Endgame calculation

Hooray, You’ve finally made it to the endgame. After N consecutive losses for sufficiently large N, you have cleared two suits, there are few hidden cards and victory is in sight. Provided you draw the right card(s) that is.

Okay I lied. This is the continuation of a game where I felt cocky after 5 wins and started by dealing 20 cards from the stock instead of the usual 10

A little thought should show we are in some trouble. We have only one hidden column. A little thought shows we can turn over a card in column 1 but this is kadoban. If we draw a bad card then game over.

With only 10 face down cards, one can enumerate the possibilities and calculate the probability the next card will 83,85,67,75. For the purists among you, let me state at the outset that counting cards is not cheating. Counting cards is second nature for serious Bridge players and tile-tracking is even allowed in Scrabble. I’ve even heard of “professional” Spider Solitaire programs that does card-counting for you (but I’ve never played them).

Captain Obvious is not always right

However, before turning over column 1 that we should check for any other less-obvious options. Options include turning over columns 6 or 9 or completing a suit.

• Column 6: clearly impossible, we would in fact need 3 empty columns
• Column 9: Impossible because of the suit breaks (we can’t shift the 9-8-7-6-5-4-3-2-A despite having a free Ten).
• Complete Clubs: It is usually impossible to complete a suit twice until all other suits are completed once and this is no exception.
• Complete Diamonds: impossible: the Nine is not visible.
• Complete Hearts: same as clubs
• Complete Spades: Oh-so-close! If we had a spare Deuce we can expose the Spade Ace in column 9, then swap with the Club Ace in column 2 and Bob Brown is Julia Gillard’s uncle. Or something like that.

Now that we have verified that there are no options we can expose a new card in column 1.

You should find the missing cards are: A447990JKK, ignoring suits.

You would immediately notice these are probably not the first ten cards you would wish to start a game with, since there is only two turnovers. As a general principle, if FOO is a set of 10 cards you don’t wanna see at the start, then EVERYTHING – FOO is probably not a set of 104-10=94 cards you wanna see at the end. Similar principles apply if we had some number of face-down cards other than 10. We have completed two suits, have one empty column and we are still in trouble.

Let us assume we play <18><19>. Assuming columns are labelled 1234567890 from left to right, you should have no trouble working out what that means.  It turns out any Ace, Ten, Jack or King spells 76,79,83,69. Any Four Seven or Nine means we are still alive.

This is bad news. Five out of 10 cards guarantee a loss, and we haven’t verified that any of the other five guarantee a win. We estimate we are a serious favourite to lose this game.

Fighting For Scraps

Actually, there is another option we haven’t considered. In the right-most column we notice the possibility of replacing 6-5-8-7 with 8-7-6-5. There might be some advantage in doing so … or there might not. Notice that we must commit to 6-5-8-7 or 8-7-6-5 before exposing a card in the leftmost column.

You might also notice the possibility of building the Spade Five onto the Spade Six in column 7. Thus we can move the Five of Clubs off somehow and then move <17><19>. This is possible, but requires us to compromise our position in some way.

The difference between these options may seem small, but at the end of the day it might allow us to snatch victory from the jaws of defeat, if you excuse the numerous clichés. In the end, I decided the obvious option is the best option.

EXERCISE: Assuming best play, what are the chances of victory in the original diagram position? This is not an easy question and I do not claim to have an exact answer. Let me know your thoughts by leaving a comment!