askerian | teamwork needs your help!

Never mind, ignore my deleted comments >> it is a few hours past my bedtime.

Anyway I'm donig the smart thing and e-mailing this question off to a friend instead of doing the dumb thing and muddling through this with no sleep while drunk.

Give me a moment, Probability and Statistics was last year and I don't have my book on hand.

Okay, order doesn't matter here, so we have 16 choose 2 for each fight, that's 120 different ways to match up participants for a battle. We have 3 members in Team Seven, so there are only 3 ways to pare 2 of them up in a fight. So that should be 3/120 or 2.5% chance that 2 of them are matched up.

With the next stage you have 8 choose 2 for a match, so that's 28 different ways to set up a match. So if we still have all of Team Seven that's 3/28 or 10.7%.

I'll re-check that logic in a bit, I just need to find my old notes.

Well here's my stab at the logic:

Each person has the potential to be matched with 15 other people. A person from team 7 has 2 out of those 15 chances of being matched with a teammate. So the odds are 13.33%.

If they all make it into the next round, they have 2 out of 7, or 28.57% chance.

Note, never took statistics, this is straight up "2 chances in 15 is 13.33%" logic.

I haven't done anything like this for a class since seventh grade, but I muddled through it on paper and came up with the same results. If that means anything. XD;;

6 / (16x15) = 6/240 = 3/120 = 3 / (15+14+13+12+11+10+9+8+7+6+5+4+3+2+1+0) = 0.025 = 2.5%
6 / (8x7) = 6/56 = 3/28 = 3 / (7+6+5+4+3+2+1+0) = 0.10714285... = 10.7etc.%

Okay! I am the slow, but at least we all seem to agree--well, except for Tephra, but I think she was coming at it from a different angle than the rest of us. With sixteen people, you can match them into 120 different pairs, three of which involve two members of Team Seven; with eight, it's 28 distinct pairings, et cetera for the 2.5% and 10.7% numbers. However, from the perspective of a member of Team Seven, there are only fifteen (or seven) people that you could be matched with, two of whom are your teammates; 2/15 and 2/7 are about 13.3% and 28.6%, respectively. So, um. I'm confused now, but I don't think anyone's wrong, exactly. @_@;;

I guess the 2.5/10.7 bit is for ALL the possible matches, and the 13.3/28.6 bit is for JUST Team Seven's matches? Oy. *headdesk*

And this is why most people find statistics confusing, you can come at it from multiple angles, get multiple answers, and all of them are still correct! xD

Okay, the 120 and 28 numbers are the possible match-ups for the first fight of that stage, the numbers will get lower as we go through and pair people off:
first match: 120 possibilities (16 choose 2)
second: 91 possibilities (14 choose 2)
third: 66 possibilities (12 choose 2)
and so forth.

Now, for the 120 possibilities for the first match-up, 3 of those involve both fighters being from Team Seven, 39 are a member of Team Seven and a member of another team, and the other 78 possibilities are match-ups of just the other participants.
So for the first fight, we have a 2.5% chance of two Team Seven members, a 32.5% chance of a Team Seven verse someone not on there team, and a 65% chance of no-one from Team Seven being in the fight.

Now depending on which one of those happens we will get different numbers to work with for the next match, I'll just chug out the numbers for the second two cases, since in the first match has two Team Seven members, the rest of the fights don't really matter.

Case I: First fight has two non-Team Seven ninja.
There are 14 fighters left to pick from, 3 are from Team Seven. There are 91 possible outcomes, 3 are Team Seven verse Team Seven (3.30%), 33 are Team Seven verses another team (36.26%), and 55 where neither fighter is from Team Seven (60.44%).
So basically every time two non-Team Seven ninja are paired for a fight the chance of two Team Seven members having to go at it goes up.

Case II: First fight has one Team Seven member and one non-Team Seven ninja.
Still 14 fighters, still 91 possible outcomes, but only 2 left from Team Seven. So there is only 1 outcome that involves two Team Seven members (1.10%), 24 outcomes for Team Seven verse another team (26.37%), and 66 are neither fighter being from Team Seven (72.5%).
So basically the first time a Team Seven member is paired of with a non-member, the chance of the other two having to fight drops.

