First and last golden ticket in a row of chocolate bars

Question

Thirty chocolate bars are placed in a row in random order, and exactly 6 of them hide a golden ticket. What is the expected position of the first golden ticket, and of the last?

Accepted Answer

The 6 golden tickets split the 24 plain bars into 6 + 1 = 7 gaps. By symmetry each plain bar is equally likely to fall in any gap, so each gap holds \frac{24}{7} bars on average. First ticket. It sits just after the first gap: E[	ext{first}] = \frac{24}{7} + 1 = \frac{31}{7} \approx 4.43. This is the general result \frac{n+1}{k+1} = \frac{31}{7} with n = 30, k = 6. Last ticket. By the same symmetry, the gap after the last ticket also holds \frac{24}{7} bars on average, so the last ticket sits that far from the end: E[	ext{last}] = 30 - \frac{24}{7} = \frac{210 - 24}{7} = \frac{186}{7} \appr…

First and last golden ticket in a row of chocolate bars

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