Little math problem(s) October 2020

[pier4r]

I still have a lot to catch up, but if I guess I can add a couple of contributios even if I missed a lot in the last years.

Months ago, before Covid 19, I got in the metro. After few stations a colleague entered the train and sat directly beside me and we didn't recognize each other until we were at the station near to work. I was busy reading and he too.

Then I thought: what are the odds that we sit, without noticing each other, near to each other?

Given that:
- train schedule is a train every 5 minutes (between stations the train use 2 minutes, when there are no problems)
- period of time where we can pick the train is like 2 hours (from 7 to 9 AM)
- my station is 5 or 6 stations before his (he takes sometime one, sometime the other)
- we work around 275 days a year, say 240 to avoid holidays/sickness (and I do not include those times where I take the bike)
- a train has 8 cars (let's assume that entering a different car means sitting away, although some trains have a lot of cars connected)
- Using an average train (it depends on which type of car is used), per car there are 48 places (sitting).

[Albert Chan]

Differences of 2 uniform random distribution variables followed triangular distribution.

Let x = time differences (minutes) between 2 person arriving to the same station,
triangular distribution curve peaked when x = 0, drops to 0 if |x| ≥ 120
(2 hours window per day where encountering is possible)

If |x| ≤ 120, y = 1/120 - |x|/120²
If |x| > 120, y = 0

Assuming 5 or 6 stations about a difference of 30 minutes:

P(|x-30| ≤ 5) = P(25 ≤ x ≤ 35) = (y at x=30) * (10 minutes) = 0.0625

Chance both will be on the same train ≈ 1/16
Chance both will be on the same car ≈ 1/8
Chance both seat next to each other ≈ 2/47 ≈ 1/25 (account for corner seats)

Chance of all of above to happen = 1/(16 * 8 * 25) = 1/3200

Assuming 240 days a year both will be using the train.
Chance of never seating next to each other (for the year) = (1-1/3200)^240 ≈ 93%

Its complement, at least once for the year seating next to each other = 1 - 93% = 7%

This might be completely wrong ...

[Albert Chan]

(10-09-2020 04:52 PM)pier4r Wrote:  Months ago, before Covid 19, I got in the metro. After few stations a colleague entered the train and sat directly beside me and we didn't recognize each other until we were at the station near to work. I was busy reading and he too.

You will have the same shock if this is another colleague.

If the other person is unspecified, chance of this rare encounter is not so rare anymore ...
My guess is it will even push over 50% chance.

[pier4r]

yes in this case I considered only one colleague as others use other routes.

If is another colleague then I can play the lotto xD

But 7% seems not that low.

[Albert Chan]

(10-09-2020 07:56 PM)Albert Chan Wrote:  Assuming 5 or 6 stations about a difference of 30 minutes:

P(|x-30| ≤ 5) = P(25 ≤ x ≤ 35) = (y at x=30) * (10 minutes) = 0.0625

Chance both will be on the same train ≈ 1/16

This assumed all trains stayed at the station for 5 minutes.
In other words, the station were never empty.

More realistic situtation is train only stay relative short time, then leave. Say 1 minute.

P(25 ≤ x ≤ 31) = (y at x=28) * (6 minutes) = 0.03833

This slight adjustment meant chance of both on the same train dropped to 1/26

Recalculate, we have daily probability of encounter = 1/(26 * 8 * 25) = 1/5200
Chance of never seating next to each other (for the year) = (1-1/5200)^240 ≈ 95.5%

At least once for the year seating next to each other = 1 - 95.5% = 4.5%