New approach: consider these three zones starting from where

the elevator leaves floor 2 going up:

|2U 3U 4U 5U 6U 7D 6D 5D 4D 3D 2D 1U |

| Zone 1 |Zone2 |Zone 3 |

An elevator has 4/6 chance of being in zone 1 and

1/6 of being in 2 and 1/6 in 3.

There is a 50/50 race between an up and a down elevator

to get to floor 2 if and only if one is in zone 2 and the other in 3.

Here are the probabilities for the location of the two elevators and the chance up will arrive first:

Elevators in Zones ways/36 Pr(Up wins)

-----------------------------------------------------

3 & 3 1/36 1

1 & 3 8/36 1

2 & 3 2/36 1/2

1 & 2 25/26 0

So Up now gets prob ( 1 + 8 + 2 x 0.5)/ 36 = 10/36 = 0.2777...,

(way better than 1/6 = 0.1666... for the single elevator)

So with 1 elevator, on floor 2 prob(up) = 1/6, prob(down) = 5/6

Two elevators, prob(up) = 5/18, prob(down) = 13/18

here is the simulationord@rabbit-3 ~ % gamow

sim: prob. up = 0.278 analytic = 0.278

lord@rabbit-3 ~ %

lord@rabbit-3 ~ % more gamow.f

program gamow

implicit none

integer iseed,i,cnt

real rat,a,b,da,db,x

iseed=time8()

cnt=0

do i=1,1000000

a=rand()*12

b=rand()*12

da=10.-a ! unneeded!

if(a.gt.10)da=10.-a

db=10.-b ! uneeded !

if(b.gt.10)db=10.-b

if(abs(da).lt.abs(db).and.da.lt.0)cnt=cnt+1

if(abs(db).lt.abs(da).and.db.lt.0)cnt=cnt+1

enddo

rat=cnt/1000000.

x=10/36.

print 1, rat,x

1 format('sim: prob. up = ',f5.3,' analytic = ',f5.3)

end