Context: This is from my home worksheet. The question originally just asked to compare the probability between the first and the last position in line, but I wanted to find a more general argument.

Consider a box of 18 balls, of which 1 is black and the rest (17) are white. "Winning" is to pick out the black ball. Suppose that there are 18 people in the line, each to pick out 1 ball, where should you stand to maximise your chance of "winning".

*How would your choice vary if there were 2 black balls out of 18? or 3,etc?

My attempt: I brute forced the probability of the first and last postion.

$Pr(X=1)$ = $\frac{1}{18}$

$Pr(X=18)$ = $\frac{17}{18}*\frac{16}{17}*...\frac{1}{2}$

But I'm still nowhere near the equation of a general solution, and I have no idea how the probability will change if there were more black balls.

  • $\begingroup$ Arrange the $18$ balls in a line... The black one has an equal chance of being in any slot, so it makes no difference where you stand in line. $\endgroup$ – lulu Mar 11 '17 at 22:58
  • $\begingroup$ To see that from your product note that every numerator is cancelled by the next denominator. The only surviving term is the $18$ in the numerator. $\endgroup$ – lulu Mar 11 '17 at 22:59

Every position has the same likelihood of winning.

The person in position 1 has a $\frac{1}{18}$ probability of winning.

The person in position 2 has a $\frac{17}{18}\frac{1}{17}=\frac{1}{18}$ probability of winning.

The person in position 3 has a $\frac{17}{18}\frac{16}{17}\frac{1}{16}=\frac{1}{18}$ probability of winning.

The same reasoning applies for every position, and in fact, for any number of people in the same situation: one ball for every person, exactly one winning ball exists.

If two or more winning balls are included, the probability of winning is much higher for those at the front, and much lower (or even zero) for those at the back.

This chart shows the probability of winning at each position in a line of ten people, for one through to five winning balls.

a chart of position versus probability, showing five lines: one is straight horizontal, one is straight decreasing, three are exponentially decreasing with different rates

Observe that

  • the horizontal straight blue line corresponds to a single winning ball,

  • the tilted straight green line corresponds to two winning balls, and

  • higher numbers have an exponential decay curve, hitting zero after the first position guaranteed to pull a winner.

  • $\begingroup$ Ok, I redid the calculations and it does seem like they are all 1/18. How would the probability change if there were more black balls though? $\endgroup$ – Frank Mar 11 '17 at 23:04
  • $\begingroup$ surely it wouldn't be (2/18),(3/18)... would it? $\endgroup$ – Frank Mar 11 '17 at 23:09
  • $\begingroup$ No, it becomes more and more unlikely that the next person will win, and the rate of increasing unlikelihood grows with the number of winning balls used. $\endgroup$ – Nij Mar 11 '17 at 23:18

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