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.

• Not an answer, but a simulating it 10,000,000 times gives an estimate of 0.00585, so you expect it to happen about once in every 170 games. Nov 30 '12 at 18:02
• Note than in Win7 (unlike WinXP) version of the game you have the option to take back dealing the final ten cards. This gives you more options. If you have made enough progress prior to that point you may be able to arrange one of the newly dealt cards to extend an existing single suit sequence. Which may or may not help your cause. Nov 30 '12 at 18:51
• @JyrkiLahtonen The question is specifically about the situation in which no matter how the rest of your cards are arranged it's impossible to do anything with the final ten, so taking back the top ten certainly won't help you. Nov 30 '12 at 20:17
• @JonathanChristensen: An example of the situation I am describing is the following. Say that there is a nine of spades among the last ten cards to be dealt. Assume that there is also a jack among the ten last cards. Now if it happens that there is a single column with a ten of spades prior to dealing those last ten cards, and if it also happens that you have made enough progress so that you have some choice about which column that ten of spades will occupy. Then if you backtrack the last deal you may be able to arrange a column with ten-nine of spades, and you can move that column on the jack. Nov 30 '12 at 22:00
• I recall having rescued some games with a trick like the above. The idea can be extended to stacks with more cards. The ability to backtrack deals is the reason why my success rate in win7 is about 80% but in winXP it was (towards the end) something like 60%. Nov 30 '12 at 22:04

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.

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.

• Please be more polite. The statement "Your assumption is wrong!" can be taken to be aggressive. Apr 21 '20 at 20:59
• That a good player can leave cards ready for the next deal was made in the comments to my question. But the fact remains that the final (or other) deal may have no cards that can go (all the card values are even, for example). Apr 21 '20 at 21:56