Hey guys, as you know the standard Wartune Dungeon invitation is: [World][Player]: Thrills await you in [Tower of Kings(Nightmare)_Activity] room 9. Join now! (Dungeon hasn't started yet) but players often edit…

# Probability of Maximum Rewards in Dragon Invasion

Hey guys, in this post we share with you calculations by Jesse from Kabam S67 (thanks!) related to dice chances in Dragon Invasion (or BI or any other similar system) in Wartune. It’s really nice to play with maths and learn more about Wartune and what you will see in this post is that there is actually a lower chance of having small rewards and a higher chance of having more.

## Calculation Assumptions and Basis

For both DI and BI the reward for clearing each stage is related to the highest dice roll of the players in the dungeon. It’s always best to have four players rolling dice. But, assuming the dice are clean, the probability of the highest dice roll for any number of players can be calculated.

For any given **number to be the maximum, two conditions must be true**.

- First, all rolls of the participants must be in the range of one to that given number. In other words, for 5 to be the maximum, all rolls must be in the range 1 to 5. It seems a bit obvious but the probability of all rolls in that range needs to be calculated to determine the probability of any given number being the maximum.
- The second condition that must be true is that at least one roll must be the given number. Again it is obvious but for 5 to be the maximum number, at least one roll must be five.

So the **probability of the given number being the maximum is the product of the two conditions**:

**Probability of Given Number is Maximum = (Probability of all rolls between one and that number) X (Probability of at least one roll equal to the given number)**

Breaking these down a little further,

### The Probability of all rolls between one and the given number

Probability of all rolls between one and the given number =

= (1 – (6 – Given Number) / 6) ^ ^{(Number of Participants)}

So for a given number of 5, the probability of all rolls are between 1 and 5 with 4 participants is:

Probability of all rolls between one and the given number =

= (1 – (6 – 5) / 6) ^ ^{4} = 0.482253

And,

### The Probability of at Least One Roll is Given Number

Probability of at least one roll is given number =

= 1 –.. | (Given Number – 1) ————————– Given Number |
^{^ Number of Participants
} |

Again for a given number of 5 and 4 participants:

Probability of at least one roll is given number =

= 1 –.. | ( 5 – 1) ————————– 5 |
^{^ 4} |
= 0.5904 |

Finally:

### Probability of Given Number is Maximum

Probability of given number is maximum =

Probability of all rolls between one and the given number × Probability of at least one roll is given number

For a given number of **5 as the maximum and 4 participants**:

Probability of given number is maximum =

0.482253 ×0.5904 = 0.284722 or **28%**

The probability of rolling each of the six possibilities for 1, 2, 3 and 4 players is calculated. The probability of rolling a maximum of 1 for 3 and 4 participants is not 0% but is so low it rounds to 0%.

Percent Probability of Any Given Number as Maximum Roll |
||||

Maximum Roll | Number of Participants | |||

4 | 3 | 2 | 1 | |

6 | 52% | 42% | 31% | 17% |

5 | 28% | 28% | 25% | 17% |

4 | 14% | 17% | 19% | 17% |

3 | 5% | 9% | 14% | 17% |

2 | 1% | 3% | 8% | 17% |

1 | 0% | 0% | 3% | 17% |

Note: you might think how come 1 player has a higher chance than 2 players on some rolls; the way to understand this table is that 2 players have a higher chance of rolling a higher number and indeed have a much lower chance to roll 1-3. |

## How about the chance of all dice being the same number?

The probability of all four people rolling the same number is (1/6)**4 or **0.07%**. This means we have to roll dice around 1400 times to get one of these to occur 🙂 hehehe

### Discuss in comments:

- Did you like this analysis?
- Did the resulting probabilities surprise you?
- Do you like having dice in Wartune?

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Wow #33 many numbers

Thanks! but this calculation very complicated for me

hhhhh it’s true,like stock market

analysis

hehe that’s OK! the good news is that we can learn anything and have fun with new knowledge 🙂

Just always believe in yourself! 🙂

– Cosmos

Thank you Cosmos for yet another informative post! I would like to make a few suggestions to edit this post to make it more clear.

First, you should set “The Probability of all rolls between one and a given number” = ((Given Number)/6)^(Number of Participants). Your formula is correct, however my formula is both correct and more simple. This would make it a little easier for some to understand. Your formula gives an indirect interpretation: “One minus the probability that all rolls are above a given number”. This is technically correct, but my formula give a direct interpretation of “The Probability of all rolls between one and a given number”. You can verify my formula with your example: (5/6)^4 = 0.482253.

