War of Legends Divination Trigram Guide

War of Legends Divination Trigram Guide by Nobodynat

1 Contents & Intro
2 Theory 1 – The Easy Theory – 50% win
3 Theory 2 – The Mathematical Approach – 41% win
4 Theory 3 – The Auto Lose Theory, Theory 3 extended – 27% win
5 Theory 4 – The Rearrangement Theory (disproved)
6 Theory 5 – The 1st Line Theory – 16.7% win
7 Theory 6 – The fixed line Theory – 68.4% win
8 Theory 7 – The improved fixed line Theory – 25% win
9 Reserved
10 Rewards & concluding comments

Hi, I am Nobodynat, public cookie giver and member of [Legacy] in the Royal Gardens server.

This is a Guide to the well known ‘money waster’ The Divination Trigram. Here I will try to show the probablility of sucess/fails with different theories I have come up with. Also if it is worth doing or not.

Please note, there may be some slight errors due to rounding off or wrong calculations. However, I am pretty confident it is enough to give a general estimate of success/failure.

The Trigram is reseted every night at 00:00 GMT and after every server update (this includes maintanace)

-Ingame Info-
Since the dawn of time, people have cast divinations to predict the future. You have 20 tries per day. Click Start to test your luck!

There are three lines per circle, if in each circle the lines are the same, there will be a prize.

In the theories, I will describe lines by the symbol – or _
– stands for the line with a break in the middle in the circle.
_ stands for the line without a break in the middle in the circle.

Please note this guide mainly deals with theoretical probability. Since each individual try is INDEPENDENT to each other try it means that each try has its own small probability of success and is no way linked to another. As such, please do NOT use this guide to calculate what you get every 20 tries. If you do, your a mathematical noob.

-Theory 1 – The Easy Theory-

This is what non-mathematicians think… So sorry to call you stupid or anything :P

The three lines have to be the same for any prize to come in the circle. If we look at it, there is only two possible lines. – and _.

Thus we can see that there is a 2/8 chance of the lines being equal in one circle considering it has a 1/2 possibility of each line.

Put that aside, there is 1/8 chance of getting three – in a circle. Since there is 4 circles, there must be 4 times as many chances, so its 4/8 = 1/2 chance! (WRONG)

hmm but even still, it doesnt seem that likely to get so many prizes! Either Jagex got it wrong (yeah go blame them) or I must be wrong! (yes you are lol)

-Theory 2 – The Mathematical Approach-

Here we go, the good stuff.

In the next few theories, I will consider this:
The – or _ prize is chosen right from the start with a 1/2 probability of either one.
This is because for some reason, if I dont consider this, the probability of winning is like 80+% O_o

The probablity of not winning anything is (7/8)^4 = 58.6%
So you have a chance of winning something at 41% good deal!

0 circles = 4C0*(7/8)^4 = 58.6%
1 circles = 4C1*(1/8)*(7/8)^3 = 38.3%
2 circles = 4C2*((1/8)^2)*(7/8)^2 = 8.2%
3 circles = 4C3*((1/8)^3)*(7/8) = 0.7%
4 circles = 4C4*(1/8)^4 = 0.02%

couple of errors but its close enough.
But it still doesnt seem right does it?

Well only if you choose 20 tries a day * probability of sucess, remeber; each individual try is independent of each other, so the lower the chance of sucess (below 50%) the fewer times of success it will seem to be.

-Theory 3 – The Auto Lose Theory, Theory 3 extended-

In this theory, I will consider this;
If – is chosen to be the prize, and there is a match of three _ in a circle, you will automatically lose (everything will turn to no match OR you get another try)

In this case you get a slight decrease in possibilities if the matching in one circle clashes with the chosen line.

0 circles = 4C0*(7/8)^4 = 58.6%
1 circles = 4C1*(1/8)*(6/8)^3 = 21.1%
2 circles = 4C2*((1/8)^2)*(6/8)^2 = 5.3%
3 circles = 4C3*((1/8)^3)*(6/8) = 0.6%
4 circles = 4C4*(1/8)^4 = 0.02%
Auto Lose = *4C1*(1/8)*3C1*(1/8)*(6/8)^2 + 4C1*(1/8)*3C2*((1/8)^2)*(6/8) + 4C1*(1/8)*3C3*(1/8)^3] + [4C2*((1/8)^2)*2C1*(1/8)*(6/8) + 4C2*((1/8)^2)*2C2*(1/8)^2] + [4C3*((1/8)^3)*[1/8)] = 13.8%

Again, there are some errors, but not the chance of success is now lower! now its at 27%!

