Kabam still not reacting. They admit that they scammed us..

Here are my thoughts about it:

Lets not think about the rounds where no knights appear, just focus on the rounds where atleast 1 knight appear:

If there are 3 chests and u choose one of it, there is a 33,3x% chance to guess it right.

I wrote a simple program which imitates the game. The program returned a number between 0-2 for x,y,z. (x=0 y=1 z=2)

Here are the average results for 10k and 1k ppl playing the game with 10k gems each:

10000x100 guesses imitated -> x=33.35 y=33.43 z=33.23

1000x100 guesses imitated -> x=33.12 y=33.51 z=33.37

Now someone will say, yea of course u get it 33.x times correct with so many guesses because it compensates at that rate.

Ok then, lets look at the non "average" guesses.

100 guesses imitated -> x=33 y= 28 z=39

100 guesses imitated -> x=35 y= 41 z=24

100 guesses imitated -> x=34 y= 34 z=32

100 guesses imitated -> x=30 y= 26 z=44

100 guesses imitated -> x=36 y= 32 z=32

100 guesses imitated -> x=43 y= 31 z=26

100 guesses imitated -> x=38 y= 35 z=27

100 guesses imitated -> x=33 y= 27 z=40

100 guesses imitated -> x=26 y= 38 z=36

100 guesses imitated -> x=37 y= 29 z=34

Ofc the developers would be stupid if they would choose that kind of algorithm, because everyone would get like 25+ knights if they spend 10k gems.

Me and everyone else would say that is way too much.

The problem here, which still many people dont want to understand is, that its a different thing if u choose between 3 chests that have atleast 1 knight which makes a 33.3x% to guess it right and if u choose between 15 items/cards to get a knight.

Lets see the results for guesses between 15 items/cards:

10000x100 guesses -> a:6.6252 b:6.6368 c:6.6721 d:6.6828 e:6.7042 f:6.6909 g:6.671 h:6.6352 i:6.683 j:6.6243 k:6.6669 l:6.6835 m:6.6501 n:6.7081 o:6.6659

non-average:

100 guesses -> a:10.0 b:5.0 c:6.0 d:7.0 e:12.0 f:8.0 g:6.0 h:6.0 i:7.0 j:5.0 k:4.0 l:8.0 m:9.0 n:5.0 o:2.0

What wee see is that "o" is the one with bad luck. While the other letters have way more, o has only 2 correct guesses.

I still think 2 correct guesses on a knight card with 100 trys is fair.

Buuuuuuuuuuutt, its a different thing if u choose from 15 items/cards which makes a 6,67%*4 chance to get a knight (since there are 4. Well some of you guys said 1 is missing? make it even worse than it already is...) and if u choose between 3 chests which are containing atleast one and sometimes both other chests contain one knight.

What im trying to say is: Even if there are 100 different items to choose from, if i choose between 3 chests which contains "atleast" 1 knight, it means that i have a 33.3x% chance to be correct.

In my case, it happen around 60% of the 60 trys, that atleast one knight appeard in one of the other chests (around 5+ times even 2 knights appeard), it will mean that i had to get some sort of this results: (keep in mind that 33.3x% is and will be 33.3x% no matter what)

60% of 60 trys = 36x knight appearance:

36 guesses imitated -> a:12.0 b:12.6 c:11.4

36 guesses imitated -> a:11.4 b:15.0 c:9.6

36 guesses imitated -> a:12.0 b:10.2 c:13.8

36 guesses imitated -> a:16.8 b:7.8 c:11.4

36 guesses imitated -> a:13.2 b:10.2 c:12.6

Since i and many others did stick to one chest 40+ times, but did not or only got 1 knight, there is only 1 reason that we didnt got them. The minigame is programed to scam us. Simple as that. The algorithm is putting the knights after u choosed a chest into one of the others. Of course there are differences between program languages and u can write youre own random algorithm, but if u stick to one chest every single time with a chance of 33.3x% to be correct, u cant miss it!. Its not possible. Even if the algorithm is choosing between 15 items while u choose a chest, if atleast one knight appear in one of the 3 chests its still 33.3x% to be correct, dont u guys get it?

