Seven players sit down for a hand of draw poker. If a player has a one-pair hand, she will call the first-round bets, thus participating in the draw; and all players drawing to one pair will want to draw three cards.
Thus if six or more players are originally dealt one-pair, the stack of careds available for drawing will run out.
What is the probability of six or more players, out of seven, each having one pair?
And how does that probability compare to $p^7+7p^6(1-p)$ where $p$ is the probability of a five-card freshly dealt hand being one-pair? (It must be somewhat greater, but by what factor?)
If we can determine the probability $p_k$ for all of $k$ people being dealt a one-pair hand, then the desired probability is $7p_6-6p_7$.
Here's code that computes the probability $p_k$ by successively choosing ranks for the one-pair hands of the players, keeping track of the depletion of the ranks and memoising the results. The complexity is determined by the number of different depletion patterns, which is $\binom{13+4}4=2380$, so this is not a big computation.
The results are:
\begin{align} p_6=\frac{12459207197065420800}{2102041714430930451179}&\approx0.00592719\;,\\ p_7=\frac{2711631680942502666240}{1066936315910442270434141}&\approx0.00254151\;,\\ p_{\ge6}=7p_6-6p_7=\frac{27997773085518424104960}{1066936315910442270434141}&\approx0.02624128\;. \end{align}
The probability $p=p_1$ of one player being dealt a one-pair hand is $\frac{352}{833}\approx0.42256903$, so the probability you wanted to compare to is
$$ p_1^7+7p_1^6(1-p_1)=\frac{7074278599678754816}{278301155710840989377}\approx0.02541951\;. $$
Thus the correlations increase the probability by about $3\%$.
The code also performs simulations that reproduce the results to five decimal digits. It could easily be adapted to similar computations (e.g. for two-pair hands).