I can go through the rest of the numbers for stage one and the numbers for stage 2 if you want, but I think I might have already lost some people. xD

I BOW IN AWE OF YOU. *bows*

I forgot math years ago. XD This is so cool. I'm singling you out because you put it best instead of throwing out numbers without reasoning. *hi-fives*

Now watch her not use it. X3

Thank you, this is one of the many (or is it few?) perks of being raised by two applied mathematicians turned teachers.

Next up, the numbers from Team Seven's perspective.

Okay, amuse the first match in stage one has at least one Team Seven member. We have a total of 42 outcomes, 3 with both fighters from Team Seven (7.14%), and 39 with just one (92.86%). Remember, this is if this is the first match, the numbers will change if non-Team Seven ninjas have been paired off already.

Now, if the second match also has a Team Seven member (and the first match only had one) we have 25 different possibilities for this fight, 1 where it's two Team Seven members (4.0%), and 24 that are Team Seven verses another team (96.0%) Again, the numbers will change a bit depending on how many non-Team Seven members have been paired up.

Basically what all these numbers mean is that for the first stage it is unlikely we will have a fight between two Team Seven members. Next post will be numbers for stage two.

Alright, first off, that should be "assume the first match in stage one" and not "amuse". This would be one of the downsides to being raised by mathematicians, my writing and spelling skills leave much to be desired.

For stage two we have 8 fighters:
first match: 28 possibilities (8 choose 2)
second: 15 possibilities (6 choose 2)
third: 6 possibilities (4 choose 2)
fourth: 1 possibility (2 choose 2)

Now, for those 28 possibilities for the first match, 3 have both fighters from Team Seven (10.71%), 15 with Team Seven verses another team (53.57%), and 10 with neither fighter from Team Seven (35.71%). See that the numbers are getting less in our favor.

Now for the second match.
Case I: the first match had no Team Seven members.
We have 6 fighters left, 3 from Team Seven.There are 15 possible match-ups, 3 with two Team Seven members (20.0%), 9 with Team Seven verse another team (60.0%), and 3 with neither fighter being a Team Seven member (20%).
Again, as more non-Team Seven members are paired off, the chance of two Team Seven members fighting goes up, but in this stage having the chance almost double is a whole lot worse then in stage one.

Case II: the first match had only one Team Seven member.
We have 6 fighters left, 2 from Team Seven, 15 possible match-ups. There is only 1 possibility for a Team Seven verse Team Seven match (6.67%), 8 are Team Seven verse another team (53.3%), and 6 where neither fighter is from Team Seven (40.0%).
So the chance goes down for two Team Seven members having to fight, but not as greatly as it did in stage one where it dropped by half.

Hmm, doesn't the second match odds assume the draw is random again? Wouldn't the outcome of the previous matches, or better the match-order influence the numbers?

Or does that unnecessarily complicate matters?

If the draw is not truly random than all this statistics just got thrown out the window.

For these numbers I have separated the fighters into two groups: Team Seven members and non-Team Seven members, the previous draws only effect how many of each group we have left to draw from for the next match.

The order you pick the names for any particular match don't matter, Steve verses Dave is the same fight as Dave verses Steve. ....Not that any of these ninja are likely named Dave or Steve.

No, I meant the order the matches are decided. After all, after the first match the only question remaining for the second match is who wins.

So if Naruto and Enemy 1 are drawn for match one and Sakura and Enemy 2 are drawn for match two, the chance that they are meeting is... well, I think we would need to know more about their enemies to say for sure, but I think you get my point?^^

Still doesn't change the numbers. For each match there are three different possibilities: two Team Seven Members, one Team Seven member, or zero Team Seven members. My numbers for stage two are the chance of each of those events happening during those 4 matches. Please note that no-one will be fighting twice in either stage.
If this is set up in a bracket system, stage one will be the first round of fights (8 matches) and stage two will be the second round (4 matches.

