Wanna make a 100g? Solve a problem!

[altarboy]

0:00 bus at B
0:04 bus at B
0:012 bus at B
0:16 bus at B
0:24 bus at B
0:28 bus at B
0:36 bus at B
0:40 bus at B
0:48 bus at B
0:52 bus at B
1:00 is a different hour. Bus is at the stop 10 times in 1 hour. 60/10 is 6minutes.

P.S. like the idea of having an eye as an icon too :D

[Kobetauren]

altarboy wrote:

0:00 bus at B 0:04 bus at B 0:012 bus at B 0:16 bus at B 0:24 bus at B 0:28 bus at B 0:36 bus at B 0:40 bus at B 0:48 bus at B 0:52 bus at B 1:00 is a different hour. Bus is at the stop 10 times in 1 hour. 60/10 is 6minutes. P.S. like the idea of having an eye as an icon too :D

It's not 4 and 8, it's 4 and 12.

[altarboy]

going on the same idea as above with 4 and 12 minutes, it completes a cycle of 24 every 3 hrs, 1440min/(24stops*8cycles)=7.5minutes average

[altarboy]

more easily writen as,

Average=60/(tripA/2 +tripB/2)
Average=60/(2+6)
average=60/8
average=7.5

[altarboy]

missed one more thing, adding 2 averages i think no longer gives you an average, so it has to be devided by 2, so 3.75 minutes.

[Sintax]

You guys do realize the answer was already given, right?

[Kobetauren]

syntax53 wrote:

You guys do realize the answer was already given, right?

You do realize without any sort of verification, he could be wrong, right? l2peerreview. And without knowing WHY, it's a pointless exercise. Knowing why is how you learn from it!

And Altar, your numbers don't make any sense to me. A 16 minute cycle doesn't have a full completion after 3 hours. It's 4 hours. And it's 30 stops. Which works out to an 8 minute average.

I don't understand how probability works into this. The question isn't "what's the chance the bus is 2 minutes away from you in either direction on either loop" (which requires probability), it's "what's the average wait time," which is just adding and dividing.

Thus us questioning it. I'm not saying Tubi's wrong, I'm saying I don't see how he's right, and I'm trying to understand why it's 5 minutes.
EDIT: I missed Camberth's post, which makes sense. I saw Tubi's post and it wasn't clear, and I thought that was what you were referring to.

[Sintax]

I assumed tubi knew the answer to be concrete. I also assume five minutes is right because Dan and I did it using our average methods and it comes out to the same as those using probability.

[Kobetauren]

syntax53 wrote:

I assumed tubi knew the answer to be concrete. I also assume five minutes is right because Dan and I did it using our average methods and it comes out to the same as those using probability.

You and Dan got 4.8, which isn't 5. You can't just magic out .2 through hand-waving rounding. Moreso, your methods don't make any sense.

[altarboy]

k, then what about...


90 trips per day: 24*60 / 16
180*0+
180*1+
180*2+
180*3+
90*4+
90*5+
90*6+
90*7+
90*8+
90*9+
90*10+
90*11
=6480
6480/1440=4.5

[Sintax]

Kobetauren wrote:

Moreso, your methods don't make any sense.

Taking each amount of possible time you could have to wait, adding them up, and taking an average doesn't make any sense? I only rounded in assumption that down to the second wasn't expected for the answer since everyone was using minutes.

If you want to really get technical about it and go down to the second, the answer would be 4 minutes and 59.8 seconds.

http://spreadsheets.google.com/pub?key=r-LqIeHi98vSMBpci8cE8rg

[Cambarth]

I can't even express how overkill that spreadsheet is. How long did that take to populate?
You're working in terms the problem isn't even given in (seconds).

Working in terms of probability simplifies the crap out of it, both in terms of execution and explanation.

If I were handing this in, it'd look something like this (For Justin as well):
The bus is either on its short loop (Event A) or its long loop (Event C).
These events are mutually exclusive because it cannot be heading towards station A and station C simultaneously.

Mutually Exclusive Events:
P(A or B or C) = P(A) + P(B) + P(C)

The stops are instantaneous (this is an assumption that must be made since they are not given in the problem), so the P(B) = 0.
That means P(A) + P(C) = 1

Total Trip = 16
P(A) = 4 / 16 = .25
P(C) = 12 / 16 = .75
P(A) + P(C) = 16 / 16 = 1

The average can be expressed as (Maximum + Minimum) / 2.

Wa = Average Wait for Event A = (4 + 0) / 2 = 2 minutes
Wc = Average Wait for Event C = (12 + 0) / 2 = 6 minutes

Because the distribution between Event A and Event C is not uniform (equal probability) then the cumulative average wait time must be a found by weighting these wait times with their probabilities.
So,
P(A)*Wa + P(C)*Wc = P(A or C)*Wtotal
or
.25*2 + .75*6 = 1*5


PS I enjoyed probability class in college.
If there's someone who doesn't understand part of this proof I'll be happy to explain it further (on here or PMs). There might be a little step somewhere I missed, since I don't actually get graded on it and this post went through a few wording revisions.

[socksnogger]

Justin,

The expected value is close to 5 minutes. It depends on how specific you want to get when figuring it out but doing it at 1 minute intervals, you get ~4.9 minutes. If you want to do 30 second intervals, 15 second intervals, 1 second intervals, you can figure that out yourself.

Pat,
Probability is king!

[Cambarth]

socksnogger wrote:

Justin, The expected value is close to 5 minutes. It depends on how specific you want to get when figuring it out but doing it at 1 minute intervals, you get ~4.9 minutes. If you want to do 30 second intervals, 15 second intervals, 1 second intervals, you can figure that out yourself.

Given the initial information the expected value IS 5 minutes, doing it on the second level is doing it in a "minutes are comprised of seconds so if I break it down more I'm getting more accurate" when in actuality it is providing false accuracy.

[Sintax]

Cambarth wrote:

I can't even express how overkill that spreadsheet is. How long did that take to populate?

All of about 30 seconds (Type 4 numbers, drag them down, type in the average formula, done).

Cambarth wrote:

You're working in terms the problem isn't even given in (seconds).

1 Minute = 60 seconds? I only converted to seconds because 4.8 minutes didn't seem to be acceptable. False accuracy? It's still accurate, moreso even. Doing it by seconds it comes down to 4.99 minutes where as doing it by minutes it comes down to 4.8 minutes.

Cambarth wrote:

Working in terms of probability simplifies the crap out of it, both in terms of execution and explanation.

If you understand probability. Your solution is entirely over-complex to me, who doesn't. I can follow your methods, but I couldn't come up with them myself.