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?
 A: You should note that order is critical here. As such, it may be harder to simplify the calculation (note that yours is slightly lacking, again due to the ordering issue - also, as there are less blue balls than red ones, it should come to attention that the probability of the blue racer winning should be less than 0.5), but I will try to explain it clearly and give  a nice formula.
It is critical to understand that the event where the blue player wins happens if the 10th blue ball is drawn before the 10th red one.
Equivalently, we are seeking the probability that the 10th blue ball was drawn after 10 to 19 turns. That is, the probability of drawing 9 blue balls and $1\le n\le 9$ red ones, and the final one being blue as well. This leads us to:
$$\sum_{n=0}^9\frac{{11\choose9}*{12\choose n}}{23\choose9+n}\times\frac{2}{23-9-n\choose1}=\frac{53}{161}\approx0.33$$
A: 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.
A: There is a quicker way involving the sum of two terms rather than ten:


*

*The probability blue is first to $10$ is the same as imagining all $23$ balls are drawn and the last four balls drawn either are $4$ reds or are $3$ reds and $1$ blue.  

*And that is the same probability that the first four balls drawn either are $4$ reds or are $3$ reds and $1$ blue, which is 
$${4 \choose 0}\frac{{12 \choose4}{11 \choose 0}}{23 \choose 4}+{4 \choose 1} \frac{{12 \choose 3}{11 \choose 1}}{23 \choose 4} = \frac{53}{161} \approx 0.329$$ 
