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[Hydrad]

Is "Roll manipulation" referring to PPS rolling three 8's and Hydrad rolling multiple 7's?  I'm not convinced this is evidence of tampering at all... it's probably just regular RNG stuff.

I feel like mine can be regular RNG but i feel like triple 8's is really unlikely.

Ah sorry I didn't explain it properly. 70% is how confident I am that there is dice manipulation. I haven't calculated how likely it is at all

8 8 8 has the exact same chance of occurring as 9 2 11.  Why is one evidence of tampering but the other not?

While that is true the chance is still pretty low which makes me fell like there is about a 70% at least that we have someone tampering with dice. Its not high enough that I am positive but I am definitely going to consider it as a power

70% sounds completely made up there.

The chance is not low, at all.  You simply cannot take one outcome of a random experiment and make deductions about the distribution.

Consider also there were 14 sets of N_i die rolls on Day 1.  Generated 14 instances of three die rolls twice.

Result 1:

(12, 2, 6)
(6, 12, 2)
(3, 1, 3)
(2, 12, 11)
(10, 10, 11)
(11, 7, 11)
(7, 3, 3)
(7, 8, 10)
(12, 7, 10)
(6, 8, 12)
(5, 1, 4)
(12, 2, 4)
(8, 5, 9)
(4, 6, 4)

Result 2:

(1, 7, 10)
(6, 9, 8 )
(9, 4, 8 )
(5, 10, 4)
(8, 7, 12)
(1, 7, 11)
(6, 4, 5)
(1, 6, 9)
(10, 8, 6)
(10, 7, 3)
(2, 10, 12)
(12, 12, 12)
(6, 8, 1)
(5, 5, 2)

[Hydrad]

Is "Roll manipulation" referring to PPS rolling three 8's and Hydrad rolling multiple 7's?  I'm not convinced this is evidence of tampering at all... it's probably just regular RNG stuff.

I feel like mine can be regular RNG but i feel like triple 8's is really unlikely.

Ah sorry I didn't explain it properly. 70% is how confident I am that there is dice manipulation. I haven't calculated how likely it is at all

8 8 8 has the exact same chance of occurring as 9 2 11.  Why is one evidence of tampering but the other not?

While that is true the chance is still pretty low which makes me fell like there is about a 70% at least that we have someone tampering with dice. Its not high enough that I am positive but I am definitely going to consider it as a power

70% sounds completely made up there.

The chance is not low, at all.  You simply cannot take one outcome of a random experiment and make deductions about the distribution.

Consider also there were 14 sets of N_i die rolls on Day 1.  Generated 14 instances of three die rolls twice.

Result 1:

(12, 2, 6)
(6, 12, 2)
(3, 1, 3)
(2, 12, 11)
(10, 10, 11)
(11, 7, 11)
(7, 3, 3)
(7, 8, 10)
(12, 7, 10)
(6, 8, 12)
(5, 1, 4)
(12, 2, 4)
(8, 5, 9)
(4, 6, 4)

Result 2:

(1, 7, 10)
(6, 9, 8 )
(9, 4, 8 )
(5, 10, 4)
(8, 7, 12)
(1, 7, 11)
(6, 4, 5)
(1, 6, 9)
(10, 8, 6)
(10, 7, 3)
(2, 10, 12)
(12, 12, 12)
(6, 8, 1)
(5, 5, 2)

Ah sorry I didn't explain it properly. 70% is how confident I am that there is dice manipulation. I haven't calculated how likely it is at all

failed quote terribly

[faust]

Hydrad, I'm still waiting for your confirmation.

[Hydrad]

Hydrad, I'm still waiting for your confirmation.

Ah yes I did target you N0

[faust]

Peculiar.

[Hydrad]

So i'm guessing you have an ability that can tell who visited or something? either way if you need it i'm willing to claim i don't think my power is a huge gamechanger or anything

[Hydrad]

one second i'm going to ask something in my QT

[faust]

To settle this argument:

chances of three of the same numbers appearing when three dice are rolled:

P(three of the same) = P(2. die = 1. die)*P(3. die = 1. die) = 1/144

chances of this happening at least once in 14 rolls:

P(at least once) = 1 - P(never three of the same) = 1 - (1 - P(three of the same)^14 = 1 - (143/144)^14 = 9.3%

[faust]

The chances of this happening twice in a row are thus (0.093)^2= 0.9%

[Hydrad]

The chances of this happening twice in a row are thus (0.093)^2= 0.9%

to be fair mine was only 2 dice though

[Witherweaver]

To settle this argument:

chances of three of the same numbers appearing when three dice are rolled:

P(three of the same) = P(2. die = 1. die)*P(3. die = 1. die) = 1/144

chances of this happening at least once in 14 rolls:

P(at least once) = 1 - P(never three of the same) = 1 - (1 - P(three of the same)^14 = 1 - (143/144)^14 = 9.3%

Should really do 14+12 = 26 rolls.

