His rage over the lack of engineer and scientist roles for himself is giving me a wild throwback to the early days of him playing among us and just never being impostor ever at all 😂 and then the first time he gets it Kara knew it was him immediately 😂
More stamps of time: 0:26 Crewmate 12:57 Scientist 15:18 Crewmate 24:41 Crewmate 37:24 Crewmate 46:39 Impostor with Kara 55:25 Crewmate That's all for this one. Enjoy, and have a great day and/or night!
25:00 My guess would be that the calculation is that it runs a 20% chance for an engineer three times to determine how many engineers there are if at all, and then it randomly selects crewmates up to the engineers amount for them to be engineers.
The lack of tomatoes is relatable. Our local sandwich shop almost always runs out of bread before the end of the day, and stop serving sandwiches. The nearest ski resort does also regularly run out of rental skis at the ski shop, too.
The odds of Jordan getting a role (Scientist, Engineer, Imposter) in any round with this setup is 29.57% The odds of him only getting a role 2 times in these 7 games is 31% (29% if you want to include the last game from the last video where roles were enabled) Edit - they lost a person midway through, and I'm too lazy to account for that, but odds would actually be slightly higher
Why do they no longer use the mod that shows who's speaking in meetings? Without it i don't know everyone's voice so im just confused most of the time. I'd watch more if they used that mod.
@@theanythingguytag3420 oh they're using discord? I thought they were doing in-game, cause there's an among us mod that also lights up the people that are speaking in the meeting
@@theanythingguytag3420 Discord overlay isn't a mod, there is a mod that highlights the person talking in the meeting. The real reason is that they're playing in vanilla.
odds of separated top and bottom: first, get the odds of 5 pairs of values being of opposite polarities; 1st pair, 5/9 2nd pair, 4/7 3rd pair, 3/5 4th pair, 2/3 5th pair, 1/1 so the total odds for those pairs is (1/1) * (2/3) * (3/5) * (4/7) * (5/9) = ~12.7% then calculate the odds for the pairs having the same alignment, which is much easier, just 2^-4 (2^-5, for top being smaller, but he was just talking about having top and bottom separate) so ~12.7 * 2^-4 = ~0.8%, or 1 in 126
So the math for the split depends on whether the top has to be 1-5 or it can be top or bottom. For the top being 1-5, it's (1/2)*(4/9)*(3/8)*(2/7)*(1/6) or 1 in 252 For either the top or the bottom being 1-5 and the other being 6-10, then the math is (4/9)*(3/8)*(2/7)*(1/6) or 1 in 126
I got the same answer, but using a different approach. I thought of this problem in terms of a set partitioning problem. That is, how many ways can you partition the set [1, 10] into subsets [1, 5] and [6, 10] without regard to order. You can partition a set of size 10 into two sets of size 5 in 10 choose 5 ways. 10 choose 5 = 252. Since only one of these is valid, it would be 1/252. This is approximately 0.397%. Now, if you don't specifically care about the order of these subsets, i.e. the top row being [1, 5] and the bottom row being [6, 10] and not accepting the opposite like above, then you just multiply by 2. So, that would be 1/126. This is approximately 0.794%.
My approach was to first separate polarity, then calculate as a simple 2^4//5 first, get the odds of 5 pairs of values being of opposite polarities; 1st pair, 5/9 2nd pair, 4/7 3rd pair, 3/5 4th pair, 2/3 5th pair, 1/1 so the total odds for those pairs is (1/1) * (2/3) * (3/5) * (4/7) * (5/9) = ~12.7% then calculate the odds for the pairs having the same alignment, which is much easier, just 2^-4 (2^-5, for top being smaller, but he was just talking about having top and bottom separate) so ~12.7 * 2^-4 = ~0.8%, or 1 in 126
I'm curious how you got that number. The chances of getting [1, 10] all in order would be 1/(10!) afaik. Which is 1/3,628,800. i.e. approximately 1 in 3.6 billion.
