
cars on the road
Here is my solution:
Let p designate the probability that in any 5 minute interval that you observe a car. Then q = 1  p is the probability that in any 5 minute interval you do NOT observe a car.
We are told that P = 609/625 is the probability that in any 20 minute interval that you observe a car.
Now, any 20 minute interval can be broken down into a sequence of 4 nonoverlapping 5 minute intervals. Assuming that what happens in each of these 4 intervals is completely independent of each other (is this what is meant by "constant default probability"?), then it must be that
P = 1  (q^4)
i.e. the probability to observe a car in 20 minutes is equal to 1 minus the probabilty that NO car is observed in all 4 of the 5 minute intervals. Doing the math yields
609/625 = 1  (q^4)
(q^4) = 16/625
q = 2/5
which implies that
p = 1  q = 3/5
So, there is a 60% chance that you observe a car in 5 minutes.
Brent Boyer
Wednesday, July 24, 2002
The answer is: 609/625.
The riddle turns on the ambiguity in the phrase "in X minutes", which can mean "within X minutes" or "X minutes from now". If the latter, assuming contant default probability, then the chances of seeing a car in five minutes
are identical to the chances of seeing one in twenty minutes.
David J.
Wednesday, July 24, 2002
609/625 :: 152.25/156.25
Brad Tilley
Wednesday, July 24, 2002
Followup comment to my original solution:
At first glance, many people are likely to intuitively guess that a solution of this problem is p (the probability for 5 minutes) should be 1/4th of P (the probability for 20 minutes):
p = P/4 = (609/625)/4 = 0.2436 = 24.36%
It turns out that this is actually a good aproximation when P is small.
To see this, note that the exact expression is
p = 1  q = 1  (1  P)^(1/4)
When P is small, the Taylor series for the fourth root may be approximated as
(1  P)^(1/4) = 1  (1/4)P + ... (terms in higher powers of P that can be discarded)
so that
p ~ P/4
which bears out your intuition.
Brent Boyer
Wednesday, July 24, 2002
David writes:
*****
The answer is: 609/625.
The riddle turns on the ambiguity in the phrase "in X minutes", which can mean "within X minutes" or "X minutes from now". If the latter, assuming contant default probability, then the chances of seeing a car in five minutes
are identical to the chances of seeing one in twenty minutes.
*****
First, I have never heard the expression "constant default probability" before, so I am not exactly sure what the original poster meant; I may be wrong simply because he was unclear.
Second, the "aha" rating of this problem has NO exclamation marks, which I thought meant that this is NOT a trick question or riddle (i.e. your attempt to dodge mathematical analysis by a semantic argument is suspect).
Third, I suspect that it is not accidental that the original poster chose the precise ratio 609/625 which so neatly turns out to yield a convenient fraction (3/5) after you do the analysis.
Finally, I do not understand your argument at all, unless you are taking "X minutes from now" to actually mean "at that POINT in time EXACTLY X minutes from now". In other words, that you are completely excluding the concept of "happening within a time interval", and only consider things happening at distinct POINTS in time. Then, I suppose that you could argue that since any point in time (whether 20 minutes from now or 5 minutes) is actually always 20 minutes ahead from some earlier point in time, and the probability is constant, then the probability is 609/625 AT ANY POINT IN TIME (whether 1 minute, 5 minutes, 9 minutes, etc from now).
But there is a small problem here: if you are assuming that time is continuous (which is what we normally assume, as opposed to thinking that it is discrete), then probabilities are always integrals of the probability density function over time and to say that any particular point in time has a probability anything other than 0 is to say that its probability density function is actually a delta function (i.e. infinite spike) at that point. If you are correct that every point in time has the probability of 609/625, then an integral over any finite interval of time yields an infinite number, which is nonsense.
All this said, I definitely agree with you that the original poster failed to use clear language. He should have posed the question something like:
"If the probability of observing at least one car on a highway during any 20 minute time interval is 609/625, then what is the probability of observing at least one car during any 5 minute time interval? (Assume that car events are completely independent of each other)."
Brent Boyer
Wednesday, July 24, 2002
I fought and argued with myself for a while, convinced that it must be a simple matter of 1/4th, until I realized what makes a difference  multiple cars. If, for example, we were dealing with one car going around an incredibly long track, then 1/4 would hold, since you can take the 20 minute probability and find the probability in the first 10 minutes and the second 10 minutes of that interval  since the chance is constant, those odds must add to the 20 minute interval.
But with the possibility of multiple cars, the odds of spotting during the first 10 minutes plus the odds of spotting during the second 10 minutes, together, are greater than the odds for the entire 20. I think it's the tendency to think of "seeing a car" as a single event that throws people into the "divide by 4" mindset immediately.
Nice one.
Sam G
Wednesday, July 24, 2002
How fast are the cars moving?
If it takes 20 minutes for a car to pass me then there is 100% probability that I will (still) see a car 5 minutes after I first see it.
William Frantz
Friday, July 26, 2002
Let us for what Constant Default Probability means, since it makes little sense to most of us!
Let us assume that the car takes 20 or more minites to pass the observer.
Also, suppose 609/625=X.
Now, no matter how large or small the interval duration, no two intervels are independent of each other unless the cars travel infinitely fast. This fact is independent of the time the cars take to cross the observer.
If the 20 minite intervels are divided into 4 5minite intervals, the probability of seeing a car in the:
1st 5 minite interval is: X/4
2nd 5 minite interval is: X/4+X/4 (Probability that you see a car in this interval + the Probability that you see a car in the previous interval)
Similarly,
3rd 5 minite interval is: 3X/4
4th 5 minite interval is: 4X/4=X
These equations are valid only if our assumtion that the cars take 20 minites or more to cross the observer is valid.
Hence, the probability of you seeing a car in a 5 minite interval is (assuming that you choose among the 4 5minite intervals uniformly)
P(5)=(1/4(X/4))+(1/4*(2X/4))+(1/4*(3X/4))+(1/4*(4X/4))
Please comment on the above.
Thank you.
Chichi
Saturday, August 31, 2002
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