Wanna make a 100g? Solve a problem!

[2b]

wrong but you're heading in the right direction there's 1 thing you're missing

[Kobetauren]

Songi wrote:

Any point of the day? Is that even calculable? The Average wait time is random.

No, the wait time is random within bounds. The AVERAGE wait time is a distinct number.

[2b]

Kobetauren wrote:

No, the wait time is random within bounds. The AVERAGE wait time is a distinct number.

correct.

[Sintax]

4.8 or 5 minutes rounded? People aren't taking into account the bus already being there?

B to A to B = 0+4+3+2+1=10
B to C to B = 0+12+11+10+9+8+7+6+5+4+3+2+1=78

78+10/18=4.8

[Danr0k]

1. B = 0
2. A->B = 2
3. B->A->B = 4
4. C->B = 6
5. B->C->B = 12
1:5 = 24
24 / 5 = 4.8?

or this which seems wrong
0+2+2+6+6 = 14/5 = 2.8

[Cambarth]

I could be wrong, but at least it hasn't been guessed yet.

The average wait time is 5 minutes.

I made the assumption that a stop takes 1 second and it's embedded in the travel time towards B for simplification. (This does not have a big impact on the answer)

There's a 16 minute loop where the bus arrives at station B twice.
The probability of you being at B when it arrives is 2/(16*60) or 0.002083333. The wait time for this is 0 minutes.

Event 1: 0.002083333. @ 0 minutes = 0 average wait for Event 1

The probability of you not being at B when it arrives is 1-0.002083333 or 0.997916667.

Separate Event:
The probability of it being on the 'A' side of its route is 4/16 or .25.
The probability of it being on the 'C' side of its route is 12/16 or .75.

Recombined:
The probability of you arriving when it is on the 'A' side of its route is .25*0.997916667 or 0.249479167.
The average wait time for this event is 2 minutes. (4 max, ->0 min)
Event 2: 0.249479167 @ 2 minutes = 0.498958333 average wait for Event 2

The probability of you arriving when it is on the 'C' side of its route is .75*0.997916667 or 0.7484375.
The average wait time for this event is 6 minutes. (12 max, ->0 min)
Event 2: 0.7484375 @ 6 minutes = 4.490625 average wait for Event 3

SUM: 0 + 0.498958333 + 4.490625 = 4.989583333 or ~5 minutes.
Probabilities: 0.002083333 + 0.249479167 + 0.7484375 = 1 (or 100%)

edit for clarification, since I missed stating something in a rewriting of the post.

[Sintax]

Cambarth wrote:

I could be wrong, but at least it hasn't been guessed yet. The average wait time is 5 minutes.

Both posts above you say 5 minutes.

[Cambarth]

syntax53 wrote:

Cambarth wrote: I could be wrong, but at least it hasn't been guessed yet. The average wait time is 5 minutes. Both posts above you say 5 minutes.

They sure look like 4.8 rounded to 5, not 5.

If I take out my .2% chance of being there the second it arrives at station B then I get 5 on the nose.

.25*2 = .5
.75*6 = 4.5

.5 + 4.5 = 5

edit: .02% is actually .2%

[Shamanananigan]

Dumb question, is point B between point C and A? And I'm assuming we have all the information required to solve this?

[Sintax]

Zexii wrote:

Dumb question, is point B between point C and A? And I'm assuming we have all the information required to solve this?

he said it's like a figure 8, so i assume so.

[socksnogger]

3 Minutes 45 Seconds

[2b]

you guys are making this way more complicated then it actually is. the correct answer was guessed but the work associated with it is wrong completely way more complex then it has to be. its a whole number. no 0.0000067 about it or things like that. like i said 2 min average for route A to B and back and 6 mins from B to C.

i'll give you guys a hint. whats more likely to happen? to wait for the A to B route or the B to C route?

i posted that and then i realized i wasnt logged in.

but i just read what was written and cambarth is correct.

the answer is 5. since its 3 times more likely to fall on the b to c route.
6/2=3

that means that for every time you catch the bus on the AB route you'll catch it for the BC route 3 times.

3*6=18+2=20
20/4=5.

5 min average wait time.

[Cambarth]

2b wrote:

you guys are making this way more complicated then it actually is. the correct answer was guessed but the work associated with it is wrong completely way more complex then it has to be. its a whole number. no 0.0000067 about it or things like that. like i said 2 min average for route A to B and back and 6 mins from B to C. i'll give you guys a hint. whats more likely to happen? to wait for the A to B route or the B to C route? i posted that and then i realized i wasnt logged in. but i just read what was written and cambarth is correct. the answer is 5. since its 3 times more likely to fall on the b to c route. 6/2=3 that means that for every time you catch the bus on the AB route you'll catch it for the BC route 3 times. 3*6=18+2=20 20/4=5. 5 min average wait time.

The first time I used the complicating factor, the second time was simplified and not the least bit complex in terms of probability problems

[Kobetauren]

Do you happen to have a write-up of this you can link to?

[2b]

nope sorry