All green out before any orange

Question

A box holds 4 green tiles and 4 orange tiles, 8 in total. You draw them one at a time until the box is empty. What is the probability that all four green tiles come out before any orange tile?

Accepted Answer

Look at the full color sequence. Picture the 8 tiles in a random row of 8 positions; only which 4 positions are green matters. Every choice of "which 4 positions are green" is equally likely, and there are \binom{8}{4}=70 of them. "All greens before any orange" is exactly one pattern: the greens take the first four positions. So P=\frac{1}{\binom{8}{4}}=\frac{1}{70}\approx 0.014. Draw-by-draw check: \frac{4}{8}	imes\frac{3}{7}	imes\frac{2}{6}	imes\frac{1}{5}=\frac{24}{1680}=\frac{1}{70}. Takeaway. With equal-sized piles the count \binom{8}{4} is large, so a perfect "all one color first" ru…

All green out before any orange

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