-Theory 4 – The Rearrangement Theory (disproved) –

This I once believed, due to the low sucess rate.
This is now disproved, as I have seen data that does not match this.
The theory goes;
Let all the lines – and _ have 1/2 chance of either – or _ showing.
Rearrange all lines such that there will be a maximum amount of circles that do not match.
— -_- –_ _–
The example wins, but you will notice there are three _ which is in three different circles, this makes sure there are three circles with no match.

In such a case, this theory gives the player the worst possible outcome.

However, instead of calculating the probablility, I have seen data which counters this. I.e. a win without the worst case scenario.

-Theory 5 – The 1st Line Theory –

This theory is based on the first line that was chosen.

As such, the line – or _ is given. There is a 1/4 chance that the first circle matches.

Then for the next 3 circles there is a probablility for having more matches, no matches and auto lose.

1 circle = 1/4*3C0*(6/8)^3 = 10.5%
2 circle = 1/4*3C1*(1/8)*(6/8)^2 = 5.3%
3 circle = 1/4*3C2*((1/8)^2)*(6/8) = 0.9%
4 circle = 1/4*3C3*(1/8)^3 = 0.05%

Win % = 16.7%

-Theory 6 – The fixed line Theory-

This Theory is based on the fact that the autolose theory is hard to program. That is, there is no autolose system.

Instead, as the lines are chosen (1/2 to be -, 1/2 to be _), all other circles have one of these lines in them as to prevent the three of the opposite lines from appearing (lines are randomly placed obviously)

As such, this makes gives the sucess chance its highest probability, as it has 1/4 chance of the last two lines to match up)

0 circles = 4C0*(3/4)^4 = 31.6%
1 circles = 4C1*((3/4)^3)*(1/4)^1 = 42.2%
2 circles = 4C2*((3/4)^2)*(1/4)^2 = 21.1%
3 circles = 4C3*((3/4)^1)*(1/4)^3 = 4.7%
4 circles = 4C4*(1/4)^4 = 0.4%

As you can see, you have a 68.4% chance of winning!

This is the theory with highest success, of course this doesnt seem to be the case right?

-Theory 7 – The improved fixed line Theory-

This is based off Theory 6, which has a high probability of success. But now it has a ‘stop’ system. It discontinues the additional circles, as in it rolls its chance on a circle, then if it does not match, the system stops and the remaining circles have randomly generated/prefixed unmatching circles

So its still 1/4 chance for first circle, but more matching circles are harder to get

0 circles = 3/4 – (1,2,3,4 circles)= 75%
1 circles = 1/4 – (2,3,4 circles) = 18.7%
2 circles = 1/4^2 – (3,4 circles) = 4.7%
3 circles = 1/4^3 – (4circles) = 1.2%
4 circles = 1/4^4 = 0.4%

So now theres a 25% chance of winnind anything… Thats funny!

-Rewards & concluding comments –

Jcred in alliance donations is equivalent to 10k Gold
This does not mean you can sell items for gold or Jcred

1 – Blue Stone – worth 5 Jcred = 50k
Use: Increases success rate of forging items between 1-4
2 – Red Stone – worth 10 Jcred = 100k
Use: Increases success rate of forging all items
3 – Oak Chest – worth 50 Jcred = 500k
Use: Contains green suit items at around level 20
4 – Old Teak Chest – worth 200 Jcred = 2M
Use: Contains green suit items at around level 50 and possibility of blue suit items at around 40
1 _ Small Cache – worth N/A Jcred = N/A
Use: contains 1k of all resources
2 _ Large Cache – worth N/A Jcred = N/A
Use: contains 10k of all resources
3 _ Coffer – worth 50 Jcred = 500k
Use: Contains valuable Jewels
4 _ Jewel Box – worth 200 Jcred = 2M
Use: Contains rare and precious stones


Looking at the rewards from above, if we spend 100k on 20 tries and even if we have a low % chance of getting anything, it seems completely worth the money.

I suggest only doing this if you find yourself at over 10k+ gold per hour as at lower level, you want to keep the gold for research and market. Perhaps some troops spending too.

It is totally worth it.

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