# Probabilities of two people driving past each other simultaneously

#### fsgregs1

##### New Member
Hi. I have a weird thing going on. I am retired and leave my home perhaps once/day in my car, to go to shopping, doctors, etc. It take 2 minutes to traverse the subdivision road my home sits on, then it reaches the main road. Obviously, I have to both leave the subdivision and reenter it, so, in a 10 hour day (600 minutes), I am on that subdivision road 4 minutes/600 minutes, or 1/150th of the day, or 0.00666 of the day.

I have a neighbor who works from home and drives in and out of the subdivision more. I am guessing but I suppose he traverses our subdivision road in and out perhaps 6 times a day, or 24/600 minutes, or 1/25th of the day, or 0.04 of the day.

Here's my problem. I encounter my neighbor on that road a LOT, at least 15% of the time, or 1/6 trips. That is weird and not statistically probable.

So, what is the probability that we will both be driving down that subdivision road simultaneously, and pass or see each other on the road?

I thought I knew. To compute the probability of two events (both driving the road) occurring simultaneously, textbooks say you simply multiply the two individual probabilities, but that can't be right. If I did so, then 0.00666 x 0.04 = 0.0002664, which is a very small and very incorrect number. To prove it, imagine that my neighbor was always driving on our subdivision road, back and forth, for 10 hours a day. His probability of being there would be 600/600 minutes or 1.0. If I multiplied his probability and mine (0.00666), I would get 1.0 x 0.00666 = 0.00666, or 0.6% probability that we would see each other on the road. That is clearly wrong. If he was always on the road, I would see him 100% of the time I drove down the road, not 0.66%. So, clearly in this case, computing the probability of encountering my neighbor on my subdivision road at the same time I was on it by simply multiplying both independent probabilities, is NOT the correct solution.

So what is the correct answer?

#### Dason

##### Ambassador to the humans
Multiplying the probabilities gives you the overall probability of both of you being in the road at the same time. Basically if somebody else were to ask at some point " are both of those people driving at this exact moment?" Then assuming the probabilities you have were correct and assuming the events are independent then the probability you calculated is the probability that the answer would be yes. It's possible that neither of you are on the road in this situation.

#### fsgregs1

##### New Member
Multiplying the probabilities gives you the overall probability of both of you being in the road at the same time. Basically if somebody else were to ask at some point " are both of those people driving at this exact moment?" Then assuming the probabilities you have were correct and assuming the events are independent then the probability you calculated is the probability that the answer would be yes. It's possible that neither of you are on the road in this situation.
OK, over the course of a 10 hour day, the probability that we will both be on the road at the same time calculates to be a very small number (.026%). That means if I went out once/day and he went out 4 times/day, the odds that we would pass each other on the road is only 0.026%, or 1/3846 trips. That is nuts! In actuality, I pass him about 1 in every six trips I make. So either he and I are somehow cosmically linked, or ...

Perhaps I need to rephrase the question. "What are the odds that I will encounter my neighbor if I drive onto that road?" As an example, If I assume I am on the road 2% of the time (0.02), but my neighbor is on the road 100% of the time (1.0), then the probability I will encounter him is 100%, not 2%. He is always on the road, so naturally, I can say with certainty that I will encounter him 100% of the time. So, multiplying probabilities does not work. So, how does the math work here?

#### Dason

##### Ambassador to the humans
Well if you say that about 1/6 of the time you make a trip that you pass them then I'd say the best guess at the probability that you pass them is about 1/6.

#### fsgregs1

##### New Member
Please ... this is driving me crazy. I do encounter this guy in his car driving past me about 1 in every six times I enter or leave my subdivision, no matter what time of day. I have no idea how many times a day he would have to drive the 2 minute road both ways to equal those odds, because I really don't understand the math. That's why I posted here. Given my logic, it is not just multiplying both probabilities. So is there another way to consider the problem?

#### Dason

##### Ambassador to the humans
Well what is your question. Are you trying to estimate the probability that you will see them? If so then using the proportion of time that you see them is the estimate you want. If you are actually interested in something else then please explain your actual question.

#### fsgregs1

##### New Member
I want to see the math that shows the probability of me seeing this guy is closer to the real figure of 15%. I don't believe that he and I are somehow "linked" so that when I go out into the road, he does also. He is not watching my house, obviously. Likewise, although I estimated he leaves his house perhaps 6 times a day in a 10 hour period, that only equals a probability of us meeting of 0.026%, if I do the math the way you suggest. Instead, we pass each other about 15% of the time, no matter what time of day I go on my errands (morning, afternoon or evening). When I try to solve this using a multiplication of probabilities, I get: (Prob-me * Prob-him = 0.15) ... = (0.00666 * prob-him = 0.15) ... = (prob-him = 0.15/0.00666 = 22.5). 22.5 is impossible. There is no probability decimal >1.0 (100%). So, no matter how many times he leaves his house, we could never reach a probability of us both meeting of 15%, using this math approach. So obviously, the math is wrong. The probabilities of us meeting is NOT my probability decimal (0.00666) x his probability decimal (0.04), or anything like it. So, my question is ... What is the correct math here (I don't understand probability math)!

#### Dason

##### Ambassador to the humans
One does not need to be "linked" in any way for you to have similar patterns in when you're on the road. One of the reasons these things might happen more than expected is because maybe you have similar schedules. I have to assume you're not rolling dice to decide when you should go on a drive. There is probably a reason and if you keep track of when you leave the house I highly doubt it would be uniformly distributed through the day. When I would drive home from work from my last job I would see one particular car maybe once every week or so. I was commuting to and from a large city. Do I think that person was "linked" to me in any way? No I just think we had a similar schedule in when we drove to and from work.

#### fsgregs1

##### New Member
Fair assessment, but I of course, considered that. The hours I might leave my home are legitimately between 10 AM and 6 PM. I have NO schedule! I may go food shopping any time between those hours ... any time. So, a 1/6 chance of seeing my neighbor is not just attributed to our overlapping schedules.

Look, I listed a good example of how the math of multiplying probabilities does not work. If the odds of leaving my house is say, 4% in a 10 hour period, but the odds of my neighbor being on the road is 100% (he stays there all day), then the probability of me seeing him on the road is 100%, not the multiplication of my odds (0.04) x his odds (1.0). That equals only 0.04, or 4%, not 100%. So PLEASE ... what is the CORRECT math approach here?