I recently found the game "98 Cards" by VdH and wondered if there is an optimal strategy to play it.

The four empty stacks with a random starting hand


  • There is only one player.
  • Cards with each of the numbers $2$ to $99$ are contained in the main stack once, in a random, unknown order.
  • There are four stacks on the field; two of them starting with $1$ (rising stack), the other two starting with $100$ (descending stack).
  • The player's hand contains the top 8 cards from the main stack.
  • Each turn, the player puts two cards from their hand onto stacks. They may be put onto the same stack, or different ones. The condition: cards can only be put on a stack if their value is higher than the current top card of that stack, if it is a rising stack, and lower if otherwise.
  • After each turn, the player draws two cards from the main stack, if there are any left. If the main stack is empty, the next turn starts without drawing.
  • Rule of Ten's: If a card in the player's hand is exactly 10 lower than the current top card of a rising stack, it may be put there. The reverse is true for descending stacks.

Win Condition:

The goal is to sort all 98 cards from the main stack into the four stacks. The game ends prematurely when there are no cards in the palyer's hand that could be put onto the stacks, and there are still cards left to sort.

My Question:

Is there an optimal strategy to play this game? The Rule of Ten's is giving me some headaches here.

Feel free to correct spelling or grammar mistakes. If this question would suit mathSE better, please migrate. I could not find any tags besides "cards" that would fit the question in my opinion. Feel free to add any suitable ones.

  • $\begingroup$ just downloaded the game, there is no track of the numbers you have put into the stacks/piles. it seems fun game but needs good track for the numbers you put, otherwise it will be hard to play game. $\endgroup$
    – Oray
    Feb 10, 2018 at 15:42
  • $\begingroup$ This has the distinct smell of bin packing all over it, my guess is that the optimal strategy is at least NP-hard. $\endgroup$
    – Bass
    Feb 10, 2018 at 19:06

2 Answers 2


To solve this game optimally,

I would be putting my computer to play it with enough time to see its statistics and improve it.

To do so,

I would put some algorithm firstly, such as;


Put the furthest number to the 50 to the proper pile first, then put another furthest accordingly. Such as if you have 97,7,20,25,57,54,89. I would put 97 and 7 first since there are the furthest. Because they will give you the most options for the next move.


If you have -10 or +10, options in your card, use Rule of Ten's rule immediately if have more options. Otherwise use 1.


Leave one descending and one ascending pile furthest to the 50 as long as possible.


If I have YX and ZX numbers, I would use them inversely whenever possible. Such as if I have 96 and 86. I would put 86 to descending pile and then 96 to have more options.


After getting out of options, play the nearest number to the pile's numbers.


Check the results and modify my algorithm accordingly with the results and logs.

I do not believe there is directly optimal methodology that you can put into words.

  • 1
    $\begingroup$ Thanks for your answer. For 2 and 4, situations can arise where it is more useful to use R.o.Ten's not immediately: Consider the hand 96, 94, 93, 92, 91, 90, 89, 86 and 95 on one of the decreasing stacks. You can empty 94 through 86 on that stack, and then use the 96 efficiently. Also random drawn cards may fall inbetween two RoT cards, so saving RoT's could allow for a better strategy. $\endgroup$
    – Orphevs
    Feb 20, 2018 at 14:19

this might be a little late, but today, after playing this game on and off for about 2years, I finally won the game twice in a row.

My strategy is as follows: Always try to "lose" as few numbers as possible. I will try to give an example down here.

TL: Top Left TR: Top Right BL: Bottom Left BR: Bottom Right

If you have on your TL a [51], in your TR a [65] and in your BR a [10]. Your BL is without any card number and in your hand you have [94, 53, 8] (I will not write all 8 numbers because it makes very difficult to try to explain my strategy). You'll maybe be tempted to put the [94] in your free card space (BL), because is the most high number you have until now and maybe you hope for getting to do a "10 combo" or something in the future, but it would be a mistake. The right decision would be to put the [8] in your BR spot and the 53 in your TL spot.

This is all about basic math and how many numbers you can stretch your limits.

In my strategy I put the number 8 under the number 10, and doing simple math: 10-8 = 2. So for now we have "lost" 2 numbers. 53-51 = 2. Sum it up and we have lost 4 numbers in two moves.

If we have put the 94 on the blank space, then doing some math: 100-94 = 6. We would spend/lost 6 numbers in just one move.

I hope you understand my explanation. The idea is to try to skip the least amount of numbers as possible with the two moves combined.

Also, it's important to try to enforce the 10 rule whenever is possible.

  • $\begingroup$ Thanks for your answer! I'll try to take a closer look at your explanation when I have time, maybe rephrase it to make your algorithm clear. I Appreciate it! $\endgroup$
    – Orphevs
    Oct 5, 2019 at 22:06
  • $\begingroup$ No problem :) Also sorry for the writing, english is not my primary language $\endgroup$ Oct 15, 2019 at 14:25

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