The Genius, by A Skymin’s Mind #6

Open, Pass

Used as Main Match of Season 1, Round 7.


Each player receives a deck of 20 cards, numbered 0-9 along with mathematical operations +, -, ×, ÷. They may spend garnets to buy additional cards from three decks: black, red, and blue, worth 1, 2, 3 garnets each, and have increasingly better cards. (For example, the black number cards are two copies of 0-3 and one copy of 7-8, while the blue number cards are three copies of 7-9 and one copy of 10. Also, black operation cards have three of subtraction and division, with two of the others, while blue operation cards invert this, having three additions and multiplications and two of the others.)

After buying cards and exchanging with other players if necessary, each player constructs a 20-card deck, which will be shuffled by the dealer. A player may ask for a reshuffle for at most three times. After a player is satisfied with their lay, the dealer begins filling a 10-slot expression with the deck. Each time, a card is placed face down at the rightmost empty space. The player may choose either to open the card, thus opening it and fixing its position, or pass the card, throwing it away without seeing its value. After ten opens (and thus at most ten passes), the expression is created. Only the leftmost number is kept across a stretch of numbers, and similarly with operations. After that, if the rightmost card in the expression is an operation, it is discarded, and if the leftmost card is an operation, a zero is appended to the left of the expression. The result is then evaluated and becomes the score. For example, -, 8, 5, 7, +, ×, 3, ÷, 5, + is converted into 0 – 8 + 3 ÷ 5 = -7.4. The winner is the person with the highest score.


What kind of strategy is there for this luck-based game? Apparently there’s plenty…

This game is essentially a game of hidden information. The players didn’t know that the colors of the backs are different; the players didn’t know the contents of each deck until they opened what they bought; the players didn’t and probably have never known that the backs are not rotationally symmetric, which gives Jinho the easy win.

…okay, let’s retrace back.

The decks are as follows:

  • Black (1 garnet): 0, 0, 1, 1, 2, 2, 3, 3, 7, 8, -, -, -, ÷, ÷, ÷, ×, ×, +, +
  • Red (2 garnets): 4, 4, 4, 5, 5, 5, 6, 6, 6, 9, -, -, ÷, ÷, ÷, ×, ×, ×, +, +
  • Blue (3 garnets): 7, 7, 7, 8, 8, 8, 9, 9, 9, 10, -, -, ÷, ÷, ×, ×, ×, +, +, +

Thus the highest score is 100000, obtained by multiplying five 10s together. This requires four multiplications, five 10 cards, and one card that goes to the rightmost square.

As you can see, more expensive decks offer higher cards and more multiplications (and to some extent additions). This is not known yet by the players, which have to buy the decks to figure out the contents.

Since the colors of the backs can be seen during shuffling, this gives a possibility for strategy. Five players followed the same strategy: each color (hopefully) has only numbers or only multiplications. Due to lack of cards and money, this turns out to be quite difficult, and some people had to mix the two together for one color or include unwelcome cards such as addition.

Poong, one of the two people not using the strategy (for not knowing the colors are different; black and blue look quite similar), doesn’t have any strategy. However, the other person, Jinho, has tricks up his sleeve: he noticed that the design is not symmetric. Rotating the card gives a different pattern, which although difficult to spot, is still there. Assuming the dealer shuffles well, no card will be rotated and thus his strategy will do, by putting number cards in one orientation and multiplications in another. This allows him to use all cards in hand, which propels him for first position.

So, what other strategy exists? I have another strategy that works even better with only buying a single blue deck. Since subtractions and divisions are clearly out of bounds, one can ask for them and they should be able to obtain them. (It might cause suspicion though; this part is left for the reader to figure out. A solution would be to buy red deck too, but that’s boring.) The trick is that after receiving the cards, they only take the red cards to mix with the deck. These red cards are essentially useless and should always be passed, but it helps propelling for a 20-card count, thus allowing one to include only high number cards (8-10) instead of having middle-high cards (6-7). Of course, one cannot include ten red cards, as the shuffle will be too unlikely to give a good lay (five out of 210 combinations work, if there are six number cards and four multiplications), but including somewhere like six red cards, along with 5-6 multiplications and thus 8-9 number cards, looks reasonable. And with the number cards as they are, only the cards numbered 8-10 go in, which gives a lower bound of 32768 already.

With some scripting I find that six red cards, six multiplications, and eight number cards have a probability of 29.60% of a good lay for a minimum score of 41472, or use five red cards instead to get 37.76% probability for a minimum score of 36864. Of course, the safest is still to go with plenty of multiplications—my script says seven multiplications and zero red cards is optimal—but searching for that many multiplications as well as high cards is hard. You are welcome to test my script here (and perhaps help check it!).

Any better trick?


2 thoughts on “The Genius, by A Skymin’s Mind #6

    • Given that one can reshuffle three times (for a total of four lays), the probability of having no proper deck among all four is roughly 25% for six red cards, or 15% for five red cards. I’m still thinking that’s a good tradeoff for a more or less guaranteed win if luck sides.


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