Losing at Spider Solitaire

There can be $2$ to $7$ different card values, and $k$ card values can occur in $\binom{14-k}k$ different value combinations. The number of ways of choosing $10$ cards with exactly $k$ different card values can be found using inclusion–exclusion: There are $\binom{8k}{10}$ ways to choose $10$ cards with values in a given set of $k$ values, so there are

$$ \sum_{j=2}^k(-1)^{k-j}\binom kj\binom{8j}{10} $$

ways to choose $10$ cards with values forming exactly a given set of $k$ values. Thus the number of combinations is

$$ \sum_{k=2}^7\binom{14-k}k\sum_{j=2}^k(-1)^{k-j}\binom kj\binom{8j}{10}=153020720928 $$

(computation) out of a total of $\binom{104}{10}=26100986351440$, so the probability for this to occur is

$$ \frac{153020720928}{26100986351440}=\frac{9563795058}{1631311646965}\approx0.00586264 $$

in agreement with Jonathan's simulations.