Probability that a walking person sees two bikers cross as they bike towards each other

[2xq1]

Hello,

What is the probability that, given a 24-hour period, a runner, R with speed S(R), and two bikers, B1 and B2 with speeds S(B1) and S(B2), pass each other on a line, L, assuming the runner and one biker will start at one end of the line and the other biker starts at the other end?




Here is my attempt and please help correct or shed light on how to think about this.


Short answer:

[math]\frac{L^4}{S(R)*S(B1)*S(B2)^2*24^4}=\frac{L}{S(R)*24}*\frac{L}{S(B2)*24}*\frac{L}{S(B1)*24}*\frac{L}{S(B2)*24}=P(R,B2) * P(B1,B2)[/math]
Meaning of the symbols:
L = Length of the line
S(R/B1/B2) = Speed of R/B1/B2, in same distance units as length of the line over hours
P(R,B2) = Probability that R meets B2
P(B1,B2) = Probability that B1 meets B2


Long answer with explanation:
I think the answer is found by multiplying the probability of two independent events:
(Probability that R and B2 meet) * (Probability that B1 and B2 meet)

Which should be the same as the following, but going to focus on the above for simplicity.
(Probability that R and B1 meet) * (Probability that B1 and B2 meet)

To break that down,
(Probability that R and B1 meet) = ( (The time it takes for R to run the path) / (Period of time) ) * ( The time it takes for B2 to bike the path) / (Period of time) )
(Probability that B1 and B2 meet) = ( (The time it takes for B1 to bike the path) / (Period of time) ) * ( The time it takes for B2 to bike the path) / (Period of time) )

(The time it takes for R/B1/B2 to complete the path) = (Length of the path) / ( R/B1/B2's speed)

For example,
L = 1000 meters
S(R) = 3600 meters / hour = 1 meter / second
S(B1) = 7200 meters / hour = 2 meters / second
S(B2) = 7200 meters / hour = 2 meters / second

[math]P(R,B1)=\frac{\frac{1000 m}{3600\frac{m}{h}}}{24h}*\frac{\frac{1000 m}{7200\frac{m}{h}}}{24}=\frac{1000 m}{3600\frac{m}{h}*24h}*\frac{1000 m}{7200\frac{m}{h}*24h}=\frac{(1000 m)^2}{2*(3600\frac{m}{h})^2*(24h)^2}=\frac{25}{373248}=0.00006697959=6.69795953 × 10-5[/math]

[math]P(B1,B2)=\frac{\frac{1000 m}{7200\frac{m}{h}}}{24h}*\frac{\frac{1000 m}{7200\frac{m}{h}}}{24}=\frac{1000 m}{7200\frac{m}{h}*24h}*\frac{1000 m}{7200\frac{m}{h}*24h}=\frac{(1000 m)^2}{(7200\frac{m}{h})^2*(24h)^2}=\frac{25}{746496}=0.00003348979=3.348979 × 10-5[/math]
The answer for the example would be P(R,B1) * P(B1,B2) so
[math]\frac{25}{746496}*\frac{25}{373248}=\frac{625}{278628139008}=2.2431331096e-9[/math]As a percentage %2.2431331e-7 which seems minuscule.


Mental aids
While trying to simplify this problem in my mind, I thought to replace each independent event with a flip of a coin. Ignoring the exact probability for a minute, replacing the runner meeting one biker with one flip of the coin helped because it showed that, in a sense, they're just independent events. Also, the other biker meeting them would be just like another coin flip landing tails just as the first coin tip landed tails. Sometimes, the one event affects the other, but that doesn't seem to be the case here.


Thanks.

Credits:
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[2xq1]

I just got a chance to read the site guidelines more carefully, and wanted to edit the question to add these:

1. From where does your question come?
I came up with this question while running on the trail yesterday morning and noticed two bikes cross each other after a long while of not seeing any bikes.

2. Post the exercise or your question completely and accurately.
(I would copy the question here)

3. Show work that you've already done (even if you think it's wrong), and try to explain why you're stuck.
(I would show my work here)

4. We may respond by asking you questions (if we think something is missing or unclear).
Questions are, of course, welcome.

