/sci/ - Science & Math

1/100

>>7910185
Use the binomial distribution for this homework question OP.

>>7910202
It can't be 1/100 because there's already a 1/100 chance the first guy gets the right seat, in which case everything would be fine. However even if he gets the wrong seat, there's a chance a guy who has his seat taken will take the initial guy's seat by chance.

>>7910203
>homework
Where do I apply to the anime university that would give me these kinds of problems?

>>7910209
The University of Highschool.

It's 1/100

Why does /k/ and /g/ have waifus, but we don't?

>>7910209
Every single person sits down somewhere regardless of where the first guys sits. So the last person sitting in the right seat is wholly dependent on the first guy sitting in the right seat

>>7910219
Do you want us to have waifus, anon? We can have waifus.

>>7910219
My waifu is pic related.

>>7910185
Every person coming in has a 1/100 chance of being your seatowner, but theres also a 1/100 chance that you sat in your own seat, since this is a story lets keep that 1/100 of sitting in your own seat at the front of our minds.

Every time soneone comes in it changes the probability that its your seatowner- for conveniences sake lets say that it statisrically would be the 49th or 50th person to board whose seat youre in. At that point the chance becomes 1 in 49 or 50 that the next person to come in has their seat taken, and for convenience's sake lets say that its the 24th or 25th person. The same logic follows to 12 or 13, then 5 or 6, then 2 or 3.

Realistically, (and im being nonprecise) the number of people who are inconvenienced is about equal to the number of times you can split 100 before reaching 1. 100-50-25-12.5-6.25-3.125-1.5625

So the chance is 5 or 6 out of 100, with an added 1/100 that you picked your own seat, so 6 or 7 out of 100 or about 6.5%

If this were a multiple choice question I would choose the answer closest to 5.5/100, under the presupposition thatbthe designer of the question was too stupid to consider the probability of sitting in one's own seat at random

I think its [math]\frac{1}{100}[/math] plus a really, really small number. That really small number is the chance that the first guy sits in the wrong seat, [math]\frac{99}{100}[/math], times the chance that the next guy doesn't randomly sit in the last guy's seat, [math]\frac{98}{99}[/math], and so on, all the way down to the chance that the 99th guy doesn't sit in the 100th guy's seat, which is [math]\frac{1}{2}[/math] because there are only two seats left to choose at that point.

>>7910233
Are there any STEM girls besides Kurisu?

>>7910235
>My waifu is pic related.
Bea Arthur for me, please!

WTF's with the new captcha???

>>7910237
Eh, no, that's wrong, because the second guy has a [math]\frac{99}{100}[/math] chance that the first guy isn't in his seat, so he'll sit in his actual seat. Yeah it's a nasty problem.

>>7910185
1/2
simple symmetry

>>7910185
really depends on whose seat you take so for (100 - nth) passenger who's seat you took where n > 0 you get 1/(n+1) chance for that person to pick his own seat and so forth untill the last one

ITT - people who didn't pass A-Level statistics

>>7910238
Yui from yuru yuri is quite the STEM gril in mangos and doujins

>>7910238
There's a few I guess.

>>7910246
This "either it happens or it doesn't xD" meme is getting old.

Assuming you AREN'T in your own seat:

All the passengers will sit in their assigned seats until the person whose seat you took comes in. That person has a chance of sitting in the seat you were assigned. If that happens, then only two people will sit in the wrong seat. If it doesn't happen, then that person reproduces the original problem, but with fewer seats. As more people come in, the chance of someone being displaced and then taking your seat grows. So the chance of the last person getting his own seat is very high: Slightly over 99/100.

>>7910304
Wrong professor, its about 94% likely. See above explanation.

If you really teach post secondary education i fear for the future of our children

>>7910324

94% sounds reasonable, but that isn't what you answered you said:

>If this were a multiple choice question I would choose the answer closest to 5.5/100,

...and thus were much further off than I was.

>>7910341
5.5/100 that they are a victim of seat thievery

Also expressable as 94.5% of not being a victim

>>7910219
With waifu you mean "anime girl associated to a board" or waifu waifu?

>>7910347

Story problems always test reading comprehension as well as math. You did the math right, but then forgot the question and chose a wrong answer that would likely be on a multiple choice test:

>If this were a multiple choice question I would choose the answer closest to 5.5/100

>>7910352

Ah, you got me!

My reading comprehension is really low for a published novelist I guess

>>7910357
SAVAGE
A
V
A
G
E

>>7910357
>>7910352
i fucking hate tripfags
and this sai guy is a faggot too
fuck you both

>>7910357
>My reading comprehension is really low for a published novelist I guess

It's true - talking is much easier than listening.

