probability of a simple children’s race game Suppose that a simple children’s race game is played as follows, using an urn containing some blue balls and some red balls, and a “track” having positions numbered consecutively from 0 to 10. A blue “racer” starts at position 0 and moves one space forward (towards position 10) each time a blue ball is drawn from the urn. Similarly, a red “racer” starts at position 0 and moves one space forward each time a red ball is drawn, and the winning “racer” is the one that reaches position 10 ﬁrst. If balls are drawn without replacement from an urn initially containing 11 blue balls and 12 red balls, what is the probability that the blue “racer” will win?
For my answer, I considered that the blue "racer" will win when get 10 blue balls, and the number of red balls is flexible (no more than 9). So I calculated the probability of 10 blue balls with 1 red ball, plus 10 blue balls with 2 red balls plus etc. The probability is 0.59.
However, I was wondering is there has any easier way to calculate the probability?

The $10th$ blue has to be be in slots $10$ thru $19$ to meet the criterion.
Suppose the desired event occurs in the $k^{th}$ slot,
then $9$ blues must be present by the $(k-1)^{th}$ slot, which yields the simple formula, 
$$ {\sum_{k=10}^{19}}\dfrac{{\binom{k-1}{9}(23-k)}}{\binom{23}{11}}= {\dfrac {53}{161}}$$

Added Explanation
$(23-k)$ is for $\binom{23-k}1$, the number of choices for the $11th$ blue after inserting the $10th$ blue.
