First we determine the total number of 5 digit numbers, which will be the
denominator of the desired probability.

We can choose the first digit 4 ways (it can't be 0).
We can choose the 2nd digit 4 ways (the others can be 0).
We can choose the 3rd digit 3 ways.
We can choose the 4th digit 2 ways.
We can choose the 5th digit 1 way.

So there are 4∙4∙3∙2∙1 = 96 5-digit numbers with digits among {0,2,4,6,8}

The denominator of the desired probability will by 96.

Now we will determine the numerator of the desired probability, the number of
"successful" 5-digit numbers from the 96. 

The thousands digit is the 2nd digit and the tens digit is the 4th digit.

Whatever digit we choose for the 2nd and 4th digits, there must be two smaller
digits on each side of it.

The smallest we can choose the 2nd digit is 4, because 0 can't be the first
digit of a 5-digit number, and there must be smaller digits to go 1st and 3rd.

If 4 is the 2nd digit, it must be of the form 240--.  Only 6 and 8 are still
available for the 4th digit, and we must put the 8 4th, so it can only be 24086.
 
If 4 is the 4th digit, it must be of the form --042 or --240.  Only 6 and 8 are
then available for the 1st and 2nd digits, and the 2nd must be larger than the
first, so they can only be 68042 and 68240.  

All the others have 6 and 8 as their 2nd and 4th digits.  They are either of the
form -6-8- or -8-6-, with 0, 2 and 4 as 1st, 3rd and 5th. 

The ones of the form -6-8- are 26084, 26480, 46082, and 46280. 
The ones of the form -8-6- are 28064, 28460, 48062, and 48260

Here they all are:

 1.  24086
 2.  68042
 3.  68240
 4.  26084
 5.  26480
 6.  46082
 7.  46280
 8.  28064
 9.  28460
10.  48062    
11.  48260

There are 11 "successful" 5 digit numbers out of 96 possible 5 digit numbers,
so the desired probability is 11/96.

Edwin