TWG 194 - Game Thread

[ShadoWolfe]

Originally Posted by Duskfall

leetic didnt do maths right btw given the info we know, assuming leetics claim is also correct the odds of there being exactly 3 Cs in the setup is actually



~5.25 percent

You were quicker than me, but you're wrong too.

The binomial probability of rolling 3 Cs out of 7, where the chance of each C is 1/10, is roughly 2.296%

I cheated tho I used a calculator
https://stattrek.com/online-calculator/binomial.aspx

[Duskfall]

Originally Posted by ShadoWolfe

You were quicker than me, but you're wrong too.

The binomial probability of rolling 3 Cs out of 7, where the chance of each C is 1/10, is roughly 2.296%

I cheated tho I used a calculator
https://stattrek.com/online-calculator/binomial.aspx

i realised i was wrong after but you are also wrong LOL

its around 1 percent because its 2 cs out of 5 since we already have 1 c and 1 v given

[leetic]

Originally Posted by Duskfall

i realised i was wrong after but you are also wrong LOL

its around 1 percent because its 2 cs out of 5 since we already have 1 c and 1 v given

But the one C and one V were rolled with the rest of the numbers, so leaving them out skews the results

[Duskfall]

Originally Posted by leetic

But the one C and one V were rolled with the rest of the numbers, so leaving them out skews the results

you are calculating the odds of mikey's claim being correct now, not pre roll, and we have more info now so not including it makes the odds of him being that role incorrect because the knowledge you have skews the odds. If you are calculating the pre roll odds of him being C then you would actually need to remove the fact you are already claimed 1 shot cop because that does not effect the result you are trying to look for and you should be account for every scenario thats has more than one C, as all those setups allow for a cop and therefore for mikey to claim cop.

[ShadoWolfe]

Originally Posted by Duskfall

i realised i was wrong after but you are also wrong LOL



its around 1 percent because its 2 cs out of 5 since we already have 1 c and 1 v given

No. It's 3 Cs out of 7. The V just gets treated as a non-C roll and you'd never take out anything because the setup was rolled out of 7

[Duskfall]

Originally Posted by ShadoWolfe

No. It's 3 Cs out of 7. The V just gets treated as a non-C roll and you'd never take out anything because the setup was rolled out of 7

what are you trying to calculate exactly?

[ShadoWolfe]

Originally Posted by Duskfall

what are you trying to calculate exactly?

The same thing leetic originally was. The probability of there being 3 Cs out of 7 rolls