Losing at Spider Solitaire Spider Solitaire has the property that sometimes none of the cards in the final deal can "go" and so you lose, regardless of how much progress you have made beforehand. You would have known that you would lose had you seen the final ten cards before the game started.  

I wonder if we can calculate the probability of this happening.  

To be clear, I want to find the probability that the final ten cards out of two packs of well-shuffled cards comprise cards no two of which are exactly one away from each other numerically (only the values matter, not the suits).  

Note: there are several variants of Spider solitaire. I'm primarily interested in the standard 104-card, four-suit game.
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&ndash;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.
I don't believe your assumption is correct. If, in a final (or any other) deal, there is no single card that can be moved to another column, it may still be moved together with one or more cards that are beneath it. So the expert player will plan to leave one or more cards of the same suit and increasing order in one of the columns, to then move them together to some matching other column.