The lone golden ticket on the fourth draw

Question

A box holds 1 golden ticket and 5 blank tickets, 6 in total. You draw tickets one at a time, without replacement. What is the probability that the golden ticket comes out on exactly the fourth draw?

Accepted Answer

With just one special item, there is a shortcut, but let us confirm it with the chained approach first. Draws 1 to 3 all blank, draw 4 golden. Blanks go 5 of 6, then 4 of 5, then 3 of 4, and the golden ticket is the 1 of 3 remaining: \frac{5}{6}	imes\frac{4}{5}	imes\frac{3}{4}	imes\frac{1}{3}=\frac{60}{360}=\frac{1}{6}. The shortcut: since there is only one golden ticket, it is equally likely to sit in any of the 6 draw positions. So the chance it lands in the fourth spot is simply \frac{1}{6}\approx 0.167. Takeaway. When exactly one special item exists, its draw position is uniform over a…

The lone golden ticket on the fourth draw

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