I got a slightly different answer. It's been about a year since the combinatorics unit/section in my discrete math class, so sorry if I'm wrong lol. First, I'll mention for comparison to what I i got that: 120/30420 = 2/507 I thought of this problem in terms of a set partitioning problem. That is, how many ways can you partition the set [1, 10] into subsets [1, 5] and [6, 10] without regard to order. You can partition a set of size 10 into two sets of size 5 in 10 choose 5 ways. 10 choose 5 = 252. Since only one of these is valid, it would be 1/252 = 2/504. This is approximately 0.397%. Now, if you don't specifically care about the order of these subsets, i.e. the top row being [1, 5] and the bottom row being [6, 10] and not accepting the opposite like above, then you just multiply by 2. So, that would be 1/126. This is approximately 0.794%. I'd be very interested in how you arrived at your number/calculation because I could be wrong. Edit: Accidentally forgot that 10 choose 5 gives the one with order between the subsets, so you multiply by 2 to get the version where you don't care. My original answer had 10 choose 5 as the not caring version and thus divided by 2 to get the caring version.
Got the fun size increasing and it extending high and being good to play around with and the crews in nice and the big time growing on nice and the Task Winning big time and the matter of time and yes makin' the scene big time and yes makin' the scene fun and the growing together amazing and yes having a good time altogether and the fun time having the scenes gone amazing and yes the fun times enjoying the scene and thr good growing
His rage over the lack of engineer and scientist roles for himself is giving me a wild throwback to the early days of him playing among us and just never being impostor ever at all 😂 and then the first time he gets it Kara knew it was him immediately 😂
"It was not me because i dont want real life trouble" 😂😂
More stamps of time:
0:26 Crewmate
12:57 Scientist
15:18 Crewmate
24:41 Crewmate
37:24 Crewmate
46:39 Impostor with Kara
55:25 Crewmate
That's all for this one. Enjoy, and have a great day and/or night!
Legend 🎉
Somebody needs to make a compilation of all of the normal crewmate raging.
The one time he got Scientist the impostors just gave up that round at the start
Jordan never gets sussed unless he’s actually the impostor haha
46:49
“the murder house”
*sprinkler noises*
25:00 My guess would be that the calculation is that it runs a 20% chance for an engineer three times to determine how many engineers there are if at all, and then it randomly selects crewmates up to the engineers amount for them to be engineers.
And they were roomates
Vine is better than ticktock
Change my mind
@@newt3546oh my god they were roommates
@@newt3546cant
@@newt3546tiktok is the asian version of vine.
The lack of tomatoes is relatable. Our local sandwich shop almost always runs out of bread before the end of the day, and stop serving sandwiches. The nearest ski resort does also regularly run out of rental skis at the ski shop, too.
the NO U with reverb killed me
42:00
It was so powerful
Jordan: They are out of tomatoes again?
Me: Well, it takes a few months to grow them, so you will have to wait.
He better have been serious about the modded thing, because now my hopes are up.
good news!
The odds of Jordan getting a role (Scientist, Engineer, Imposter) in any round with this setup is 29.57%
The odds of him only getting a role 2 times in these 7 games is 31% (29% if you want to include the last game from the last video where roles were enabled)
Edit - they lost a person midway through, and I'm too lazy to account for that, but odds would actually be slightly higher
Kara has played 108 days of among us 😂
I miss the mod thing displaying names of who is speaking during meetings. I only recognize about half of them by voice alone. D;
32:26 The chance of this happening is around: 0.793%
Lol I was thinking about how Twitter loves making duo names so I propose boos and Jardon become edging duo
I beg your pardon
Missed these among us videos bro
Why do they no longer use the mod that shows who's speaking in meetings? Without it i don't know everyone's voice so im just confused most of the time. I'd watch more if they used that mod.
Discord changed how names work, so now it’s hard to get that overlay to work without leaking the user’s full discord name
@@theanythingguytag3420 oh they're using discord? I thought they were doing in-game, cause there's an among us mod that also lights up the people that are speaking in the meeting
@@theanythingguytag3420 Discord overlay isn't a mod, there is a mod that highlights the person talking in the meeting. The real reason is that they're playing in vanilla.
This is fully vanilla, when Kara and co. play modded they still use it.
Let the record show: tomatoes are, in fact, gross.
odds of separated top and bottom:
first, get the odds of 5 pairs of values being of opposite polarities;
1st pair, 5/9
2nd pair, 4/7
3rd pair, 3/5
4th pair, 2/3
5th pair, 1/1
so the total odds for those pairs is (1/1) * (2/3) * (3/5) * (4/7) * (5/9) = ~12.7%
then calculate the odds for the pairs having the same alignment, which is much easier, just 2^-4 (2^-5, for top being smaller, but he was just talking about having top and bottom separate)
so ~12.7 * 2^-4 = ~0.8%, or 1 in 126
Someone needs to send Jordan a picture of ketchup salad.
We love you Jordan, your videos are amazing ❤❤
So the math for the split depends on whether the top has to be 1-5 or it can be top or bottom.
