# probability paradox with two queues

There are two cash decks, and two queues of similar lenghts. You choose one of two queues as you wish, and join it. Paradox claims that whatever queue you choose, the other queue has > 0.5 probability to move faster that your queue. (I think this is funny).

I recalled this recently. But was not able to find any reference. Anybody can recall or guess what is explanation of this paradox ?

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If you repeat the experiment often, and measure the time spent in queues, you will spend more time in the slower queue. It's the same as the old joke that all bicyclists know there are more uphills than downhills: They spend more time on the former. Anyway, this is about the only “explanation” I can think of, but it seems rather … lame. – Harald Hanche-Olsen Jul 12 '12 at 20:34
If @Harald is right, then the "paradox" should be stated differently: If you pick some point in time uniformly from all time intervals that you spend standing in such queues, then the probability is $\gt0.5$ that you're standing in the slower queue. In the situation as described in the current version of the question, the probability of picking the slower queue is exactly $0.5$. – joriki Jul 12 '12 at 20:39
This reminds me of an annoying fact I observed many years ago. Suppose you are stuck in stop-and-go traffic on a two-lane road. Being greedy, if you see the other lane moving faster than yours, you start changing lanes. But this is not instantaneous; it takes some finite time $T$. Now if the difference between the speeds of the lanes goes as $\sin(\pi t/T)$, you will spend 100% of your time in the slower lane. – Rahul Jul 13 '12 at 1:05
The waiting time paradox is discussed in Feller's probability book, volume 2 I think. – Michael Chernick Jul 13 '12 at 1:38
@Michael Chernick ... Bingo, I still have Feller 2-volume in my boolshelf ... Looking ... – Andrei Jul 13 '12 at 18:22