Driving cars and meeting people

Breakerrr

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Jan 4, 2017
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So in 1991 mitsubishi released, for sale in the USA, exactly 2000 galant vr4's. After I was spotted in my gvr4 at an intersection by another gvr4 owner, I pondered on what the odds were of running into another gvr4 within the same model year, numbered series, at a traffic light in the U.S., at the same time.

I'm hoping someone who has better equation writing skills and an understanding of statistics can help.

I collected two other necessary variable values.

# of cars in the us: 250,000,000
# of intersections with traffic lights: 350,000
We are also assuming that all 250,000,000 cars are currently at 1 of 350,000 intersections, and that out of 250,000,000 2000 cars are 1991 mitsubishi galant vr4s.
(To simplify further we could assume there are 714.3 vehicles at each intersection, but I'd rather not, if possible)

I know that it's complex but it's eating a hole in my brain.
Please help, and thanks for the time spent reading my post.
 
A lot depends on exactly how accurate you want to be. The more assumptions you make, the easier the calculations, but the less accurate the answer. In absence of any actual data about the distribution of gvr4's, I think it's reasonable to assume that they're uniformly distributed across the country (and further that all 2000 are in the US in the first place). The population of the US is approximately 319 million. You can find the population of the city where you live and get the percentage of the country that lives there, then assume that that percentage of the 2000 cars is also in your city. Here's some other data you might find helpful in reducing the number of assumptions you need to make:

  • In April 2015, AAA published the findings of a survey. Among other things, they found that the average driver spends 46 minutes per day driving.
  • The Telegraph reports that British motorists spend a fifth of their daily drive at a stoplight. It seems reasonable to assume that US drivers spend a similar percentage at stoplights.
  • A man living in the Minneapolis-Saint Paul area of Michigan logged several days of driving and found that he spends an average of 17 minutes at stoplights each day. He lives in a fairly big city, so I'd think the figure of 10 minutes (1/5 of 46, rounded up to a "nice" value) seems more accurate for a country-wide sample.
  • According to Taylor Forbush, a traffic engineer in Orem, Utah, the average traffic light's cycle is 2 minutes.
 
I am not looking for an extremely accurate solution, if for no other reason than the probability is getting extremely small. As you stated taking the time into consideration reduces the probability to .001 of the original value. That said I suppose one could argue for the fact, and you touched on this a bit as well, that distribution of these cars was not uniform...or even random for that matter.

Back to my point.

I don't really care how accurate the answer, as long as it was arrived at using good data to get there. Ajectives vary in meaning from person to person but a number doesn't. I want to be able to say, "these are the odds, if these conditions existed....1/xxxxxxxxx" right now I can only say, "very low" or "not good". That's too much grey area.
 
I don't really care how accurate the answer, as long as it was arrived at using good data to get there....
If you really "don't really care how accurate the answer" is, then you can just make stuff up! On the other hand, if you're wanting a value whose accuracy you can justify, you'll need to start setting parameters; that is, you'll have to make some assumptions. The "goodness" of the data will depend upon those assumptions. ;)
 
I'm sorry. I must have made a mistake. I thought this was a place where one could get help with questions dealing with math.

I'm 100% certain that I laid out all of the necessary variables that I give a crap about in my first post. I DON'T NEED help with the QUESTION. I NEED help with the SOULTION.

Also, i stated that I don't care about accuracy, then immedimmediately qualified that with the remarks about the data. Here's one equation I don't need help with:

Good data + good data = accurate solution

In retrospect my statement is an oxymoron but valid nonetheless. If all the data to solve the problem is qualified, and I trust whomever does the work...no I don't care about the accuracy because it's validity has been guaranteed by inspecting the quality of ingredients if you will.

All that aside, my point was that I am aware that, as with engine optimization, you only get out of it, from a performance standpoint, what you put in. One could literally narrow down the exact probability by researching specifics that eliminate variables or "assumptions" until it made them crazy. I'm not willing to do that so again I say all of the required information about my question has already been listed.

What I am gathering is that:
A. nobody here knows how to help, so instead of offering solutions, you feel like picking apart my scenario and the specifics therein
B. You just enjoy talking down to those who know less about math, like some kind of game.

Hopefully neither is the case and someone who is willing to accept the task can come up with something useful that will help me because as of now this has been simply a waste of time.

Someone from another forum sent me this.

