View Full Version : betting on the street car

05-22-2002, 06:39 PM
i am not quite so good with advanced math, but a particular scenario has my interest. the streetcar in my neighborhood is scheduled to arrive at any stop every 15 minutes on a weekday.
assuming that the street car is precisely on time,what are the odds that it will show up within one minute of my arrival to the stop and what are the odds of its arrival for each minute after that?
since there is no timetable printed for the route, i could possibly take bets on its arrival. what kind of odds should i give for each minute?

Church Key Kid
05-22-2002, 08:04 PM
If nobody knows what time it will show up, the odds it will show up in the 1st minute are 1/15. If it doesn't show up then, the odds that it will show up in the next minute are 1/14, and so on. If the car still does not show up after 14 minutes, the odds that it will appear in the next minute are 1/1.

Of course, that's all assuming that you actually do show up in the 1st minute (ie, immediately after the car has just left).

I am not a mathematician... so I have no idea how to calculate it if you showed up around say the 7th minute (and assuming that you don't know that it's the 7th minute.)

05-23-2002, 02:23 AM
actually fatdave your analysis is correct for the situation where you don't know which minute it is, and it could be the first or the seventh or whatever.

If you DO get there in the 'first minute' then obviously the chance of the streetcar arriving exactly one minute later is exactly 0.

Which (in my neighborhood) is also pretty much equivalent to the probability of public transport arriving 'exactly on time' But I digress...

05-23-2002, 10:05 AM
Public transport has a strong tendency towards vehicle clusters. In theory, there should be one car every 15 minutes; in practice, you'll sometimes have to wait 30 minutes until one arrives, and immediately afterwards, often just driving behind, you have another car of the same line. Murphy's Law in action.

And the strange thing: There is even a logically waterproof explanation for this.

05-23-2002, 02:40 PM
And the strange thing: There is even a logically waterproof explanation for this.Which is thus: Suppose that the cars drift slightly off schedule, such that two are a little closer together than they're supposed to be. The one in front has, on average, more people to pick up at each stop, since those folks have been waiting longer. Further, it is less likely for the front one to have a stop with nobody waiting there, for the same reason. The more people you need to pick up, the longer each stop takes, so the front car is, on average, slower, and the back car catches up.

But, of course, the OP is stipulating the assumption that the cars are all exactly on time, so it's a moot question. Of course, also, if they're exactly on time, you could construct a schedule yourself, and always know exactly when they'll show up, so nobody's likely to take your bets.

05-25-2002, 12:28 PM
i think the odds of the streetcar arriving on the 14th minute would be roughly the same as the 1st minute. wouldn't i have a 1/1 chance at the 7th minute?

05-25-2002, 12:45 PM
mangomerlot: If you know the car will arrive in 15 minutes (this is the extreme case, if you came to the stop immediately after the departure of the last car) or less, the situation is the following:

Bus will certainly arrive within 15 mins. The odds whether I came to the stop in the 1st, 2nd, 3rd, 4th.....or 15th minute after departure are the same, so probability for the event "car will arrive within the next minute" is 1/15.

After one minute, you know for sure the car will arrive within 14 mins - 15 is impossible, because since you've already waited 1 min this would mean time between two cars is 16 mins, which contradicts the assumptions. The odds for every single minute are the same, and 14 minutes are possible, so the event for "bus will arrive within one minute" is 1/14.

And so on; with every minute, the number of remaining possible minutes decreases by 1, until the 15th minute of me waiting at the stop when the number of possible arrival minutes is 1, so the odds of "car will arrive within one minute" are 1. That's fatdave's result.

In order to avoid confusion, it is important to keep in mind that "first minute" here means "first minute of myself waiting at the stop", not "first minute after departure of last car".

I suppose hat you mean with 1/1 odds for the 7th minute, mangomerlot, is that the odds are equal that you've started waiting at the stop within the first 7.5 or the second 7.5 minutes after departure of the last car. This is true, but it doesn't mean the probability of the event "car will arrive within one minute" is even if you've already waited 7 mins.

05-25-2002, 12:52 PM
Would this be a streetcar you desire? :D

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