The Wisdom Shrine - My Method

So, Wisdom. Two very important things to start:

The Wisdom shrine will frustrate you.

The Wisdom shrine will burn your cards.

As with any game of chance, being lucky trumps any possible strategy, and taking any approach you like may well work with a little luck, but I have some tips from my approach that may well help you in the long run.

Disclaimer: All of the following is my methodology, after discussing it a little with others and thinking it through in my head, applying my logic. While in some situations I have done the math, it may not be the case that I am entirely correct. Also, please note that this is still heavily reliant on luck, which can skew the results of any methodology/approach.

First off, just a reminder that we are looking for a unique three- or four-digit solution, using the Q value of the cards you submit. This code is in a particular order, and every user has a unique code. This means you are not competing for who gets lucky first or watching everyone else guess in the hopes of sniping at the very end once whittled down.

Your game is your own, and you should take your time to work on it, as there is no time pressure.

Note that, depending on your luck, an average run at wisdom could cost from as little as 20 cards up to around 40. Of course, you could also hit it on the first try, or it could be the last number you try in every category.

Sequential search

Firstly, I will address the idea that the most efficient path may be to work on each digit individually. The merit of this is that when you find a correct digit, you will instantly know the correct position, since the information the shrine gives back to you is limited. This is a stress-free approach.

However, it is highly inefficient in card usage, even if it may seem cheap at first. This is because once you have found the first digit, in order to test the second digit, you have to enter a card into the first slot every time, and then the first two for the third, and then the first three to check the 4th. This is a lot of wasted cards.

If you were to perfectly guess the card first time, and your number was only three digits, you need to submit 1 card for the first guess, 2 for the second, and 3 for the third, totalling 6 cards. Sounds good, but if you also guessed it right the first time and just entered all three, it would only have cost three cards.

If you had a four-digit solution and hit every wrong number before the right one, this would cost 90 cards. It could even cost 54 cards for a three-digit solution. 

If it took you 5 guesses for each of the four numbers, it would still cost 50 cards.

This means that in the sequential method above, for a three-card solution, you would average burning around 30 cards, up to a max of 54. For a four-card solution, you would average burning around 50 cards, up to a max of 90.

In conclusion: valid approach, easy to use, should average to an okay card efficiency, but has the potential to spiral out of control.

Parallel search

I recommend opening with four cards on every guess until you hit a non-zero% from the shrine. If the % is 33/66, you know it’s three digits and can drop to three-number submissions going forward. 

The reason I would not start with three-number suggestions is that if it takes you however many guesses to get a percentage and discover it is 4 due to getting a 25/50/75% result, you will need to check all 10 possibilities for the 4th digit, which is expensive due to always having to submit cards in the first three positions to check it.

Note: I haven't done the exact math on this, and starting with either 3 or 4 could be more efficient depending on luck, but this is the logic I will proceed with.

As much as this game is an exercise in finding the correct numbers. It is more so an exercise in eliminating incorrect numbers, and a parallel search is much more efficient at this.

By starting by submitting 4 numbers into Wisdom, every 0% eliminates a single number from each of the four positions, reducing your odds of getting one correct on your next spin to 1/9, then ⅛, etc.

Once you hit a percentage result (25/33/50/67/75), you are able to determine if your solution is three digits (33/67) or four digits (25/50/75), and you should proceed by submitting the number of cards that matches your digits.

Once you hit a number, this is where a parallel search can require more thought than a sequential search. This is because you do not know which number is correct.

There are a couple of approaches from this point, where you can test numbers alone or continue to make further guesses to get other ‘partially correct’ answers and then cross-reference them using guesses that bridge two partially correct answers.

I recommend the pragmatic approach by resubmitting the first digit of your guess alone in an isolation test. By isolating this number, you are either locking it in as correct or ruling it out as incorrect. You can then test the second number and third number by entering numbers into slots 1 and 2 that you know are already incorrect (from previous guesses) until one of them is confirmed. If 1, 2 and 3 are not confirmed in the four-digit number, you know 4 is correct. (1 and 2 for a three-digit number)

Once you have identified which number is correct, you can then continue with parallel searches using either the correct number or a placeholder in the slot where the correct digit is, and then repeat the process once a correct number is found.

Okay, that sounds a bit more complicated. Is it worth it? Well, in short, yes.

Using the parallel search method, the absolute worst case for a four-digit code is approximately 80 cards, which is 10 less than the sequential search; however, due to the chance to hit multiple right numbers in a single guess (50%, 75%), the possibility to find the right numbers faster, and the small chance of hitting a perfect 100, the expected average costs sits below the 50 cards of the sequential method, at around 40 cards.

An example in practice

Whole lotta waffle, Prael, just show me how it’s done already.

Fine…

So let’s start with a few submissions, where I am only using unique numbers for each position:

1 2 3 4 = 0%

3 5 6 7 = 0%

6 1 5 3 = 0%

10 9 4 1 = 25%

Okay, I hit a percentage, and I now know that this code is four digits long. I have also learned that the first number cannot be 1, 3 or 6, the second cannot be 2, 5 or 1, etc.

Let’s isolate.

10 = 0%   - I now know the first digit is not 1, 3, 6 or 10. Using this:

1 9 = 0%   - Likewise, the second digit is not 2, 5, 1, or 9.

1 2 4 = 25% - Okay, we can lock in digit three as 4.

Resuming search (I can use any number other than 4 in position 3, and I know it will return 0%, or use 4 and know that it will return 25%, I will mark this as X, and use a number other than 4 in the example below.)

8 4 X 6 = 0%

7 7 X 2 = 50% - two numbers found, time to isolate.

7 = 25% - lock in 7

7 7 = 25% 

Now we know that the second digit is not seven. Since our submission of 7 7 1 2 is 50% correct and we know the second and third digits are incorrect, then we know that 2 is the fourth digit.

So now we know that the answer is 7 _ 4 2

And we know that the second digit is not 2, 5, 1, 9, 4 or 7. This leaves 3, 6, 8 and 10 as possibilities. Since we are only looking for digit two:

X 3 = 0%

X 6 = 0%

X 8 = 25%

That’s all four digits, we can now submit:

7 8 4 2 = 100%

This cost 43 cards, which is above average for this method, but below the average for the sequential method. In this example, we had moderate luck, and I think it is a fair representation of how a run can go.

Bonus:

Managing the above method is often times confusing and will likely need something to keep track of eliminated and possible numbers. I just use a notepad to do so.

While the order in which you test each number doesn’t really matter, since they all have an equal chance of coming through, you may wish to try to uniform your guesses to make it easier to track, if you have the cards available.

For example:

1 1 1 1

2 2 2 2

….

If you go for several runs at the wisdom shrine, doing this may see your results converge towards the average by making your approach consistent.

If you’re doing the above, if your unique solution is 1423, you may find it in just ~25 cards, whereas when your code is 6 9 4 10, it may take a spend of ~60 cards. But it should average out over multiple runs.

So there we have it, a consistent approach to a game still entirely based on luck that hopefully minimises your spend. However, the reality is that it could get expensive, even with the best method.

But, please note the example above is a four-digit code, and your costs drop dramatically should you see a 33% or 67% pop.

And finally, remember that if you get Secret Kkyusol to drop, send it to me, thanks.