Second, you should restate your assumption that all rolls are between one and the given number in “The Probability of at Least One Number is Given Number” section. If you do not, people may be mislead to think that the sample space is all rolls from 1 to 6. But your formula, which is correct, gives the probability that at least one number is the given number, ASSUMING THAT ALL ROLLS ARE BETWEEN ONE AND THE GIVEN NUMBER. This is not an easy post to understand for many, so restating your assumption will help to clarify your points and remove ambiguity. In addition, some may skip over the first few paragraphs because of laziness and go right to the formulae. 😛

If you would like to include the probability of rolling “straight numbers”, here it is. First, there must be 4 players rolling dice. If this condition is satisfied, the highest roll must be 4, 5, or 6. For each of these rolls, the other 3 rolls must be 1, 2, and 3 values under the highest roll. Therefore, the probability of rolling “straight numbers” = (4!)*(3/6)*(1/6)^3 = 5.56%. The “!” refers to factorial and must be included to account for all possible permutations of rolls. In other words, the order of the rolls does not matter. For example, “3,5,6,4” and “4,5,3,6” will both count as “straight numbers”.

If you would like to include the expected number of loot piles, here it is. The number of loot piles is twice the maximum roll. Therefore, simply multiply the probability of having a specific maximum roll with its corresponding number of loot piles. Then, add them all together. For 4 players, include “straight rolls” and “4 of a kind”. For example, the expected number of loot piles for 4 players is (0.52*12)+(0.28*10)+(0.14*8)+(0.05*6)+(0.01*4)+(0*3)+(0.0556*1)+(0.0046*1) = 10.56. Note that if the maximum roll is 1, this guarantees that 4 1’s are rolled, thus giving an extra pile for a total of 3 piles. For 3 players, the expected number of piles is 9.86. For 2 players, the expected number of piles is 8.96. For 1 player, the expected number of piles is 7.00.

Thank you for the additional analyses. There is no end to the fun of probability calculations!

The odds of 4 people rolling the same number is 0.46%, or approximately 1/216 rolls. While it is tempting to calculate this rate at (1/6)^4, but this actually give us the rate of rolling a specific four of a kind. The chance of rolling ANY four of a kind is (6/6)*(1/6)^3 because the 1st roll can be any number and then the rest of the rolls need to match.

I was hoping somebody else would spot that mistake 😉

Rat is indeed correct on the calculation of four players rolling the same number. Four players rolling the same number is 0.46% and four players rolling a specific number is 0.07%. Has anyone ever seen a roll of four 1’s?

i have actually seen this a couple times, if i have it happen again i will screen shot it for you

You are under the misapprehension that the dice are true. This is R2 so they are bound to be rigged

good job! stochastic is not an easy kind of stuff and you have done it in the right way! congratulations!!!

1 Q: in 2 years i did a 4-six(es).

what the probability??

i did get quite a few 1-4/2-5/3-6 straight…but NO all-4s :S :S 🙁

I get at least 5% of the times the 1,1,1,2 roll….

Just a general input but I think there might be a pattern to learn for how the dice will end up iv noticed that I can predict who gonna get sixes by watching the order of who rolls dosen’t work all time but some times find my self holding back from rolling so I can time a 6 might be a idea to look in too thanks for the facts guys keep up good work

Hey Jesse, nice first post and glad to see someone bringing in additions like this. I had wanted to do one which empirically proves that the dice (in DI in particular) are not fair, but I haven’t gotten to collecting data for it.

A nice addition would have been a column for P(rolling at least one number greater than or equal to x | x in {1:6}). So, P(at least one 6) ~ .52, P(rolling at least a 5 or greater) ~ .52 + .28 = .80, etc. From personal experience, it’s clear that we don’t get at least one 6 half the time, or at least a 5 80% of the time.

All that said, thanks for the theoretical breakdown. Looking forward to more.

lol this is wrong, the game “wartune” uses Korean type random for everything, and it also uses server clock generator to generate the pseudo random function events used in all “probable” things that the game need, it is the same type of randomness as galaxy online 2 uses. I wont say more, but If u know how that algorithm works u can exploit it to get rewards.

weeeeeeeeeell…

tell us how its done then.

if that seems fairly “easy”. we coult test things out..

u cant just throw the ball n hide ur hand 🙂 show us how to!

What are the odds of rolling (9) 6’s in a row for the same player? That’s my best streak.

that has to do with the “maximum late of a number” or at least thats the ill-translation from italian language 😀 not sure how to translate,….but still i wanted to say it s something TOTALLy different 🙂

those are not “probabilities” but “late(s)” 🙂

1+1 = 3 ?