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Red never trails blue

ProbabilityHard

An urn holds $3$ red and $3$ blue balls. You draw all $6$ balls one at a time at random without replacement. What is the probability that, after every draw, the number of red balls drawn so far is at least the number of blue balls drawn so far? Give a fraction in lowest terms.

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Hint

All color sequences are equally likely; count the sequences that never go below the line, as in a ballot problem.

Solution

Each of the $(3 6) = 20$ color sequences is equally likely.
A valid sequence never has fewer reds than blues in any prefix; these are counted by the Catalan number for $3$ pairs.
That number is $(3 6) \div 4 = 5$ , and the sequences are RRRBBB, RRBRBB, RRBBRB, RBRRBB, RBRBRB.
The probability is $\frac{5}{20}$ , which reduces to $\frac{1}{4}$ .

Answer: $\frac{1}{4}$

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