CHOOSE YOUR OWWN IV - Game Thread

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  • Probability exactly 4 of the remaining 47 drawn cards are Vanilla Townie (hypergeometric):
P1=(8/4)(133/43)/(141/47)≈0.1718382592
  • Given 4 Vanilla Townie appear among the 47 slots, probability they occupy 4 distinct other players (choose 4 of the 13 other players, pick one of their 3 slots each):
P2=(13/4)* 3^4/(47/4)≈0.3246993525
  • Combined probability:
P=P1×P2≈0.1718382592×0.3246993525≈0.0557957715

Or 5.58% chance that everyone who is claiming they used "Vanilla Townie" for their affiliation is telling the truth.
 
  • Probability exactly 4 of the remaining 47 drawn cards are Vanilla Townie (hypergeometric):
P1=(8/4)(133/43)/(141/47)≈0.1718382592
  • Given 4 Vanilla Townie appear among the 47 slots, probability they occupy 4 distinct other players (choose 4 of the 13 other players, pick one of their 3 slots each):
P2=(13/4)* 3^4/(47/4)≈0.3246993525
  • Combined probability:
P=P1×P2≈0.1718382592×0.3246993525≈0.0557957715

Or 5.58% chance that everyone who is claiming they used "Vanilla Townie" for their affiliation is telling the truth.
Thank you Janet
 
  • Probability exactly 4 of the remaining 47 drawn cards are Vanilla Townie (hypergeometric):
P1=(8/4)(133/43)/(141/47)≈0.1718382592
  • Given 4 Vanilla Townie appear among the 47 slots, probability they occupy 4 distinct other players (choose 4 of the 13 other players, pick one of their 3 slots each):
P2=(13/4)* 3^4/(47/4)≈0.3246993525
  • Combined probability:
P=P1×P2≈0.1718382592×0.3246993525≈0.0557957715

Or 5.58% chance that everyone who is claiming they used "Vanilla Townie" for their affiliation is telling the truth.
Zenge pls numbers is villains
 
  • Probability exactly 4 of the remaining 47 drawn cards are Vanilla Townie (hypergeometric):
P1=(8/4)(133/43)/(141/47)≈0.1718382592
  • Given 4 Vanilla Townie appear among the 47 slots, probability they occupy 4 distinct other players (choose 4 of the 13 other players, pick one of their 3 slots each):
P2=(13/4)* 3^4/(47/4)≈0.3246993525
  • Combined probability:
P=P1×P2≈0.1718382592×0.3246993525≈0.0557957715

Or 5.58% chance that everyone who is claiming they used "Vanilla Townie" for their affiliation is telling the truth.
Yes exactly, burn all the vanilla towny claims.

Who claimed vanilla townie for aff?
 
  • Probability exactly 4 of the remaining 47 drawn cards are Vanilla Townie (hypergeometric):
P1=(8/4)(133/43)/(141/47)≈0.1718382592
  • Given 4 Vanilla Townie appear among the 47 slots, probability they occupy 4 distinct other players (choose 4 of the 13 other players, pick one of their 3 slots each):
P2=(13/4)* 3^4/(47/4)≈0.3246993525
  • Combined probability:
P=P1×P2≈0.1718382592×0.3246993525≈0.0557957715

Or 5.58% chance that everyone who is claiming they used "Vanilla Townie" for their affiliation is telling the truth.
i didn't expect to see the word hypergeometric in WW but here we are
 
checking blood and oil GIF
 
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  • Probability exactly 4 of the remaining 47 drawn cards are Vanilla Townie (hypergeometric):
P1=(8/4)(133/43)/(141/47)≈0.1718382592
  • Given 4 Vanilla Townie appear among the 47 slots, probability they occupy 4 distinct other players (choose 4 of the 13 other players, pick one of their 3 slots each):
P2=(13/4)* 3^4/(47/4)≈0.3246993525
  • Combined probability:
P=P1×P2≈0.1718382592×0.3246993525≈0.0557957715

Or 5.58% chance that everyone who is claiming they used "Vanilla Townie" for their affiliation is telling the truth.
Too much math, must be mafia
 
I don't recall having a strong read on him day 1 very often. I know it's happened a time or two, when he hit specific tonal points that I've never seen him replicate as scum, but in addition to my being more cautious about that kind of thing in CYO given history, he just wasn't hitting those points in the posts I was reading. samac can maybe confirm because I think we read him from similar perspectives (or at least we rarely end up in disagreement, I think).

