Guest Post: Alternative Memory System

Editor’s Note: After a reassuring amount of consistency in the leaderboard data over the past couple weeks, it appears that once again the integrity of the data is suspect. Apparently Sunil is currently the #1 ranked player and has helpfully written another guest article with what appears to be an oversimplified memory system but perhaps its trivial ease of use will be an advantage for some.

I find myself atop the leaderboards once again, and will use this opportunity to share some more pearls of wisdom with the wodka-strategy.uk.to readership.

Previously, Jack talked about his memory system (https://wodka-strategy.uk.to/index.php/2020/06/27/memory-a-simple-system/). However, this requires maintaining multiple lists between which numbers frequently move. This is an unwieldy, and dare I say, inefficient system.

My approach requires you to only remember a single number. Firstly, we assign a different prime number to the highest three values:

  • A = 2
  • K = 3
  • Q = 5

For each of these cards in the deck, we multiply the corresponding value into a ‘memory-value’ (which starts at 1, the multiplicative identity). Since we start with four of each A, K, Q, our ‘memory-value’ starts at 2^4 * 3^4 * 5^4 = 810,000.

Whenever an A, K or Q is played, we simply divide our memory-value by 2, 3 or 5 respectively.

Similarly, if we want to know how many A, K or Qs are yet to be played, we just see how many 2, 3 or 5s are factors of the memory-value. For example, if the memory-value is 45,000 and we want a simple way to know how many Aces are in the deck we keep dividing by 2 until we hit an odd number (45,000 -> 22,500 -> 11,250 -> 5,625). We divided 3 times, and so there are three Aces still to be played.

Clearly, this is much simpler and less confusing than rembering how many Aces are still in the deck directly. In this case, rembering 3 (the number of Aces still to play) is problematic as it could get confused with other things – such as the number of Kings, or the number of points you will win with a successful Wodka. 45,000 has no chance of getting confused with other important Wodka numbers.

Furthermore, it’s easily extensible. For each additional piece of information you wish to remember, you only need to assign a unique prime.

For example, if we wish to include the 0 and the 3s, we could use:

  • A = 2
  • K = 3
  • Q = 5
  • 3 = 7
  • 0 = 11

Exercise for the reader (assuming this extended system):

The memory value is 1,980,825.

Q: Will my K win a trick?

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