Okay, this should be the last part unless you need me to expand on anything, stage two battle numbers that Team Seven cares about.

Okay, first match in stage two assuming there is at least one Team Seven member in the match, we have 18 possible outcomes, 3 have both fighters from Team Seven (16.67%), and 15 with only one Team Seven member (83.33%).

Now if the second match also has a Team Seven member (and the first match had only one) there are 9 outcomes, 1 with two Team Seven members (11.11%), and 8 with only one (88.89%).

Like with stage one, these numbers will change a bit if non-Team Seven members have already been matched-up.

So what this is all summing up to is that for stage two Team Seven needs to cross their fingers and hope their names get pulled before too many other names are matched up.

I really want to help.

And the math is hurting my brain.

erm, lurker here, very much a fan of your fics (namely teamwork, yup), i suck at maths, but i believe what you have here is a problem of combinations/permutations combined with probability?

so (don't take my word for granted, it's been years since i worked with this stuff too), if you divide "number of combinations with any 2 of the 3 team 7 members" by "total number of possible combinations", you should have your probability, right?

so since you have 16 people, divided into groups of two, (you said "paired", right?) if i'm doing the maths correctly, it means you have 120 possible combinations. (formula i'm using: 16! / 2!(16-2)!=120)

If you do the same thing with Team 7, you find you have 3 possible pairs. So to calculate the probability of any 2 of the team ending up together you go for 3/120, which is 1/40, or 2,5%.

Doing the same thing for the next stage, you have 28 possible pairs, so 3/28, which is roughly 11%.

i just hope that's the right answer... XD *goes back to lurking*

Je dirais que ça dépend de à quel moment Kakashi fait le commentaire ! :D Avant le début du tirage au sort, chaque concurrent a 15 adversaires potentiels, il a donc 1/15 de chance de rencontrer une personne en particulier (et donc, chaque membre de ta Team a 2x1/15 chance de rencontrer un coéquipier). (2/15, c'est 13,3%)

Mais dès que le tirage commence, ça devient plus complexe : A personne a tjs 1/15 chance de rencontrer B au premier tour, et ainsi de suite, tant que des "paires" ne sont pas fixées ! Au fur et à mesure que les places se remplissent, et se de façon aléatoire, le calcul change.

Mettons qu'il y a 8 adversaires, donc 8 places, répartie en couple (1-2), (3-4), etc.

A -> 1/15 de chance de rencontrer F ; A tire la place n°1
B -> 1/15 de chance de rencontrer F aussi (car B peut rencontrer A, qui n'a pas encore d'adversaire attitré) ; B tire la place n°3
C -> 1/15 de chance de rencontrer F (idem) ; C tire la place n°2
D -> 1/13 de chance de rencontrer F (puisque A et C sont déjà les adversaires de l'autre, ça les retire du compte)
Donc, le calcul doit être complexe puisqu'il prend en compte le résultat de chaque tirage au sort. Ca doit aussi prendre en compte dans quel ordre les concurrents prennent leur papier dans le chapeau : plus ils tirent tôt, moins ils ont de chance de rencontrer un adversaire en particulier/
(Là, honnêtement, c'est largement au delà de mes souvenirs de stats v.v)

C'est aussi complexe au 2e tour (... je crois) (ou au contraire bcp plus simple) parce que l'ordre de passage a été fixé au premier tour : donc, là, l'ordre entre en ligne de compte, on n'est plus dans un cas de hasard complet. (Le plus simple, alors, je dirais, est que tu fasses un tableau en définissant qui rencontre qui ! :D) Parce que la probabilité de rencontrer un coéquipier au 2e tour inclut aussi quelle probabilité ledit coéquipier avait de gagner au premier tour. Ce n'est pas 1 chance sur 2 ! Donc... fais ton tableau !

I think Rag Doll Witch is the mathematician you need. ^^
All of what they said flew right over my head while still making sense.

Here's another way to look at it:

Assuming a standard 16 person single elimination bracket and that Team 7 wins all matches against non-Team 7 people, the longest you can delay a T7 v T7 match up is the first fight of the third round (game 13 out of 15--which is actually a lot better than I first thought). The winner of that round will then face off with the other T7 person for the final fight.

