Odds of picking matching 2 digits (each digit 0-9) picked randomly, order doesn't matter So someone randomly picks two digits, each 0-9, so for example 23, 05, 50, 99, 00, etc. 
So, if I pick 15, and the random selection is 15 or 51, I win. 
I calculated the odds of winning as 3%. There are 100 possibilities (10*10). 10 of these possibilities (e.g., 55, 66) have no "counterpart" to help you win like how, say, 31 has 13. The odds of a random selection hitting a double number like 55 or 66 are 1/10 (10 out of 100 possibilities). In that world, you have a 1/10 chance of winning. You have a 90% chance of not hitting a double number and in this world you have a 1/45 chance of winning (90 remaining possibilities / 2 since you have mirror image helpers). 
(1/45) * .9 + (1/10)*(1/10) = 3% 
But when I run a model in excel with 500,000 rows, with random number generators I'm getting something closer to 1.9%. Is the flaw in my math logic or my excel model logic? I have a feeling I'm not thinking about something quite right in my math logic. 
 A: The number of possible winning pairs is the "stars and bars" problem with $2$ balls and $10$ buckets. The solution is $$
\binom{2+10-1}{2}=\binom{11}{2}=55
$$
Therefore there are $55$ possibilities for the winning numbers. The value $\frac{1}{55}$ is about $1.8\%$, so your simulation seems to assume that each of the pairs is equally possible.
However, by the way you described the problem, not every pair of winning numbers has equal chance of being the winning numbers: for instance $00$ can only be the winning numbers if a $0$ is picked and then a $0$ is picked. However, $12$ can be the winning numbers if $1$ is picked and then $2$ is picked or if $2$ is picked and then $1$ is picked. Therefore, $12$ (or $21$) has twice the chance of $00$ of being the winning numbers.
A: Let's say the first person picks a two-digit number at random (including numbers with a leading zero). Each number is equally likely, with a probability of $0.01$.
Then, the second person does the same. Ten of the numbers ($00$, $11$, $22$, ... $99$) will have probability $0.01$ of matching, while all of the others will have probability of $0.02$ of matching (since there are two numbers with the same set of different digits).
Adding up all of the probabilities gives $0.01(10\cdot 0.01 + 90 \cdot 0.02) = 0.019 = 1.9\%$, which is what you got from your simulation.
A: The error is your model.
You say that if you draw a double pair you use the phrase "In that world, you have a 1/10 chance of winning. " 
You seem to be assuming that if YOU drew a pair, then the other person had to ALSO draw a pair.
And when YOU don't draw a pair you say " and in this world you have a 1/45 chance of winning" implying you are assuming the other person ALSO drew a mismatch.
But the other person draw is independent of yours.    There are, with respect to order, $100$ things she could have drawn and that is true no matter what you drew.   So in the world where you drew a pair she has a $\frac 1{100}$ probability of matching it.   And in the world were you didn't draw a pair, she has a $\frac 2{100}=\frac 1{50}$ probability of matching it.
So the probability of a match is
$\frac 1{10}\frac 1{100} + \frac {9}{10}\frac {2}{100} = \frac {19}{1000} = 1.9\%$.
=====
Alternatively:
Probabilty that you both drew a pair:  $\frac 1{10}\frac 1{10}= \frac 1{100}$.  If so probability they match: $\frac 1{10}$.
Probability that you both drew a mismatch: $\frac 9{10}\frac 9{10}=\frac {81}{100}$.  If so probabitilty they match: $\frac 1{45}$.
Probability that you drew a pair and she didn't: $\frac 1{10}\frac 9{10} = \frac 9{100}$.  If so probability they match: $0$.
Probability that you didn't draw a pair and she did: $\frac 9{10}\frac 1{10} =\frac 9{100}$.  If so probability they match: $0$.
So probability of match:
$\frac 1{100}\frac 1{10} + \frac {81}{100}\frac 1{45} + \frac 9{100}\cdot 0 + \frac 9{100}\cdot 0= \frac 1{1000} + \frac {81}{4500} = \frac 1{1000} + \frac 9{500}=1.9\%$