2,000/250,000,000 + (249,998,000/250,000,000)*(2,000/249,999,999) + (249,998,000/250,000,000)*(249,997,999/249,999,999)*(2,000/249,999,998) = 2.3999808096510464646131716665246e-5
 
Oh, alright. I understand now. It wasn't immediately clear to me from your original post exactly what you were asking. If you just want a fully worked solution, then, yes, you'll have to look elsewhere, as that's not what we do here. As I understand it, being a student is not strictly a requirement to ask a question here, but students are our primary focus. As such, we try to help people understand what's going on, how and why the math works, and to help them arrive at the solution on their own. You've gotten a solution from somewhere else. That's great. Now, it's time to show what you know. Do you understand how that solution was derived? Can you come up with even a very rough "back-of-the-envelope" calculation to see if their answer seems reasonable? If it does, then you're good to go. But if the answer seems suspect, why do you think it's wrong? What might you do to fix it?

If you're unsure of how to even begin estimating what the answer might be, I think a good place to start is to read up on the probability of independent events. Then, maybe start making a list of the separate events that would need to happen in order to produce the coincidence you describe. Try working out the probabilities of each of them individually happening and then go from there. If you get stuck, that's perfectly alright, but you gotta work with us a little. If you're willing to share with us any and all work you've done on this problem, we'll try to help you figure if you've gone wrong and where, or give you hints as to what to do next.
 
I am a heavy equipment mechanic and master fabricator for a large construction company in Colorado. My shortest work week is a minimum of 55 hours I am a father of six, my oldest will graduate high school this spring and my youngest is 22 months old and is hands down the most independent and determined to do it by herself 2 year old of all my children. My spouse prepares meals for 9 people twice a day then bartends a few nights a week and I keep the vehicles that are used by our family running as my hobby. I get about 4 hours of sleep a night and lately even that amount is not guaranteed.

I unfortunately don't have a lot of spare time to devote trying to solve the problems that I think about. While my hands are busy but my mind has some freedom to daydream I occasionally get stuck on a detail such as this one and I don't think that I ever give my attention fully to other items that I am working on.

Anyway what I came up with is first to solve for possible combinations of cars at stoplights
(N/r)(n-r)=(249999999!/714!)(249999999!-714!)=(249999999!/714!)(248999486!)
Then the probability of 1 of 1999 being at 1 of 350000 stoplights.
 
Hm.. okay. It looks to me like you're trying to calculate the probability of your car being at any one of the 350,000 stoplights. At first glance, that seems like a good place to start, but I'm not really sure that's actually necessary. One of the assumptions you said you're making is that all the cars in the county are always at a stoplight. You also mentioned being next to another gvr4, which to me implies that the hypothetical situation is one of one car idling next to another. There may be 712 cars queued up waiting at any particular stoplight in these two lanes, but they're not particularly relevant for this problem. My reasoning would then suggest that you being at the stoplight in question is a given. In other words, it doesn't matter if you're at a stoplight in Pittsburgh, or Reno, or New York City. It only matters whether the car next to you is a gvr4 or not. Would you agree?

You don't explicitly state this, but I believe another of your assumptions is that all cars are treated equal, meaning that any car has an equal chance to be at any given stoplight. The car next to you could be any one of the 250 million cars in the US. The question then becomes, how likely is it that the car next to you will be one of the 1999 gvr4's. Incidentally, this is why I was trying to prod you to maybe refine your assumptions a bit and consider more data, because the answer I get by using your bare bones assumptions is certainly far far more likely than the actual odds. If you're happy with this answer though, that's totally fine. :)
 
Yes my question is pretty simple. What are the odds that while in my gvr4 i would be at one of 350000 estimated intersections in the usa and at the same time as another gvr4. I guess it could read 1 in {the correct value} times 2. maybe.
 
Yes my question is pretty simple. What are the odds that while in my gvr4 i would be at one of 350000 estimated intersections in the usa and at the same time as another gvr4. I guess it could read 1 in {the correct value} times 2. maybe.

I'd square the value instead, but otherwise that seems right to me. If you've got one event that happens with, say, a probability of 1 in 10, and another event that also happens with 1 in 10 probability, the probability of both happening would be (1/10)(1/10)=(1/10)2=1/100, not 2*(1/10)=2/10=1/5. That value would be the probability of one or the other happening (assuming the events are independent of course; if the probability of one event changes based on whether the other happened, then the calculations will be different).
 
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