Also a lurker, and not sure if this is relevant, I am only remembering a traumatic experience involving a stats class.

Do we know the drawing method? Do we know how the original chunnin games worked? I don't remember clearly, but I think there were two pools, at some point. Presumably, this is not the case in asuka's exam, but if it is, the chances of one member of team seven facing a member in the opposite pool increase significantly. The only reason the percentages are so low is that "sakura vs. mook" and "mook vs. sakura" are two different and valid options affecting completely random selection. Not so in a two pool system.

...I think. This should probably be checked, it's entirely possible I'm just having a flashback to a horribly frustrating test.

Also: hi! I used to be vhasbls. Hilarious offline shenanigans have prompted a change. Such fun! Also, I have friended you in an attempt to stop lurking. *waves*

Didn't have time to read through other comments, but:

1/16 x 1/16 = 0.00390625% chance.

Based on my high school stats course. I enjoyed it.

Assuming viewpoint of Naruto (just because) for a moment:

1/15 chance he will meet Sakura, 1/15 Sasuke. Combined 2/15 chance he will fight someone in Team 7 first round.

Next round: 1/7, 1/7, combined 2/7.

So, in other words, 13.34% chance in first round and 28.57% in second round.

Also, I didn't read all the other comments, but I can't believe how many different answers there are... wow.

hah the way to do this is to find the opposite: what is the probability that 2 members of team seven do have fight each other?
We assume that a person has an equal chance of being picked as any other person. (aka we are picking names out of a hat) and that every person is distinct

Round 1:
How many ways can we pick matches such that 2 members of team seven fight each other?

first we pick the two team seven members that have to fight each other (3 choose 2)
then we pick the match they will fight in (8 choose 1)
then we pick the people that fight in the first match that is not the special match (14 choose 2)
then the second non special match (12 choose 2)
etc.
till the last match (2 choose 2)

so math:
(3 2) * (8 1) * (14 2) * (12 2) * (10 2) * (8 2) * (6 2) * (4 2) * (2 2) =
3* 8 * 91 * 66 * 45 * 28 * 15 * 6 * 1 = 16 345 929 600

to get the probability, we compute the total number of possibilities which is first match (16 choose 2)* second match (14 choose 2) * ... *last match (2 choose 2)

(16 2) * (14 2) * (12 2) * (10 2) * (8 2) * (6 2) * (4 2) * (2 2) =
120 * 91 * 66 * 45 * 28 * 15 * 6 * 1 = 81 729 648 000

so the probability of 2 members of team seven having to fight each other is
16 345 929 600 / 81 729 648 000 = .2 or 20%

Thus for round 1 the probability of them not having to fight each other is 80%

For round 2, if all 3 members of team seven advanced:

How many ways can we pick the matches so that 2 members of team seven have to fight eachother?

(3 choose 2) - which 2 members fight
(4 choose 1) - which match they fight in
(6 choose 2) - first non special match
(4 choose 2) - second non special match
(2 choose 2) - last match

(3 2) * (4 1) * (6 2) * (4 2) * (2 2)
3 * 4 * 15 * 6 *1 = 1080

total ways to match up all 8 people for 4 matches:
(8 choose 2) * (6 choose 2) * (4 choose 2) * (2 choose 2) = 28 * 15 * 6 * 1 = 2520

so the probability of 2 members of team seven having to fight each other is
1080/2520 = 0.428571429 or 43%

Thus for round 2 the probability of them not having to fight each other is 53%

Continuing in this manner (because I was interested...)

Round 2 if only 2 members of team seven advanced:

(2 choose 2) - which 2 team 7 members fight
(4 choose 1) - which match they fight in
(6 choose 2) - first non team 7 match
(4 choose 2) - second non team 7 match
(2 choose 2) - third non team 7 match

= 360

so 360 / 2520 = 0.142857143,or 14%, so there is a 86% chance that the 2 members of team 7 would not have to fight eachother.