>>7910185
It's 1/100

They kick you off the plane because you snuck on without showing a boarding pass?

>>7910351
Board waifu.

K has Korbo, G has Maki, etc

>>7910463

What is a "tripfag"? It's true that I think watching girls trip and stumble is cute. (As long as they aren't really hurt.) But I don't think that is what you mean.

>>7910856
>It's true that I think watching hairy men trip and stumble is cute, i masterbait to it.
FTFY

>>7910185
What is the number of permutation of a set 100 elemnts. What is the number of permutations that don't move a chosen element. What is the ratio of those numbers.

(1/100)!

100 people with a 1/100 chanced are shuffled in the airplane, shuffling gives you a factorial.

(1/100)! = 0.99432585119%

>>7911005
>>7911005
No, think of the problem from the perspective of the last passanger. For him to be in his seat, every other person should have chossen a seat other than his. The probability that the first person has chosen a seat other then the seat of the last person is 99/100, for the second it is 98/99, for the fird it is 97/97 and so on. So the probability is 99/100 *98/99 *... *1/2=1/100

>>7910185
its 1/2, either someone sits in the original guys seat or they sit in yours. If the former comes first you will get your own seat, if not, you don't.
>inb4 two options doesn't mean its 50/50
shut the fuck up and think about it retards

>>7910185
50%.

http://www.teamten.com/lawrence/puzzles/airplane_seating.html

>>7911082
What gets posted:
>the correct answer
+
>shut the fuck up and think about it retards

What it means:
>I heard this one before and had the answer explained to me, and though I struggled with it, eventually I understood it, so I no longer respect it as a challenge, even though I couldn't get the answer myself.

>>7910219
Kurisu is /sci/

>>7910238
What more do you need?

It is 1/2, no matter the number of seat

1/100

First, you have a 1% chance of sitting in the right seat.

If you sit in the someone else's seat, some other passenger will be displaced with probability 1. If this is not the last passenger, the displaced passenger will displace someone else, which means that there is always one displaced passenger.

At n=100, there is still one displaced passenger, if you sat in the wrong seat at the beginning.

>>7910185
>>7910219
>>7910233
>>7910235
>>7910238
>>7910263
Go back to your containment board, pedophile degenerates.

Weeaboos are a burden on society.

>>7910270
but he's correct. the final passenger either sits in the left or in the right seat

0/100
no survivors

>>7912747

>>7911956
You neglect the fact that it is possible for a passenger to sit in the person who displaced them's seat.

>>7911985

>>>/reddit/

>>7911985
Are you even putting any effort in your bait anymore?

Easy 1/2 explanation. Let's start at 2 seats. 1/2 chance you chose right, and the one seat game the last guy plays depends on your 50/50 one-to-one. With a 3 seat game, if you pick the last guys seat, then he will not pick the right seat. Also, if you pick your own seat, then he will pick his seat correctly. If you take the second guys seat, it now becomes the 2 seat game where he either sits in your seat or the last guys seat. Totalling up to 50/50 overall. Two more iterations left, don't worry. For a four man game, we again eliminate you picking his seat and picking your own as equally likely, but now when we investigate you picking the second guys seat, it becomes the three seat game, which was 50/50. If you pick the third guy's seat, then he will have a two seat game, which was 50/50. This means that all n-seat games will be 50/50 since they break down into the n-1 seat game, then n-2 seat game, and such and so forth which are all 50/50.

>>7912740
And choosing a marble from a randomly chosen bag of 3 bags of randomly distributed marbles also results in either a 1/2 chance of me either getting the marble I wanted or not.

The point is that you must account for the previous actions in order to understand the full question: after everyone has been seated, what is the probability that the remaining seat is their seat or not their seat.

For the previous example, you have to account for the fact of which bag you chose and the probabilities of getting that specific marble you wanted from that specific bag.

Am I right, /sci/ statsfags?

This is a Bayes Theorem application, isn't it?

>>7913307
>>7913314
who do you believe he is trying to bait?
weebs are too beta to make any sort of retaliating reply

>>7913332
I thought the 1/2 meme was present here until I read your post.

Excellent job, Anon. Induction is a miraculous process.

>>7910185
Guy who made op's pic here, its 1/2, generally for the reasons >>7913332 lists.
A similar explanation is that there are only two seat choices that decide the solution, and they are randomly chosen between, since people who sit on a different seat just push the 1/2 to someone else.

