TWG CXV: Just a Numbers Game

[Toby]

No cause we could say that since there's 9 players I can only choose a number divisible by 9. But then perhaps you could only choose a number (divisible by 9) + 1. E.g. I can choose 9,18,27,36 etc. and you can choose 10, 19, 28, 37 etc.

There's no middle range the wolf can pick to guarantee he wolfs me there. And even if the wolf thinks to choose one of mine to get me, it's far more likely he won't wolf me rather than will.

I just figure why not turn something that's unlikely into something impossible

[E. Gadd Industries]

Didn't realize you're from Scotland, huh

Anyway, just to make sure I'm following, you're saying that each person gets a set of multiples to choose from, rather than a specific range? To give an example, if there were 10 people in this game, say that Toby, you'd get all the numbers ending in 1 (1,11,21,31,...) and I'd get all the 2s (2,12,22,32,...) and THC would get all the 3s (3,13,23,33,...); am I understanding correctly?

The way I initially interpreted it (and maybe THC also seems to be interpreting? Correct me if I'm wrong, obvs) is that you're saying we each get a range. Following the same example above, you pick from 1-10, I pick 11-20, THC would pick 21-30. That would clearly allow the wolf to pick and choose, so I could see where he has a problem with that strategy.

[Toby]

Yeahh I'm from Scotland 🏴󠁧󠁢󠁳󠁣󠁴󠁿

Yes as you explained in your example above. Not specific ranges, but set multiples.

If we did it as multiples of [player count] + (a number from 0 to [player count]). If that makes sense

So there are 9 players so I'd do multiples of 9, the next person would do multiples of 9 then +1, and next person multiples of 9 but then + 2.

I get this game is already difficult for wolf to hit 2 people on the same number but I'd prefer impossible

[TheZeldaPianist275]

No cause we could say that since there's 9 players I can only choose a number divisible by 9. But then perhaps you could only choose a number (divisible by 9) + 1. E.g. I can choose 9,18,27,36 etc. and you can choose 10, 19, 28, 37 etc.

There's no middle range the wolf can pick to guarantee he wolfs me there. And even if the wolf thinks to choose one of mine to get me, it's far more likely he won't wolf me rather than will.

I just figure why not turn something that's unlikely into something impossible

The problem is less that that strat allows the wolf to pick and choose *which* player he goes after and more that that it actually *increases* the chance of getting multiple wolfings. If two players are restricted to a group of ten, and the wolf picks the number straddling the line between those tens, then it's much more likely that two people die than in a situation where everyone is picking a number scattershot. THC is right that everyone picking in an uncoordinated fashion already makes multiple wolfings extremely unlikely. To my eye, the most dangerous thing we could have happen is have multiple people saying "huh, looks like if I pick 0 or 100 I'm only vulnerable from one side, I guess I'll pick one of those." Having everyone swing toward the extremes would be dangerous and we should avoid that.

No cause we could say that since there's 9 players I can only choose a number divisible by 9. But then perhaps you could only choose a number (divisible by 9) + 1. E.g. I can choose 9,18,27,36 etc. and you can choose 10, 19, 28, 37 etc.

This is better, but still doesn't make it "impossible," as you say. Say in this situation I am assigned numbers 9n, you are assigned numbers 9n+1, E. Gadd has 9n+2, and THC has 9n+3. In this situation my numbers are (9, 18, 27, 36, 45, 54...) and E. Gadd's are (11, 20, 29, 38, 47, 56...). The wolf could pick a number exactly in between our number sets (such as 10, 15, 19, 24, 28,...) to try to tag both of us. (same thing for you and THC in this example but I don't feel like typing that out). It is still very unlikely, but I think my point is that it still gives the wolf *some* set of information to work off of.

Ultimately it's star-aligning chance that would ever result in multiple wolfings unless we make it easy for the wolf.

[A# Minor]

This is better, but still doesn't make it "impossible," as you say. Say in this situation I am assigned numbers 9n, you are assigned numbers 9n+1, E. Gadd has 9n+2, and THC has 9n+3. In this situation my numbers are (9, 18, 27, 36, 45, 54...) and E. Gadd's are (11, 20, 29, 38, 47, 56...). The wolf could pick a number exactly in between our number sets (such as 10, 15, 19, 24, 28,...) to try to tag both of us. (same thing for you and THC in this example but I don't feel like typing that out). It is still very unlikely, but I think my point is that it still gives the wolf *some* set of information to work off of.

Ultimately it's star-aligning chance that would ever result in multiple wolfings unless we make it easy for the wolf.

^ This is what I was trying to write out a minute ago. That and it's still very possible to pick someone else's number because multiples do line up, you know.


OK bye

[ThatHiddenCharacter]

Not specific ranges, but set multiples.

For example, tonight each player could have one number from 1-10, one number from 11-20, one from 21-30... so on so fourth.

You only changed your idea after I called you out on it. Your original suggestion was ranges. It wasn't just my understanding, it was your exact words.