Round 3: (semi-finals)
obviously if all 3 team 7 members are here, 2 will have to fight eachother.

if only 2 team 7 members made it however,

(2 choose 1) - which match they fight in
(2 choose 2) - other match
2 * 1
2

total ways to match up 4 people into 2 matches:
(4 choose 2) * (2 choose 2)
6 * 1
6

so 2/6 or 33% chance that 2 members of team seven fight so 67% chance that they won't have too :)

Um that of course is if they randomized the matches every round...

If they didn't:

Round 2:
what we want to find is which bracket the 3 members of Team 7 are in. There are 4 brackets: match 1:2, match 3:4, match 5:6, and match 7:8

The total number of ways to arrange 16 people into 4 distinct groups is
(16 choose 4)(12 choose 4)(8 choose 4)(4 choose 4) = 63 063 000

The number of ways the three members of team 7 can be in 3 separate brackets:

(4 choose 3) - choose the brackets they are in
(3!) - permute team 7 across those 3 brackets.
(13 choose 4) - 4 people for the bracket team seven is not in
(9 choose 3) - 3 people for one of the brackets team 7 is in
(6 choose 3) - ""
(3 choose 3) - ""
= 28 828 800

so assuming Team 7 won every match, the probability that they would not have to fight in round 2 would be 28 828 800/ 63 063 000 = 0.457142857 or 46%

Please ignore my previous post, I made a mistake.

1) It's easier to figure out the probability of non of them meeting. So, lets start with UN. UN will be paired up with 1 of 15 people, 2 of which are teammates, so the probability of him pairing up with neither is 13/15.
2) Now that UN isn't going to fight a teammate we are left with US and HS. US will be paired up with 1 of 13 people, 1 of which is a teammate. So US has a 12/13 chance of not fighting HS.
3)So the probability that non of them will fight each other is (13/15)*(12/13)=0.8
4) The opposite of non of them fighting is that two of them fight, so the probability of that happening is 1-0.8=0.2

It's the same for the second round. 1-((5/7)*(4/5))=0.4285714

I think that you're definitely on to something. This is a smarter, creative way to look at it. I'm reasonably sure you're right, although I have no statistical knowledge, just high level mathematics. But I'm going to second this idea.

Thank you! It's the same answer as the person above me, but a little shorter. My stats teacher liked to show us shortcuts like that.

All I know for sure is that the odds are better than them having a kyuubi induced orgy that results in pregnancy and a long term three way relationship....so I wouldn't worry about the odds of stuff happening too much.

Lol, I'm a long time lurker here, but I found this post amusing since I just had a stats midterm today (too much stats...arg). You seem to have a few different responses, but I have to agree with the 2.5% and 10.7% ones.

So I'm assuming that Kakashi is telling this to them right before anything is chosen and, if so, it's just the 2.5% and 10.7% values that you'd be interested in. If, however, two ppl are chosen at a time, then those two are fighting, then *after* that another two are chosen (ie. they're not all paired up at the start) and for some reason you decide that Kakashi is going to tell them how the odds have changed after every single match (don't know why you'd do this--maybe Kakashi is trying to be funny), then go with the subsequent values rag_doll_witch put down.

First round: 16 ppl, 120 different ways to pair them (16 choose 2--> this only means that order doesn't matter. Picking A from a hat then B is the same as B then A). There are 3 ways the trio can be paired (N&Sas, N&Sak, or Sas&Sak). So probability is 3/120 * 100% = 2.5%
This would be the value Kakashi says before they start pairing people in round one.

Second round: By now all 16 have fought and 8 are left, 3 of which are team 7. 28 different ways to pair them up (8 choose 2) and 3 ways the trio can be paired up. Probability is then 3/28 * 100% = 10.7%
This would be the value Kakashi tells them after round one but before round two started.

Whoops, I just realized that I probably misread your question (Just to add more confusion to all the different answers! :P) The numbers I gave above were the probabilities for just the *first* match of each round. I realized you probably want the *total* probability that any teammate meets another in the *whole* round (not just a match), which would give different numbers. Lol, you might not even make Kakashi say anything, but I just don't want to give someone the wrong answers :).