For the top being 1-5, it's (1/2)*(4/9)*(3/8)*(2/7)*(1/6) or 1 in 252
For either the top or the bottom being 1-5 and the other being 6-10, then the math is (4/9)*(3/8)*(2/7)*(1/6) or 1 in 126
The math shouldn't change between the two as if its 1-5 on the top the bottom has to be 6-10.
@@ChiefArug no if it's 1-5 on the top or 1-5 on either the top or bottom.
I got the same answer, but using a different approach.
I thought of this problem in terms of a set partitioning problem. That is, how many ways can you partition the set [1, 10] into subsets [1, 5] and [6, 10] without regard to order. You can partition a set of size 10 into two sets of size 5 in 10 choose 5 ways. 10 choose 5 = 252. Since only one of these is valid, it would be 1/252. This is approximately 0.397%.
Now, if you don't specifically care about the order of these subsets, i.e. the top row being [1, 5] and the bottom row being [6, 10] and not accepting the opposite like above, then you just multiply by 2. So, that would be 1/126. This is approximately 0.794%.
My approach was to first separate polarity, then calculate as a simple 2^4//5
first, get the odds of 5 pairs of values being of opposite polarities;
1st pair, 5/9
2nd pair, 4/7
3rd pair, 3/5
4th pair, 2/3
5th pair, 1/1
so the total odds for those pairs is (1/1) * (2/3) * (3/5) * (4/7) * (5/9) = ~12.7%
then calculate the odds for the pairs having the same alignment, which is much easier, just 2^-4 (2^-5, for top being smaller, but he was just talking about having top and bottom separate)
so ~12.7 * 2^-4 = ~0.8%, or 1 in 126
11/10 title
Wish he would do town of us
Hey Jardoon! All this is great but where is my fricky towers?
YOO YOU KNOW BAABLU AND BOOSFER!!!
1 out of 10 billion chance to get 1-10 in a row
I'm curious how you got that number. The chances of getting [1, 10] all in order would be 1/(10!) afaik. Which is 1/3,628,800. i.e. approximately 1 in 3.6 billion.
@@lincolnsand5127 10 to the 10th power, which, isn’t including the fact that there is two, so not exactly. That’s just the basic average
Kara grinds!
1:05 Yeah you and dumbdog are like super focused on taskes and I love it but am sad y'all never met yet 😢
Historians are among us
What the fork bro
I agree with cap Tomatoes are good especially for a salad
boos kinda has a valorant accent
These are fun but i miss all the roles
I’m too early for imposter time stamps 😞
Me too
Like where is the g.o.a.t
Mogus
34:00
I believe the odds of it being split top and bottom like that are
120/30,420 or 0.39%
I got a slightly different answer. It's been about a year since the combinatorics unit/section in my discrete math class, so sorry if I'm wrong lol.
First, I'll mention for comparison to what I i got that:
120/30420 = 2/507
I thought of this problem in terms of a set partitioning problem. That is, how many ways can you partition the set [1, 10] into subsets [1, 5] and [6, 10] without regard to order. You can partition a set of size 10 into two sets of size 5 in 10 choose 5 ways. 10 choose 5 = 252. Since only one of these is valid, it would be 1/252 = 2/504. This is approximately 0.397%.
Now, if you don't specifically care about the order of these subsets, i.e. the top row being [1, 5] and the bottom row being [6, 10] and not accepting the opposite like above, then you just multiply by 2. So, that would be 1/126. This is approximately 0.794%.
I'd be very interested in how you arrived at your number/calculation because I could be wrong.
Edit: Accidentally forgot that 10 choose 5 gives the one with order between the subsets, so you multiply by 2 to get the version where you don't care. My original answer had 10 choose 5 as the not caring version and thus divided by 2 to get the caring version.
12:20 welp demonetized again..
The fact the title isn’t roommates are crew mates smh
Got the fun size increasing and it extending high and being good to play around with and the crews in nice and the big time growing on nice and the Task Winning big time and the matter of time and yes makin' the scene big time and yes makin' the scene fun and the growing together amazing and yes having a good time altogether and the fun time having the scenes gone amazing and yes the fun times enjoying the scene and thr good growing
W
Let's go
10 min gang
9th!
is capitan like dating with kara or someting?
You missed the title that says roommates??
@@starlight5765yeah idk how to read thanks for making fun of me now I will go and unalive myself
@@starlight5765in a historical way. ^^
Mogus