B4. Bob and Jane hold identical decks of twelve cards, three of each colour: red, green, yellow, and blue. Bob and Jane shuffle their decks and then take turns dealing one card at a time onto a pile, with Jane going first. Find the probability that Jane deals all her red cards before Bob deals any of his red cards.
Give your answer in the form of a fraction in lowest terms.
Solution: We may assume the decks consist of 3 red and 9 black cards. We view each shuffled deck as a binary sequence of length 12 , with 1 's and 0 's indicating red and black cards, respectively. We wish to compute the probability that none of the 1's in Bob's sequence come before the rightmost 1 in Jane's sequence. A valid pair of sequences can be modelled by a single sequence of length 12 containing either 5 or 6 ones. In both cases, the first three 1's correspond to Jane's red cards while the last three ones correspond to Bob's. (The combined sequence of length 12 will contain 5 1's in case when Bob's first red cards goes into the pile right after the last Jane's red card.)
There are $\binom{12}{5}+\binom{12}{6}=\binom{13}{6}$ valid pairs of sequences out of $\binom{12}{3}^{2}$ pairs altogether, giving a desired probability of $\binom{13}{6} /\binom{12}{3}^{2}=\frac{39}{1100}$.
Alternative solution: Consider sequences of 24 positions numbered by odd numbers $(1,3,5, . ., 23)$ for Jane's moves and by even $(2,4,6, \ldots, 24)$ for Bob's. We look at the most right position of Jane's 1 and find the number of possibilities for her other two 1's to the left of it and Bob's 1's (among his 12 possible positions) to the right of it. Then we sum up all these cases to get:
$$
\binom{2}{2}\binom{10}{3}+\binom{3}{2}\binom{9}{3}+\binom{4}{2}\binom{8}{3}+\binom{5}{2}\binom{7}{3}+\binom{6}{2}\binom{6}{3}+\binom{7}{2}\binom{5}{3}+\binom{8}{2}\binom{4}{3}+\binom{9}{2}\binom{3}{3}=\binom{13}{6}
$$

Or by explicit calculation of the coefficients:
$$
1 \times 120+3 \times 84+6 \times 56+10 \times 35+20 \times 15+21 \times 10+28 \times 4+36 \times 1=1716 \text {. }
$$

The total number of possibilities is $\binom{12}{3}^{2}=220^{2}=48400$.
So the probability is
$$
\frac{1716}{48400}=\frac{429}{12100}=\frac{39}{1100} .
$$

Answer: $\frac{39}{1100}$.