28
$\begingroup$

Kent Haines describes the game of Integer Solitaire, which I find to be excellent for young kids learning arithmetic. I'm sure they will be motivated by this game to get a lot of practice.

Kent asks a question about his game, which I find very interesting, and so I am asking here, in the hopes that Math.SE might be able to answer.

The child draws 18 cards from an ordinary deck of cards, and then regards the cards to have values Ace = 1, 2, 3, ..., Jack = 11, Queen = 12, King = 13, except that Black means a positive value and Red means a negative value.

Using 14 of the 18, the child seeks to find solutions of four equations:

Target equations

For example, a successful solution would look like:

Successful play of Integer Solitaire

Question. Does every set of 18 cards admit a solution?

Kent Haines says, "I have no idea whether all combinations of 18 cards are solvable in this game. But I have played this game for five years with dozens of students, and I have yet to see a combination of 18 cards that is unsolvable."

Follow up Question. In the event that the answer is negative, what is the probability of having a winning set?

For the follow up question, it may be that an exact answer is out of reach, but bounds on the probability would be welcome.

$\endgroup$
4
  • 8
    $\begingroup$ First... it can be simplified considerably by moving all minuses to the other sides as plusses. You have then two lines of the form ___ + ___ = ___ and two lines of the form ___ = ___ + ___ + ___ $\endgroup$ – JMoravitz Feb 12 at 20:23
  • $\begingroup$ the probability problem is probably NP-hard (it looks like a variant of the packing problem) $\endgroup$ – Jasen Feb 13 at 9:35
  • 2
    $\begingroup$ I'm a little confused by the instructions. It says, "using 14 of the 18 [cards]," (implying that each card is used once), yet the solution diagram repeats 2 of Diamonds and Queen of Spades. It doesn't matter in this case (can replace the repeat with the like-colored suit), but for the answer, should we assume that there are no repeated cards, or do we allow repeats? If we do allow repeats, can we repeat entire equations? If so, by @JMoravitz's reduction, we have two equations, not four. $\endgroup$ – Charles Baker Feb 18 at 2:14
  • 1
    $\begingroup$ @CharlesBaker the image was poorly made, yes. It seems clear from context that cards may not be repeated. The image can be corrected as you say by replacing by the offsuit of the same color. If we were allowed repeats it would be quite trivial otherwise and there would have been no reason to have the third and fourth equation to fill. $\endgroup$ – JMoravitz Feb 18 at 2:58
31
+100
$\begingroup$

Unsatisfyingly, a counterexample is (all black):

$$(5,5,6,6,7,7,8,8,9,9,10,10,J,J,Q,Q,K,K)$$

which does not satisfy the last two equations, since

$$\_+\_+\_ \ge 5+5+6 =16>13 = K$$

Extending this result, we need at least $22$ cards to guarantee a solvable $14$-tuple since we have the $21$-card counterexample

$$(3,4,4,5,5, \dots , K, K)$$

where $3+4+4+5+5+6 = 27 > 26 = 2K$, so the last two equations cannot both be satisfied. I do not know whether a counterexample to $22$ cards exists at this moment.

$\endgroup$
0
19
$\begingroup$

28 card counterexample:

$$ black: K, K, J, J, 9, 9, 7, 7, 5, 5, 3, 3, A, A $$ $$ red: K, K, J, J, 9, 9, 7, 7, 5, 5, 3, 3, A, A $$

cannot satisfy __ + __ = __ because 2 odds make an even (whether added or subtracted), and there are no evens in the set.

Edit: it's 28 cards, not 26.

The 29th counterexample is easy: with only one additional even, it isn't enough to satisfy both of the top two equations. So, 2 evens are needed to be added.

$\endgroup$
1
  • $\begingroup$ The same analysis can be carried out for the sets of cards in the form of $5k\pm1$ and $5k \pm2$, and would help establish a lower bound for the probability in the Follow up Question. $\endgroup$ – player3236 Feb 13 at 7:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.