George Gamow and Marvin Stern occupied offices on the second and sixth floors of a seven-story building, and noted that when either took the elevator to the other's floor, it was going the wrong way. It's apparent why: there were ten segments of the elevator's 12-segment cycle (6 going up and 6 going down in a continuous cycle) where the first elevator arrival would be going the wrong way and only two segments where it would be going the desired direction the next time it passed the boarding floor.

But what if a second elevator were placed in the building. What would the probability be that the next elevator to arrive would be going the wrong way? Ignore stops along the way, as they do not affect the distance that need be traveled and probably have more of them for longer trips. The two elevators move independently of each other.

Gamow himself did not get the correct answer for the two-elevator case, but the correct answer was found by Donald Knuth.

w/ one elevator: the elevator just keep chugging along, independent of passengers. Whenever the door opens on the 2nd floor, there is a 50/50 chance it is going up, since for every round trip it makes, it opens the 2nd floor door once going up and once going D.

But, if you arrive at the 2nd floor lobby at a random time there is a 1/6th chance that when you arrived, the elevator was in its 2D 1U phase and not its 2U 3U 4U 5U 6U 7D 6D 5D 4D 3D phase. So there is a 1/6 chance the next arrive is going up.

With a second elevator the probabilities don't change:

There is a 1/36 chance they both are in the nice 2D 1U phase, 5/36 that just one is, and 20/36 that they are both is the longer phase.

So when the door opens, there is only a 1/6th chance for "up".

The wait time for the next arrival has halved, however.

*Edited on ***July 27, 2020, 7:08 pm**