 

[Cubist]

(Probability that R and B2 meet) = ( (The time it takes for R to run the path) / (Period of time) ) * ( The time it takes for B2 to bike the path) / (Period of time) )

What if...
(Period of time) = 24h
(The time it takes for R to run the path) = 24h, this uses my own running speed
( The time it takes for B2 to bike the path) = 1h

Your equation gives (Probability that R and B2 meet) = (24/24) * (1/24) = 1/24. This is obviously incorrect because they must pass each other (with probability 1) IF the runner and bike 2 complete the trail within the same 24h period.

However, before you think any more about this problem, I recommend that you read the following excellent article should rare events surprise us

 

[2xq1]

I never realized how helpful it can be to run 1000m in 24 hours until I read the last post

That test and reading the article helped me to come up with a new answer, which happens to pass the test.

Short answer

[math]P(R,B1,B2) = P(R,B1) * P(R,B2) \\ P(R,B1)=\sum_{i=1}^{n=C(R)}{(\frac{C(R_i,B1)}{C(B1)}*\frac{R_i}{C(R)})} \\ P(R,B2)=\sum_{i=1}^{n=C(R)}{(\frac{C(R_i,B2)}{C(B2)}*\frac{R_i}{C(R)})}[/math]
Symbols meaning:
P(R,B1,B2) = Probability that R meets B1 and B2
P(R,B1) = Probability that R meets B1
P(R,B2) = Probability that R meets B2
C(R_i,B1/B2) = count of how many segments of B1/B2 could be in [imath]R_i[/imath], i.e. one segment of R, e.g. for the runner, there are 24 one hour segments, for the bikers, there are 48 half-hour segments, so there are two B1 segments for every one R segment, and similarly, there are two B2 segments for every one R segment
R_i = one segment of R
C(R/B1/B2) = count of segments in R/B1/B2, e.g. for R there are 24 segments of 1h each

So, for the example of [math]Period\ of\ time\ =\ 24h, Runner\ time\ = 1h, Biker\ time = \frac{1}{2}h[/math][math]P(R,B1,B2) = P(R,B1) * P(R,B2) = \frac{1}{24^2}\\ P(R,B1)=\sum_{i=1}^{n=C(R)}{(\frac{C(R_i,B1)}{C(B1)}*\frac{R_i}{C(R)})} = \sum_{i=1}^{n=24}{(\frac{2}{48}*\frac{1}{24})} =24 * {(\frac{2}{48}*\frac{1}{24})} = \frac{2}{48} = \frac{1}{24} \\ P(R,B2)=\sum_{i=1}^{n=C(R)}{(\frac{C(R_i,B2)}{C(B2)}*\frac{R_i}{C(R)})} = \sum_{i=1}^{n=24}{(\frac{2}{48}*\frac{1}{24})} =24 * {(\frac{2}{48}*\frac{1}{24})} = \frac{2}{48} = \frac{1}{24} \\[/math]




Long answer (trying to verify the short answer)
I think Cubist's test helped eliminate a bunch of wrong answers in my head. If the test is applied to the equation for probability that R meets B1, it looks like the answer is 100% certainty that they will meet. Plugging in the numbers, the answer is 1 for probability that R1 meets B1.

[math]P(R,B1) \\ P(R,B1)=\sum_{i=1}^{n=C(R)}{(\frac{C(R_i,B1)}{C(B1)}*\frac{R_i}{C(R)})} = \sum_{i=1}^{n=1}{(\frac{48}{48}*\frac{1}{1})} = 1 * 1 * 1 = 1[/math]
Similarly, P(R,B2) is also 1. Multiplying P(R,B1) and P(R,B2) the result is also 1.

If the runner's time was halved, seems the result would be the same as a coin flip whether R and B1 meet.
[math]P(R,B1)=\sum_{i=1}^{n=C(R)}{(\frac{C(R_i,B1)}{C(B1)}*\frac{R_i}{C(R)})} = \sum_{i=1}^{n=2}{(\frac{24}{48}*\frac{1}{2})} = 2 * \frac{24}{48} * \frac{1}{2} = \frac{1}{2}[/math]
Phew. I think this is it but I'm almost posted an entirely incorrect answer one time before and only was able to get to this point by applying the "what if R runs the 1000m at a pace of 24 hrs" logic.

Lastly, I tried to simplify the answer, but ended up getting a wrong probability and thought to not get ahead of myself and see if this answer makes sense before optimizing.
Thanks for your time!

 

[2xq1]

I just realized this doesn't make sense since there are units of hours leftover and doesn't work when Runner completes the length in 12 hr will try again soon