Aaron i want an explanation from u!

Here are my thoughts about it:

Lets not think about the rounds where no knights appear, just focus on the rounds where atleast 1 knight appear:

If there are 3 chests and u choose one of it, there is a 33,3x% chance to guess it right.

I wrote a simple program which imitates the game. The program returned a number between 0-2 for x,y,z. (x=0 y=1 z=2)

Here are the average results for 10k and 1k ppl playing the game with 10k gems each:

10000x100 guesses imitated -> x=33.35 y=33.43 z=33.23

1000x100 guesses imitated -> x=33.12 y=33.51 z=33.37

Now someone will say, yea of course u get it 33.x times correct with so many guesses because it compensates at that rate.

Ok then, lets look at the non "average" guesses.

100 guesses imitated -> x=33 y= 28 z=39

100 guesses imitated -> x=35 y= 41 z=24

100 guesses imitated -> x=34 y= 34 z=32

100 guesses imitated -> x=30 y= 26 z=44

100 guesses imitated -> x=36 y= 32 z=32

100 guesses imitated -> x=43 y= 31 z=26

100 guesses imitated -> x=38 y= 35 z=27

100 guesses imitated -> x=33 y= 27 z=40

100 guesses imitated -> x=26 y= 38 z=36

100 guesses imitated -> x=37 y= 29 z=34

Ofc the developers would be stupid if they would choose that kind of algorithm, because everyone would get like 25+ knights if they spend 10k gems.

Me and everyone else would say that is way too much.

The problem here, which still many people dont want to understand is, that its a different thing if u choose between 3 chests that have atleast 1 knight which makes a 33.3x% to guess it right and if u choose between 15 items/cards to get a knight.

Lets see the results for guesses between 15 items/cards:

10000x100 guesses -> a:6.6252 b:6.6368 c:6.6721 d:6.6828 e:6.7042 f:6.6909 g:6.671 h:6.6352 i:6.683 j:6.6243 k:6.6669 l:6.6835 m:6.6501 n:6.7081 o:6.6659

non-average:

100 guesses -> a:10.0 b:5.0 c:6.0 d:7.0 e:12.0 f:8.0 g:6.0 h:6.0 i:7.0 j:5.0 k:4.0 l:8.0 m:9.0 n:5.0 o:2.0

What wee see is that "o" is the one with bad luck. While the other letters have way more, o has only 2 correct guesses.

I still think 2 correct guesses on a knight card with 100 trys is fair.

Buuuuuuuuuuutt, its a different thing if u choose from 15 items/cards which makes a 6,67%*4 chance to get a knight (since there are 4. Well some of you guys said 1 is missing? make it even worse than it already is...) and if u choose between 3 chests which are containing atleast one and sometimes both other chests contain one knight.

What im trying to say is: Even if there are 100 different items to choose from, if i choose between 3 chests which contains "atleast" 1 knight, it means that i have a 33.3x% chance to be correct.

In my case, it happen around 60% of the 60 trys, that atleast one knight appeard in one of the other chests (around 5+ times even 2 knights appeard), it will mean that i had to get some sort of this results: (keep in mind that 33.3x% is and will be 33.3x% no matter what)

60% of 60 trys = 36x knight appearance:

36 guesses imitated -> a:12.0 b:12.6 c:11.4

36 guesses imitated -> a:11.4 b:15.0 c:9.6

36 guesses imitated -> a:12.0 b:10.2 c:13.8

36 guesses imitated -> a:16.8 b:7.8 c:11.4

36 guesses imitated -> a:13.2 b:10.2 c:12.6

Since i and many others did stick to one chest 40+ times, but did not or only got 1 knight, there is only 1 reason that we didnt got them. The minigame is programed to scam us. Simple as that. The algorithm is putting the knights after u choosed a chest into one of the others. Of course there are differences between program languages and u can write youre own random algorithm, but if u stick to one chest every single time with a chance of 33.3x% to be correct, u cant miss it!. Its not possible. Even if the algorithm is choosing between 15 items while u choose a chest, if atleast one knight appear in one of the 3 chests its still 33.3x% to be correct, dont u guys get it?

Aaron i want an explanation from u!

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