>>7913282
I stand corrected

>>7913365
Except he did it wrong. It's 1/100. First you have a (N-1)/N chance of sitting in the wrong seat. They guy after you has a (N-2)/(N-1) chance of sitting in someone else's seat etc. This leads to

[math] \prod_{k=1}^{N-1} \frac{N-k}{N-k+1} = \frac{1}{N} [/math]

>>7914229
>"Given how the other passengers are choosing their seats, you have a 50 percent chance of finding your assigned seat empty. The solution may seem counterintuitive — at first glance it seems like there will be a growing wave of wrong-seat-sitters, mayhem will ensue, fights will break out, and the odds that you get your assigned seat will plummet. But really, there are only two seat assignments you care about: yours and the one meant for the world’s worst passenger, the guy who boards first. These are the only two seats that can be the final unoccupied seat. This can be proven by contradiction: If the seat belonging to the Nth passenger (where N isn’t you or the first passenger) is the final open seat, then it was also open when the Nth passenger boarded, and she would’ve taken it then, so it can’t be the final open seat. Q.E.D. The two potential final seats — your and the first guy’s — are otherwise identical, so when you board, there’s a 50-50 chance yours is the one open at the end.

>"Another tack is a sort-of induction argument, starting with a smaller plane and seeing what happens to the solution when it gets larger. Imagine the plane has just two seats. The world’s worst person will clearly take your seat half the time, leaving you with your assigned seat half the time. Now imagine the plane has three seats. The world’s worst person will take your seat a third of the time. Another third of the time he’ll take the other passenger’s seat, in which case that other passenger will take your seat half the time. Your seat is taken ⅓+(⅓*½)=½ the time. Again, you get your assigned seat half the time. And so on and so forth. The answer is 50 percent regardless of how large the plane is."

http://fivethirtyeight.com/features/can-you-decode-the-four-secret-messages/

>>7910185
It should be 1/100. Let n=100.
The total number of possible configurations in seating is n! , the number of configurations where the last person gets their own seat is (n-1)! Assuming all configurations are equally likely then the probability is (n-1)!/(n)! = 1/n.

>>7914483
>Assuming all configurations are equally likely
This is not the case, since persons other than the first only pick a random place if theirs is taken.

>>7910185
Its 1/2 ypu beta cuck faggots

>>7914489
Im not sure, but I dont think it would matter.
Since the configurations are agnostic, and only the result of the last person matters

https://ideone.com/rS0D3J

You first off have a 1/100 chance of picking your own seat (lucky you)
If you pick a seat that is not yours the last passenger has a 1/100 chance (if the flight is fully booked) that not one single person (in the 98 between you and him) has sat in their seat. This of course ignores that people don't behave like that, everyone else will have their boarding pass(how did you get onboard without one?)
If everyone behaves like a robot the chances are 2%, if everyone behaves like people then about half way through boarding somebody notices you took their seat, sits nearby, and the last one to board has their own seat with at least 3 people sitting in the wrong seat.

>>7910185
It's 50 50

Think of the last two and not the 1st one

50%

When the last passenger enters the plane only your seat or his seat can be empty. Both have an equal chance of being empty because of symmetry.

>>7914845
>When the last passenger enters the plane only your seat or his seat can be empty.
THIS
I wrote a little Perl program to brute-force the answer (got 50%), then saw this post and added a check for his idea.

use strict;

my ($rightSeat,$tries);

$tries = 100000;
foreach my $x (1..$tries) {
if (tryItOnce()) {
$rightSeat++;
}
}
print ($rightSeat / $tries) . "\n";


sub tryItOnce {
my ($s, $p, @seats, $correctSeat);
$s = int(rand(100))+1;
$seats[$s] = 1;
print "You sat in seat $s\n";

foreach $p (2..100) {
if ($p == 100) {
if ($seats[1] && $seats[100]) {
die "Anon is wrong!\n";
}
}
$s = $p;
while ($seats[$s]) {
$s = int(rand(100))+1;
}
if ($seats[$s]) {
# should never happen
print "WTF?\n";
}
$seats[$s] = $p;
if ($s == $p) {
$correctSeat = 1;
}
else {
$correctSeat = 0;
}
}
if ($correctSeat) {
print "The last passenger got the right seat.\n";
}
else {
print "The last passenger got the wrong seat.\n";
}
return $correctSeat;
}

>>7914526
Perhaps I worded it incorrectly. The point is that not all n! configurations are allowed, since the passengers do not allways pick a random seat. The configurations where the first person gets the correct seat and some others do not, for instance, are not allowed.

>>7913356
What the fuck are you even talking about? Changing the subject to some alpha/beta debate literally means you don't know how to reply to being BTFO.

>>7910352

On a multiple choice test the answer they were looking for would be glaringly obvious in the choice of answers. He got it right dick bin, let it go.