y\big)P\left(\min_\limits{j\ne k}(X_j)\ge y\right)\\ &=\frac{1}{6}\left(\frac{7-y}{6}\right)^3 \end{align}
Likewise, if $\ x>y\ $, then the equations $\ X_k=x, Y=y\ $ hold if and only if $\ X_k=x\ $ and $\ \min_\limits{j\ne k}(X_j)= y\ $. Therefore, \begin{align} P\big(X_k=x,Y=y\big)&=P\left(X_k= x,\min_\limits{j\ne k}(X_j)= y\right)\\ &=P\big(X_k=x\big)P\left(\min_\limits{j\ne k}(X_j)= y\right)\\ &=\frac{1}{6}\left(\left(\frac{7-y}{6}\right)^3-\left(\frac{6-y}{6}\right)^3\right)\ , \end{align} and the joint probability mass function of $\ X_k\ $ and $\ Y\ $ is given by \begin{align} P\big(X_i=x,Y=y\big)=\cases{0&if $\ x<y$\\ \frac{1}{6}\left(\frac{7-y}{6}\right)^3&if $\ x=y$\\ \frac{1}{6}\left(\left(\frac{7-y}{6}\right)^3-\left(\frac{6-y}{6}\right)^3\right)&if $\ x>y$} \end{align} This gives \begin{align} \mathbb{E}\left(\frac{X_k}{Y}\right)&=\sum_{y=1}^6\sum_{x=1}^6\left(\frac{x}{y}\right)P\big(X_i=x,\,Y=y\big)\\ &=\sum_{y=1}^6\left(\frac{1}{6}\left(\frac{7-y}{6}\right)^3+\sum_{x=y+1}^6\frac{x}{6y}\left(\left(\frac{7-y}{6}\right)^3-\left(\frac{6-y}{6}\right)^3\right)\right)\\ &=\frac{20371}{8640}\\ &\approx2.35775\ , \end{align} close to the value you obtained from your simulation. This coincides with the value obtained by the different method used in my earlier less complete answer retained below.
As it happens, in this case the law of total expectation isn't much help in evaluating $\ \mathbb{E}\left(\frac{X_k}{Y}\right)\ $. While you can do it that way, it requires you to calculate the joint probability mass function $\ P\big(X_k=x,Y=y\big)\ $ first, divide it by $\ P(Y=y)=$$\,\left(\frac{7-y}{6}\right)^4-\left(\frac{6-y}{6}\right)^4\ $ to obtain the conditional probability $\ P\big(X_k=x\,\big|\,Y=y\big)\ $, and then sum $\ xP\big(X_k=x\,\big|\,Y=y\big)\ $ over $\ x\ $ to obtain the conditional expectation $\ \mathbb{E}\big(X_k\,\big|\,Y=y\big)\ $. Now to get $\ \mathbb{E}\left(\frac{1}{Y}\mathbb{E}\big(X_{k}\big| Y\right)\big)\ $, you have to divide $\ \mathbb{E}\big(X_k\,\big|\,Y=y\big)\ $ by $\ y\ $ and multiply it by the factor $\ P(Y=y\,)\ $ for each $\ y\ $ and sum over $\ y\ $. Since the factor $\ P(Y=y\,)\ $ is the same one you'd originally divided $\ P\big(X_k=x,Y=y\big)\ $ by, it's simpler to avoid this unnecessary division and multiplication by working directly with the joint probability mass function.
For what it's worth, the correct expression for $\ \mathbb{E}\big(X_k\,|\,Y=y\big)\ $ is $$ F(y)^{-1}\left(\frac{y}{6}\left(\frac{7-y}{6}\right)^3+\sum_{x=y+1}^6\frac{x}{6}\left(\left(\frac{7-y}{6}\right)^3-\left(\frac{6-y}{6}\right)^3\right)\right) $$ where $\ F(y)=P(Y=y)=$$\,\left(\frac{7-y}{6}\right)^4-\left(\frac{6-y}{6}\right)^4\ $.
