# 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?

• Note that the game ends when 10 balls of the same colour are drawn. So order matters here - in each of the events you mentioned, the last ball drawn must be blue - otherwise the event is meaningless in this game (the 10 blue balls were drawn even before). Commented Oct 13, 2017 at 21:52

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:

$$\sum_{n=0}^9\frac{{11\choose9}*{12\choose n}}{23\choose9+n}\times\frac{2}{23-9-n\choose1}=\frac{53}{161}\approx0.33$$

• You are quite right. One's grey cells aren't exactly firing smoothly at 2 am. I have adjusted properly what I was working at. Of course, you posted the correct result first, thumbs up +1 ! Commented Oct 14, 2017 at 7:12

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,

$${\sum_{k=10}^{19}}\dfrac{{\binom{k-1}{9}(23-k)}}{\binom{23}{11}}= {\dfrac {53}{161}}$$

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

• Perhaps I am misunderstanding, but I think order is critical here - i.e., the game stops after the 10th blue balls was drawn in our events. So it is more correct to consider the situation of when the 10th blue ball was drawn (see my answer) Commented Oct 13, 2017 at 22:24
• I have corrected a typo that should take care. Commented Oct 13, 2017 at 22:26
• I think that Studentmath is correct. This approach yields $0.59$, the same probability the OP got. Which can't be correct, since there are more red balls than blue in the Urn. Commented Oct 13, 2017 at 22:28
• Actually, it still leads to $0.59$ probability. The point is, I think, in that method of counting we are also considering the following as different possible events: 10 blue balls drawn first, and only then 2 red ones, 10 blue balls drawn first, and only then 1 red one, etc.. where the game would end after 10 turns in that case. Commented Oct 13, 2017 at 22:29
• I have made the slight adjustment that was needed, and expanded the explanation for greater clarity. Commented Oct 14, 2017 at 7:16

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

$${4 \choose 0}\frac{{12 \choose4}{11 \choose 0}}{23 \choose 4}+{4 \choose 1} \frac{{12 \choose 3}{11 \choose 1}}{23 \choose 4} = \frac{53}{161} \approx 0.329$$