And it didn't happen twice.  Hydrad's was two 7's on day 2.

[faust]

To settle this argument:

chances of three of the same numbers appearing when three dice are rolled:

P(three of the same) = P(2. die = 1. die)*P(3. die = 1. die) = 1/144

chances of this happening at least once in 14 rolls:

P(at least once) = 1 - P(never three of the same) = 1 - (1 - P(three of the same)^14 = 1 - (143/144)^14 = 9.3%

Should really do 14+12 = 26 rolls.

And it didn't happen twice.  Hydrad's was two 7's on day 2.

Okay, well, slightly bigger then. We're still talking 1 - 3% territory.

[Witherweaver]

No, 1-(143/144)^26 = 16.57%.

Chance of getting two of the same is 1/12.  Very likely (90%ish) in 26 chances.

[faust]

No, 1-(143/144)^26 = 16.57%.

Chance of getting two of the same is 1/12.  Very likely (90%ish) in 26 chances.

We're talking of it happening twice on two separate days.

[Witherweaver]

No, 1-(143/144)^26 = 16.57%.

Chance of getting two of the same is 1/12.  Very likely (90%ish) in 26 chances.

We're talking of it happening twice on two separate days.

That's cherry picking details.  If 3X and 2Y both occurred on Day 1 or on Day 2 you wouldn't be making this argument?

[Beyond Awesome]

Okay, in regards to e/yuma, I buy his story for one reason. Box lost two dice, and I lost 6 dice!!!! And, it sounds like PPS lost a similar amount of dice.

I do find this theory of a dice manipulator interesting. I could very well see that being a roll in this game.

[faust]

No, 1-(143/144)^26 = 16.57%.

Chance of getting two of the same is 1/12.  Very likely (90%ish) in 26 chances.

We're talking of it happening twice on two separate days.

That's cherry picking details.  If 3X and 2Y both occurred on Day 1 or on Day 2 you wouldn't be making this argument?

That's right.

[Beyond Awesome]

Also, we can verify if yuma is telling the truth tomorrow when he targets another player, and they state they only lost two dice.

[Witherweaver]

Also, we can verify if yuma is telling the truth tomorrow when he targets another player, and they state they only lost two dice.

It could easily be random.

[Hydrad]

Also, we can verify if yuma is telling the truth tomorrow when he targets another player, and they state they only lost two dice.

It could easily be random.

Interesting I didn't even consider that.

[Beyond Awesome]

Also, we can verify if yuma is telling the truth tomorrow when he targets another player, and they state they only lost two dice.

It could easily be random.

If it is random then why did PPS and I lose a crap ton of dice while Box only had two stolen.

[Witherweaver]

No, 1-(143/144)^26 = 16.57%.

Chance of getting two of the same is 1/12.  Very likely (90%ish) in 26 chances.

We're talking of it happening twice on two separate days.

That's cherry picking details.  If 3X and 2Y both occurred on Day 1 or on Day 2 you wouldn't be making this argument?

That's right.

Alright, I mean I agree that it's possible, but I think {Theif/Destructor}+{Reroller} roles are the more likely scenario. 

[Witherweaver]

Also, we can verify if yuma is telling the truth tomorrow when he targets another player, and they state they only lost two dice.

It could easily be random.

If it is random then why did PPS and I lose a crap ton of dice while Box only had two stolen.

We only have one outcome of Yuma's claim.  Box lost two dice on Night 0. 

[Witherweaver]

Is this the first time you claimed to have lost 6 dice?

[Hydrad]

Also, we can verify if yuma is telling the truth tomorrow when he targets another player, and they state they only lost two dice.

It could easily be random.

If it is random then why did PPS and I lose a crap ton of dice while Box only had two stolen.

maybe the thief roles a 6 sided die to steal things?
not really sure.