Just because it'd take too long to show it, the prob that any teammate would meet another in Round 1 (so for 16ppl and 8 matches) is 29.3%. The prob that they would meet in Round 2 (8ppl, 4 matches) is 42.9%

To show why the numbers can all be different, take 4ppl A,B,C, and D and A,B, and C are on the same team (* means teammates). The different match-ups can be *AB*/CD, *AC*/BD, or AD/*BC*. In the *first* match, there's a 50% chance of the teammates fighting each other. Member A, for example, has a 2/3 or 67% chance of fighting a teammate. AND we can see that there's a 100% chance (total prob) that *any* teammate will be paired with another teammate. So the new numbers I gave above are the *total* prob, not just for the first match.

That's what I wanted to know! ^o^~ Thank you very much.

Hey,

It's most likely does not matter, but the correct answer to the question is: it's 1/7 chance in the first round and 1/3 chance in the second.

There is a nice graphical way to explain it, maybe somebody as boring as Kakashi would go for it :)

Let's arrange the 16 fighters in 2 rows like that, to represent the fighting pairs:
XXXXXXXX
XXXXXXXX

Because we have 3 people in team 7, either first row or second row will have 2 or 3 Team 7 members in it, right? So let's take that row, put it on the top, and place the first 2 Team 7 members on the left side. We'll represent them with O:

OOXXXXXX
XXXXXXXX

So, now the probability of the 3rd member of Team 7 to be paired with first or second is: 2 places (2 X in the second row on the left) devided by 14 (total number of places) = 2/14 = 1/7.

Same schema works for the second round, which looks like:

OOXX
XXXX

So there the result is 2/6 = 1/3

Cheers :)

Hm, the first solution sounds to me like it ought to mean the chances for, say, Naruto to meet either Sasuke or Sakura -- I want the chances that either a naruto/sakura, a naruto/sasuke, *or a sakura/sasuke* match happen. Then again I lose the logic line very fast in cases like this, so maybe that's what you meant and I got confused about it.

I think I explained badly in the opening post. Mrgh.

The problem you've put up in the original post is crystal clear. It's my explanation that sucks. My calculation is aimed to show the chance to have 2 Team 7 members paired (no matter whom exactly), but I probably just can't explain it correctly.
Anyway even a shinobi could make a mistake, especially if there is nobody to catch them :)

And I have to apologize again, I made an arithmetic mistake. If there are N participants in a tournament (N is a power of 2, like 8, 16, etc) and we want to calculate the probability of not having 3 members of Team 7 to fight each other, the formula is the following:

(3! * (N-1)!!) / (N-3)!!

If N = 16 and K = 3, the answer is 1/5, or 20%.
If N = 8 and K = 3, the answer is 3/7, or around 42%.

Will not bother you with this any more :)

XD;;;;; That's roughly the same numbers I got when I tried to figure it out with my basic logic, so i'm going to pretend it wasn't a coincidence on my part but confirmation that we are both awesome at math! Somehow.

It wasn't a bother at all, why you sayin' that. Thank you very much for the help. ^__^

Lots of these make sense. :)

I just wanted to poke in and be another reminder (because others have mentioned this) that these wonderful stats only count if opponent selection is random - if officials choose who battles specifically ("Leaf-nin vs. Mist-nin because we set it up that way") and not randomly, then these are not valid.

You would have to know what criteria they use (e.g. from different villages, same gender, similar stature, whatever) to figure out the statistics in specific cases versus random. :)

While I suck at math and can't even begin to calculate it, I have to say that even though it's only "for the sake of two lines of dialogues that I might also rewrite", it WOULD add a lot to the story. A "whoa" factor if you will. Statistics are always impressive. So you go math people! :D

See, this is why I never took statistics... ever... I hate this kind of stuff where you have to apply circular logic...

Stick me with the trigonometry... I was good at that in high school... Or you know what? I'll sit here with the fan banner and look pretty... I like that option better...

teamwork needs your help!

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Gah, stats everywhere...

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Literary Sense

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