Question

A class of 114 students and 1 teacher all toss a coin. If a student flips a different side than the teacher, than the student must sit. The class continues to flip until all students are sitting. What is the probability that half the students will sit in any given round and how many round will pass until all students are sitting?

Answer 1

HEEEEEEEEEELLLLLLLLLPPPPPPPP MEEEEEEEEEE!!!!!!

Answer 2

I need help

Answer 3

yuk!

Answer 4

round 1: teacher flips coin, you might expect 1/2 of 114 to have the same outcome, regardless of whether it's heads or tails.

Answer 5

rounds that will pass until all students are sitting could be infinite, since a student can keep flipping and still never match the teacher

Answer 6

I think each round is the same... the coin doesn't 'remember"

Answer 7

so it's 50/50 every time?

Answer 8

but yes, that is just the probability... there is no guarantee that half WILL sit down every time.

Answer 9

(1/2)^(n/2) is the probability that half the students will sit
where n is the amount of students remaining

Answer 10

ok... is there an analytical way of expressing how many rounds are needed to sit them all down?

Answer 11

so for n = 114 at the start, P = (1/2)^(57) is the probability of sitting half, or 57, down?

Answer 12

P = (1/2)^(57) is the probability of 57 people sitting down

Answer 13

ok... 'Exactly" 57, right?

Answer 14

yep

Answer 15

I was thinking of the expected number of matches out of 114 tries... would be expected to be 57. But expected value is different than Pr of "exactly 57"

Answer 16

this question is very ambiguous

Answer 17

yep... hence my first comment, "yuk"

Answer 18

Sorry this was actually an activity that my class did and one of the questions I had to answer in response was to mathematically predict how many tosses before all students sat down.

Answer 19

what level class is it?

Answer 20

Chemistry Honors

Answer 21

So I guess, the second question is asking for the expected value

Answer 22

Chemistry? Interesting.

If I had to guess (terrible start, I know), I would guess that whoever thought of the question thought that it would be "half sit each time, how many "rounds" until 114 drops off to 0?" so 114, 57, 28.5 (up or down), 14.25, 7.125, 3.0625, 1.whatever, maybe one more, then zero

Answer 23

however, that is not correct math for something like this.

There are simulation techniques that seem unlikely for a Chem Honors class that are used to run repeated sim runs through probability scenarios to get answers like this.. I am not good at them, so I don't know how well that would fit here.

Answer 24

my first question was not actually one of the questions. I just didn't want to ask for the answer to the second one

Answer 25

the real flaw is that you can't guarantee that a 50% chance of heads will give you a run of H, T, H, T.... etc.

Real world trials generate runs of heads and runs of tails... over the long run, the perfect coin flip will trend toward 50/50 heads vs. tails, but

Answer 26

but you can have long runs of heads or tails in the middle... just part of a random process

Answer 27

ugg, laptop flakey

Answer 28

real coin flip trials produce runs of heads and tails... nothing wrong there.

most runs will be just a few in a row, but some could be quite long.

Answer 29

I thought that because the probability of one person flipping the wrong side was 50% and there are 114 students, it would just 114*1/2

Answer 30

or n*1/2 for any given round

Answer 31

yes, if you did this trial 1 million times, you would probably find an average of 57 people sitting after round 1

Answer 32

the point is that there is no guarantee in a single trial in class that you will actually get 57 who have to sit

Answer 33

And even if you did, you wouldn't be guaranteed on round 2 of getting exactly half of the 57 to sit (not even worrying about the odd number for now).

Answer 34

I think because this is all from a probability standpoint, we can safely say that 1/2*50 works

Answer 35

In fact, probability wise, it would be pretty unusual to get exactly 50% sitting on every round.

If you don't believe that, flip a coin 10 times and see if you get exactly 5 heads. If you do, try it one more time... there's practically no way you will get 5 out of 10 heads every time

Answer 36

sorry, 1/2 * 50 ? What is that?

Answer 37

sorry 1/2*114
this is all theoretical stuff so anything flies
in practice, the results will be different of course

Answer 38

I just don't know how I or you got 50%

Answer 39

yes... I think you could answer by saying, theoretically, that if you were to do this classroom experiment a "very large" number of times, you would expect ON AVERAGE to eliminate half the students on every round, and then you could count the rounds until 114 was "chopped' down to 0 students.

But this is only true on average over the long run, not a guaranteed outcome on the first experiment run.

Answer 40

true

Answer 41

Thanks for your help

Answer 42

It was a bit of a mess, but hope it was helpful or at least interesting :) I am not sure your honors chem teacher intended to go down this complicated probability path :)

Answer 43

No, it was an interesting and mind-stretching conversation.

Answer 44

I agree :) glad it didn't end on "yuk" also :)