Earlier less complete answer
If $\ Z_i=\frac{X_i}{\min_\limits{j=1,2,3,4}(X_j)}\ $, then $\ Z_i\ $ are identically distributed discrete random variables which can only assume one of the values $\ 1$,$\,\frac{6}{5}$,$\,\frac{5}{4}$,$\,\frac{4}{3}$,$\,\frac{3}{2}$,$\,\frac{5}{3}$,$\,2$,$\,\frac{5}{2}$,$\,3$,$\,4$,$\,5$, or $\ 6$. For $\ z=\frac{6}{5}$,$\,\frac{5}{4}$,$\,\frac{4}{3}$,$\,\frac{5}{3}$,$\,\frac{5}{2}$,$\,4$,$\,5$, or $\ 6 \ $, $\ Z_i\ $ can only assume the value $\ z\ $ if $\ X_i\ $ assumes the value of the numerator of the fraction and $\ \min_\limits{j\ne i}(X_j)\ $ assumes the value of the denominator: \begin{align} P\left(Z_i=\frac{p}{q} \right)&=P\big(X_i=p,\min_\limits{j\ne i}(X_j)=q\big)\\ &=\frac{1}{6}\left(\left(\frac{7-q}{6}\right)^3-\left(\frac{6-q}{6}\right)^3\right) \end{align}
For the other possible values of $\ Z_i\ $, we have: \begin{align} P\big(Z_i=1\big)&=\sum_{k=1}^6P\left(X_i=k,\,k\le\min_\limits{j\ne i}\big(X_j\big)\right)\\ &=\frac{1}{6}\sum_{k=1}^6\left(\frac{7-k}{6}\right)^3\\ &=\frac{1}{6}\sum_{i=1}^6\left(\frac{i}{6}\right)^3\\ P\left(Z_i=\frac{3}{2}\right)&=P\big(\big\{X_i=6,\min_\limits{j\ne i}\big(X_j\big)=4\big\}\cup\big\{X_i=3,\min_\limits{j\ne i}\big(X_j\big)=2\big\}\big)\\ &=\frac{1}{6}\left(\left(\frac{1}{2}\right)^3-\left(\frac{1}{3}\right)^3\right)+\frac{1}{6}\left(\left(\frac{5}{6}\right)^3-\left(\frac{2}{3}\right)^3\right)\\ P\big(Z_i=2\big)&=P\big(\big\{X_i=6,\min_\limits{j\ne i}\big(X_j\big)=3\big\}\\ &\hspace{3em}\cup\big\{X_i=4,\min_\limits{j\ne i}\big(X_j\big)=2\big\}\\ &\hspace{6em}\cup\big\{X_i=2,\min_\limits{j\ne i}\big(X_j\big)=1\big\}\big)\\ &=\frac{1}{6}\left(\left(\frac{2}{3}\right)^3-\left(\frac{1}{2}\right)^3\right)+\frac{1}{6}\left(\left(\frac{5}{6}\right)^3-\left(\frac{2}{3}\right)^3\right)+\frac{1}{6}\left(1-\left(\frac{5}{6}\right)^3\right)\\ &=\frac{1}{6}\left(1-\left(\frac{1}{2}\right)^3\right)\\ P\left(Z_i=3\right)&=P\big(\big\{X_i=6,\min_\limits{j\ne i}\big(X_j\big)=2\big\}\cup\big\{X_i=3,\min_\limits{j\ne i}\big(X_j\big)=1\big\}\big)\\ &=\frac{1}{6}\left(\left(\frac{5}{6}\right)^3-\left(\frac{2}{3}\right)^3\right)+\frac{1}{6}\left(1-\left(\frac{5}{6}\right)^3\right)\\ &=\frac{1}{6}\left(1-\left(\frac{2}{3}\right)^3\right) \end{align}
Here are these values aranged in a table \begin{array}{c|c|c|} z&P(Z_i=z)&\text{value}\\ \hline 1&\frac{1}{6}\sum_\limits{i=1}^6\left(\frac{i}{6}\right)^3&\frac{49}{144}\\ \hline \frac{6}{5}&\frac{1}{6}\left(\left(\frac{1}{3}\right)^3-\left(\frac{1}{6}\right)^3\right)&\frac{7}{1296}\\ \hline \frac{5}{4}&\frac{1}{6}\left(\left(\frac{1}{2}\right)^3-\left(\frac{1}{3}\right)^3\right)&\frac{19}{1296}\\ \hline \frac{4}{3}&\frac{1}{6}\left(\left(\frac{2}{3}\right)^3-\left(\frac{1}{2}\right)^3\right)&\frac{37}{1296}\\ \hline \frac{3}{2}&\frac{1}{6}\left(\left(\frac{1}{2}\right)^3+\left(\frac{5}{6}\right)^3-\left(\frac{1}{3}\right)^3-\left(\frac{2}{3}\right)^3\right)&\frac{5}{81}\\ \hline \frac{5}{3}&\frac{1}{6}\left(\left(\frac{2}{3}\right)^3-\left(\frac{1}{2}\right)^3\right)&\frac{37}{1296}\\ \hline 2&\frac{1}{6}\left(1-\left(\frac{1}{2}\right)^3\right)&\frac{7}{48}\\ \hline \frac{5}{2}& \frac{1}{6}\left(\left(\frac{5}{6}\right)^3-\left(\frac{2}{3}\right)^3\right)&\frac{61}{1296}\\ \hline 3&\frac{1}{6}\left(1-\left(\frac{2}{3}\right)^3\right)&\frac{19}{162}\\ \hline 4&\frac{1}{6}\left(1-\left(\frac{5}{6}\right)^3\right)&\frac{91}{1296}\\ \hline 5&\frac{1}{6}\left(1-\left(\frac{5}{6}\right)^3\right)&\frac{91}{1296}\\ \hline 6&\frac{1}{6}\left(1-\left(\frac{5}{6}\right)^3\right)&\frac{91}{1296}\\ \hline \end{array}
Computing $\ \mathbb{E}\big(Z_i\big) $ from the figures in this table gives \begin{align} \mathbb{E}\big(Z_i\big)&=\frac{20371}{8640}\\ &\approx 2.35775\ , \end{align} close to the value you obtained from your simulations.