I don't think that visiting the same street three times is all THAT probable. Pick any combination of three journeys out of eleven. The probability of them being to the same street are 1/2717 * 1/2717 (which is not a lot, about 0.000014%). There are many combinations of three and so many chances. How may? The answer (see http://en.wikipedia.org/wiki/Combinations) is C(11, 3), or (11! / (3! * (11-3)!)), which is 165. The probability of visiting a street three or more times is therefore 165*1/2717*1/2717, which is 0.000022 or 0.0022% (1 in 44740). For two it's rather more likely. C(11, 2) is 55 and the probability of any two journeys being the same is 1/2717 - so the probability of visiting the same street twice is 55/2717 or about 2%. For neighbouring streets it's a little more likely. Suppose that every street has three neighbours. The probability of the second of a combination of two journeys being to the same or a neighbouring street is 4/2717. 55*4/2717 is 8.1%, which gives the probability of going to any 'neighbourhood' of four streets twice. For three visits the corresponding result is 165*4/2717*4/2717, or 0.036%. This all assumes a variety of things: that visit probabilities are independent, that streets are all the same size, that you always do 11 visits and so on. Breaking just about any of these puts the probabilities up. If there are 2716 one house streets and one mega-street then you'll visit Mega Street many times a day with high probability. If you're routed to streets near to your previous call with higher probability than more distant ones then it goes up. If the probability of medical emergency changes in different areas (eg, because all the old people or poor people live in one place) then it goes up. If the probability of a visit are not independent (eg, because all the people live in one place and the weather increases the risk of emergency for people more than it does for everyone else) then the chances go up. etc. Xelah