First we find the number of $ 4 $ digit palindromes. There are ten palindromes for every distinct thousandth digit from $ 1 $ to $ 9 $ because there are $ 10 $ numbers from $ 0 $ to $ 9 $ we could pick for the second and third digit. This gives us a total of $ 9 \cdot 10 $ palindromes. Next, we can get that all palindromes are multiples of $ 11 $. The divisibility rule for $ 11 $ tells us that for a number $ abcd $ to be divisible by $ 11 $, then $ a-b+c-d $ is divisible by $ 11 $. Since $ a=d $ and $ b=c $, $ a-b+c-d $ is always divisible by $ 11 $ so all four digit palindromes are divisible by $ 11 $. Now we want to find now many of these palindromes are divisible by $ 9 $. For a number to be divisible by $ 9 $, the sum of the digits must be divisible by $ 9 $. It's impossible for the sum of the digits to be equal to $ 9 $ or $ 27 $ because it must be an even number (the sum is $ a+b+c+d=2(a+b) $). We find the number of palindromes whose digits add up to $ 18 $. Since $ a+b+c+d=2(a+b)=18, $ we get that $ a+b=9 $. There are $ 9 $ possible answers, where $ a $ goes from $ 1 $ to $ 9 $ and $ b=9-a $. We then find the number of palindromes whose digit add up to $ 36 $. There is only one four-digit number that does so, $ 9999 $. Therefore, we have that there are $ 9+1=10 $ four-digit palindromes that are divisible by $ 99 $. Since there is a total of $ 90 $ palindromes, the probability that it is divisible by $ 99 $ is $ \frac{10}{90}=\boxed{\frac19} $.