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And now people know that they should just trust me instead.That is exactly why I jumped on the chance to cause doubt. It was unplanned, but when opportunity strikes…
Eric > Erik
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And now people know that they should just trust me instead.That is exactly why I jumped on the chance to cause doubt. It was unplanned, but when opportunity strikes…
The math doesn’t mathAnd now people know that they should just trust me instead.
Eric > Erik
Any post of yours longer than a single sentence should result in you being shot in game.The math doesn’t math
The only known is 2•Eric > Erik
So the only thing we can say with certainty is that Eric > 0.5•Erik.
Could Eric be greater than Erik? Possibly, but we cannot prove that with the data provided
question: how to memes/gifs fit into this restriction?Any post of yours longer than a single sentence should result in you being shot in game.
Can test this out over the coming games.question: how to memes/gifs fit into this restriction?
Challenge acceptedCan test this out over the coming games.
Why is the two Doctors thing sus in the first place? It's really not that remarkable, and she didn't even have to lie about being able to protect others.
Dimmerwit, sounds like we need some stats on how often we get two of the same role.Rarely get two of the same roles, and even more rare get two double roles (dead body was a seer as well)
Yes it happens, but odds are less likely so as a math person I play the odds
We once had 3 Eric/ks with the exact same roleDimmerwit, sounds like we need some stats on how often we get two of the same role.
I bet it's a lot more likely than you think, Quagmire. Have you ever heard of the birthday paradox/problem? It's very similar in concept: how likely is it that two people in a group will share a birthday? There are 365 (technically 366 but whatever) possibilities, so you'd think a group of twenty or thirty people almost never would. However, you only need 23 people for the chance to be 50% that at least two people will share a birthday. An explanation of the math: https://en.m.wikipedia.org/wiki/Birthday_problem
We have far fewer role possibilities, so even with fewer people, I bet the likelihood of a double role is quite high. Somebody less lazy than me can do the math if they like.
Mine. First we wanted Ants to look guilty because he wouldn’t be able to prove himself, then it was pointed out that Ants attacks first, so we decided to have 2 townies die. FR would be proven and we didn’t know his role. We were concerned he was the last strong roleHey, whose idea was it to kill Fools the same night Ants said he'd shoot him? That was hilarious.
Kind of mine. I suggested it to Christina in the lovers chat and she passed it along to wolf chat that Fools was like an unprotected strong villager.Hey, whose idea was it to kill Fools the same night Ants said he'd shoot him? That was hilarious.
Independent lines of thinking. I suggested killing him first then Christina supported it. I don’t remember who brought up the idea he might be the final strong villagerKind of mine. I suggested it to Christina in the lovers chat and she passed it along to wolf chat that Fools was like an unprotected strong villager.
It’s his fault for being an alarm clock. I always silence my alarm clockIt was also just the right move. If you know he's about to be 100% confirmed it means you have one less confirmed townie out there.
Sad state of affairs though when you survive all day 1 and can't post only to be killed night 2 lol. Still, got some use out of the role.
Dimmerwit, sounds like we need some stats on how often we get two of the same role.
I bet it's a lot more likely than you think, Quagmire. Have you ever heard of the birthday paradox/problem? It's very similar in concept: how likely is it that two people in a group will share a birthday? There are 365 (technically 366 but whatever) possibilities, so you'd think a group of twenty or thirty people almost never would. However, you only need 23 people for the chance to be 50% that at least two people will share a birthday. An explanation of the math: https://en.m.wikipedia.org/wiki/Birthday_problem
We have far fewer role possibilities, so even with fewer people, I bet the likelihood of a double role is quite high. Somebody less lazy than me can do the math if they like.
Dimmerwit, sounds like we need some stats on how often we get two of the same role.
I bet it's a lot more likely than you think, Quagmire. Have you ever heard of the birthday paradox/problem? It's very similar in concept: how likely is it that two people in a group will share a birthday? There are 365 (technically 366 but whatever) possibilities, so you'd think a group of twenty or thirty people almost never would. However, you only need 23 people for the chance to be 50% that at least two people will share a birthday. An explanation of the math: https://en.m.wikipedia.org/wiki/Birthday_problem
We have far fewer role possibilities, so even with fewer people, I bet the likelihood of a double role is quite high. Somebody less lazy than me can do the math if they like.
Town | |
Z1 | 1 |
Z2 | 1 |
Z3 | 2 |
Z4 | 2 |
Z5 | 1 |
Z6 | 1 |
Z7 | 1 |
Z8 | 0 |
Z9 | 2 |
Z10 | 1 |
Z11 | 1 |
Z12 | 1 |
Z13 | 2 |
Z14 | 2 |
Z15 | 0 |
Z16 | 1 |
Z17 | 2 |
Z18 | 0 |
Z19 | 0 |
BW2 | 0 |
BW3 | 1 |
Z20 | 1 |
KatWolf | 0 |
Z21 | 1 |
BW4 | 0 |
Ben1 | 0 |
PPT | 0 |
BW5 | 0 |
BW6 | 0 |
BW7 | 1 |
I feel vindicated. I wonder if benzine is generating his roles in some way which makes duplicates less likely.
Town Z1 1 Z2 1 Z3 2 Z4 2 Z5 1 Z6 1 Z7 1 Z8 0 Z9 2 Z10 1 Z11 1 Z12 1 Z13 2 Z14 2 Z15 0 Z16 1 Z17 2 Z18 0 Z19 0 BW2 0 BW3 1 Z20 1 KatWolf 0 Z21 1 BW4 0 Ben1 0 PPT 0 BW5 0 BW6 0 BW7 1
Only 6 games out of over 30 had more than one duplicate role though. This game already had duplicate roles in aura seer before the duplicate doctor game.I feel vindicated. I wonder if benzine is generating his roles in some way which makes duplicates less likely.
I use Excel to generate lists of each type of role in the correct number (so if there are four wolves, I randomly pick from the possible wolf roles four times). If an invalid combo comes up (like two jailers), then I reroll until it's valid. Then I generate a random number for each selected role, then order the roles by that number, to mix them up. Then I assign them in that order to people in alphabetical order.How would one generate roles?
I went down the list on mine, assigning people to each unique role to ensure each role got used, then rolled for which one was to be duplicated
The numbers are the numbers of pairs (or triples, which has happened once).
The odds are a bit lower in newer games since new roles have been added, though the run of zeros is interesting. I'll check that manually. Edit: seems correct.
Surely wolf day 5 is coming soon
The most communist compound
communist