[Toby]

Isn't there like an 11% chance that 2 people are gonna choose the same number tonight. Why not make that 0%.

My plan would never have 2 people picking the same number so if anyone is interpreting that then I may have not explained correctly.

And sure if the wolf really wanted say e gadd and Zelda and they were picking from 9. 18, 27... or 10, 19, 28 retrospectively, the wolf could target those numbers, but at moment that's still only like a 11% chance to get either of them, which is the same probability as it would be for 2 people to pick the same number. And then I guess what % chance is it that they're the ones picked lol.

I mean guaranteeing 1 kill is better than 2 imo. And the wolfs kill is still a shot in the dark cause in the above example he could pick 27 and neither egadd or Zelda picked a number in the 20s.

We could even arrange a list specifically so 2 people we wanted kept alive had a lower possibility of choosing numbers close to eachother. Say we didn't want the wolf to focus egadd and zelda, then we could have egadd do 9n+1 numbers and zelda do 9n+8. Just lowers that possibility but further


Omg @BDS what happens if the numbers are the exact same distance away? Say someone picks 6 and someone picks 8 and the wolf picks 7, who dies ?

[Toby]

You only changed your idea after I called you out on it. Your original suggestion was ranges. It wasn't just my understanding, it was your exact words.

I understand the confusion it wasn't explained the best, but the intention was not ranges. As I said each player could have one number from 1-10, one number from 11-20.

I didn't say they choose one number from that range, I meant they have a number for their selection.

Saying it goes up in multiples was a better way to put it, so I adopted that phrase. As my plan suggested we were still spreading our number choose maximally, and still keep it hard for the wolf to target a specific player. Those things wouldn't align if the plan was each player selected from a range, hence there must have been a misinterpretation due to my unclear explanation

[TheZeldaPianist275]

In case of a tie, all tied players die.

This is my objection to your plan Toby--multiple wolfings are much more likely to happen via equidistant number picks to the wolf's number than via exact same number picks

[ThatHiddenCharacter]

Isn't there like an 11% chance that 2 people are gonna choose the same number tonight. Why not make that 0%.

I'm fairly confident that math is incorrect. Out of a group of two people, the chance of them picking the same number from 1-100 (let's leave 0 out for the sake of making the math a bit easier, just assume the final answer here is more likely than the actual answer for our situation) is 1/10,000. Now, with a group of nine people, the likelihood increases, but definitely not from 0.01% to 11%. I don't know the exact number, but I would definitely say it's still less than 1%.

[Toby]

This is my objection to your plan Toby--multiple wolfings are much more likely to happen via equidistant number picks to the wolf's number than via exact same number picks

Ohh ok I think I agree lol

I assumed the only way for multiple night kills was from picking the same number, but equal distance apart higher and lower also works rip

[Toby]

I'm fairly confident that math is incorrect. Out of a group of two people, the chance of them picking the same number from 1-100 (let's leave 0 out for the sake of making the math a bit easier, just assume the final answer here is more likely than the actual answer for our situation) is 1/10,000. Now, with a group of nine people, the likelihood increases, but definitely not from 0.01% to 11%. I don't know the exact number, but I would definitely say it's still less than 1%.

I don't think I agree with my plan anymore but I'm sure the easiest way to look at the probability of above would be what's the chances of player 8 choosing 7 of the numbers already picked from 1-100. Isn't it like 7% then? If we count 8 humans picking numbers

[TheZeldaPianist275]

I'm fairly confident that math is incorrect. Out of a group of two people, the chance of them picking the same number from 1-100 (let's leave 0 out for the sake of making the math a bit easier, just assume the final answer here is more likely than the actual answer for our situation) is 1/10,000. Now, with a group of nine people, the likelihood increases, but definitely not from 0.01% to 11%. I don't know the exact number, but I would definitely say it's still less than 1%.

Not quite! The likelihood of two players picking a SPECIFIC same number from 1-100 is 10,000. However, since there are 100 different options for two players to pick the same number, and also eight human players, the odds are quite a bit higher.

Also

Ultimately it's star-aligning chance that would ever result in multiple wolfings unless we make it easy for the wolf.

I am changing my opinion here because I posted this without actually crunching any numbers

Assuming pure random distribution, the likelihood that human player picks a unique number is (1)(100/101)(99/101)(98/101)(97/101)(96/101)(95/101)(94/101) = 0.752499. So that's almost a 25% chance that two people pick the same number if we're all firing randomly, and that's even before you get into the probability that two people could pick equidistant numbers from the wolf's pick (a much more intensive calculation). Still unlikely, but very far from star-aligning chance.

[TheZeldaPianist275]

If anyone doubts my math here, I invite you to take a look at the birthday problem. It's a famous combinatorics result that says that in a pool of randomly chosen people, you only need to have 23 before you have better than even odds that two people have the same birthday. Doesn't seem plausible, but that's the way the numbers shake out.

Something similar here--with only two people picking numbers, the odds are infinitesimal that they'd pick the same number, but with eight people the chances are actually quite a bit higher.

[XiaoMigros]

scotland forever!!!