And I don't think I've ever arrived at a scum read of him d1.
I do agree, there’s a certain tone I can often catch in his interactions when he’s village and I think I’ve gotten fairly good at reading him. It usually takes me more than a cycle to see it tho
 
  • Probability exactly 4 of the remaining 47 drawn cards are Vanilla Townie (hypergeometric):
P1=(8/4)(133/43)/(141/47)≈0.1718382592
  • Given 4 Vanilla Townie appear among the 47 slots, probability they occupy 4 distinct other players (choose 4 of the 13 other players, pick one of their 3 slots each):
P2=(13/4)* 3^4/(47/4)≈0.3246993525
  • Combined probability:
P=P1×P2≈0.1718382592×0.3246993525≈0.0557957715

Or 5.58% chance that everyone who is claiming they used "Vanilla Townie" for their affiliation is telling the truth.
HEY I ASKED FOR THIS

thanks Zenges
 
  • Probability exactly 4 of the remaining 47 drawn cards are Vanilla Townie (hypergeometric):
P1=(8/4)(133/43)/(141/47)≈0.1718382592
  • Given 4 Vanilla Townie appear among the 47 slots, probability they occupy 4 distinct other players (choose 4 of the 13 other players, pick one of their 3 slots each):
P2=(13/4)* 3^4/(47/4)≈0.3246993525
  • Combined probability:
P=P1×P2≈0.1718382592×0.3246993525≈0.0557957715

Or 5.58% chance that everyone who is claiming they used "Vanilla Townie" for their affiliation is telling the truth.
I read this and was like ok I think my panic attack might have been justified
and then pointed out to myself that I checked none of the math because I'm lazy, so theoretically you could have made numbers up and I'd be like mmm yeah seems legit :laugh:
 
I read this and was like ok I think my panic attack might have been justified
and then pointed out to myself that I checked none of the math because I'm lazy, so theoretically you could have made numbers up and I'd be like mmm yeah seems legit :laugh:
idk because I feel like doing math is well within zenge’s scum range
 
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additional bedtime thought: there might be too many people in my village reads?

I had this thought last night, but then realized the roster was larger than I expected and grew slightly more comfortable.
If multiball don’t we expect more scum than usual? For example good pWWace was 8 scum plus sandbag. Probably not gonna be that wild but I would expect more than usual if we expect scum on scum killing.

Sorry if this was said in the last 15 pages
 
idk because I feel like doing math is well within zenge’s scum range
I think a lot is within his scum range, but I also feel like his behavior at vote close yesterday was town 9 times out of 10, so I'm more willing to go with the vibe of it being either null or slightly towny to do the math.
 
If multiball don’t we expect more scum than usual? For example good pWWace was 8 scum plus sandbag. Probably not gonna be that wild but I would expect more than usual if we expect scum on scum killing.

Sorry if this was said in the last 15 pages
I honestly don't know how the ratio is calculated for the CYO games - it's not the 20-25% scum that a more traditional setup has, I think I remember AM saying that it rerolls if it's not at least 50% village, so there may be more scum than normal, but I think it's less likely to be at the nearly 50/50 ratio that good place was.

And scum/scum killing may be wishful thinking, based on my experiences in the last 2 CYOs. There wasn't scum crossfire in either, which was deeply tragic lol.
 
If multiball don’t we expect more scum than usual? For example good pWWace was 8 scum plus sandbag. Probably not gonna be that wild but I would expect more than usual if we expect scum on scum killing.

Sorry if this was said in the last 15 pages
Not necessarily, since alignments are selected.

We can reasonably expect (n/2)-4 as an upper max with n being the number of players just because if it's more than that we'd presumably get a reroll based on the OP
 
CYO 3 had 10 villagers in a game of 16 players, I don't have time to look back at the others, but I don't recall feeling like there were massively more scum than usual. In theory my number of Village reads is probably fine, but in reality I'm just not that accurate, so I'm saying that I personally probably have too many village reads.
 
  • Probability exactly 4 of the remaining 47 drawn cards are Vanilla Townie (hypergeometric):
P1=(8/4)(133/43)/(141/47)≈0.1718382592
  • Given 4 Vanilla Townie appear among the 47 slots, probability they occupy 4 distinct other players (choose 4 of the 13 other players, pick one of their 3 slots each):
P2=(13/4)* 3^4/(47/4)≈0.3246993525
  • Combined probability:
P=P1×P2≈0.1718382592×0.3246993525≈0.0557957715

Or 5.58% chance that everyone who is claiming they used "Vanilla Townie" for their affiliation is telling the truth.
What i gained from this is that people are liars 👀
 
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I think I'm just going to need to accept that I'm probably going to be permanently behind on the thread. I guess the rarely seen real timing exclusive PSV will be making an appearance.
trust when I say I care much more about this game than theriogenology, but alas