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 Mark Perakh's Web Site

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Title Author Date
question on example Spencer , Steven Feb 08, 2004
Hi TalkReason,

Improbable Probabilities

Assorted comments on some uses and misuses of probability theory

First posted on June 22, 1999; last update September 2001.

By Mark Perakh

http://www.talkreason.org/articles/probabilities.cfm
The first problem is as follows. Imagine that you watch buses arriving at a certain stop. After watching them for a long time, you have determined that the interval between the arrivals of any two sequential buses is, on the average, one minute. The question you ask is: How long should you expect to wait for the next bus if you start waiting at an arbitrary moment of time? Many people asked to answer that question would confidently assert that the average time of waiting is 30 sec. This answer is wrong. It would be correct if all the buses arrived at exactly the same interval of 1 min. However, the situation is different in that 1 min is just the average interval between any two consecutive bus arrivals. This number 1 min, is a mean of a distribution wherein the interval varies between zero and a certain maximum which is larger than 1 min. Therefore, the average waiting time, regardless of when you start waiting, is 1 min rather than 30 sec.

========================================================
Ok, I have a problem with this.
I am going to switch to a daily watch rather than each minute, for thinking simplicity.

Take a train that leaves daily.. 7AM time is planned, but it varies a bit, leaving an average interval of 24 hours even though sometimes it is 22 other times 23, 24, 25.... but it generally leaves around 7AM.

You observe if for a few weeks and say, look the average interval is 24 hours.

Now I can mosey over at anytime of day..
Let's take 24 points, from midnight to 12PM to midnight.

At Midnight we have approximately a 7hr wait, then 6-5-4-3-2-1-0- then 24-23-22- etc.

Leading to the 12 hours, 1/2 day average.

(moving around a couple of hours won't make a big difference.. 6AM 7AM 8AM.

How do you get that the average time of waiting would be 24 hours ???????

Thanks.

Shalom,
Steven Spencer
Queens, NY

Schmuel@bigfoot.com
http://groups.yahoo.com/group/Messianic_Apologetic/

Related Articles: Improbable Probabilities

Title Author Date
question on example Perakh, Mark Feb 08, 2004
Dear Steven:

Thank you for your comment. It is not new, though, and was suggested more than once before. The explanation also was given more than once before. In particular, when I was teaching statistical physics, the first few weeks usually were spent on probability theory, and more than once some students asked a question similar to yours. The reason for many people being puzzled is that the mistake is subtle and therefore understandable.
Here is the explanation: You have changed the problem by imposing an additional restriction which was absent in the original problem, namely you restricted the variations of the time interval to a small fraction of the average interval. In the original problem there is no such restriction, so the interval between buses may vary in a wide range between zero and some X which is larger than the average interval. The value of X is not restricted, but has the sole limitation: the arithmetic mean of the interval must have a certain value Y<X. If the problem were formulated in this way (and this is how it is formulated for buses) your train would depart at various times throughout the day, the actual interval between two consecutive trains varying between zero and some period of time which might be anything up to a very long period, with one and only one limitation - the AVERAGE interval must be 24 hours. For example, trains could be departing at intervals varying between zero and X hours where X can be any number, say 48 hours, or 35 hours, or 53 hours, etc, but the arithmetic mean of the interval must be some Y<X, for example Y=24 hours. If you formulated the problem in such a way, it would be analogous to the original problem with buses and the answer would be Y hours. Since your problem is formulated differently, no wonder the answer is also different (X hours). Inversely, if the original problem was formulated with the additional condition that the buses arrive every minute plus-minus just a few seconds, then it would be analogous to your problem and the answer would be similar to your case. Your confusion is one between unrestricted variations (which can exceed the mean value) and small fluctuations (which are less than the mean value). Hopefully this will clarify the matter for you.

Best wishes,
Mark Perakh
Related Articles: Improbable Probabilities