Location is Everything

Probability Level 3

In an industrial town, there are 3 bakeries X, Y, Z which sell the same bread at distinct prices. Everyday, their prices are randomly ordered such that each of the 6 possible orderings (3!=6) is equally likely. You can't know the prices until you visit the bakeries. After work, while people walk along the route Factory → X → Y → Z → Home , \text{Factory} \to \text{X} \to \text{Y} \to Z \to \text{Home}, they buy bread for the next morning at one of the 3 shops. The following are their behavioral pattern observed for long:

  • To get a lower price, they can go back 1 shop ( ( e.g. Y → \to X or Z → \to Y ) , ), but not 2 shops ( ( e.g. Z → \to X ) ) because it's too far.
  • They each try to maximize the probability of getting the cheapest price on their bread.

Which bakery has the best location?

List X, Y, Z in order from the most visited to the least visited shop.

XYZ XZY YXZ YZX ZXY ZYX

Jimin Khim by Jimin Khim

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1 solution

Jimin Khim Staff
Oct 23, 2017

Checking the price at X and buying bread there without further investigation can't be optimal because they can always come back to X after checking with Y. So, let's see what happens when they check the price at X, and then check with Y.

  • Case 1: Price is cheaper at X than at Y (for short, X < Y) with probability 1 2 . \frac12. ( \Big( Comparing only X and Y, this probability is 1 2 , \frac12, not 1 3 . ) \frac13.\Big)

    • Then what's happening must be one of the following three: (1) X < Y < Z, (2) X < Z < Y, (3) Z < X < Y.
      That is, given the information X < Y, the probability of X being the cheapest is 2 3 , \frac23, while that of Z being the cheapest is 1 3 . \frac13.
    • So, in Case 1, to maximize their probability of getting the cheapest price, they will go back to X.

  • Case 2: Price is cheaper at Y than at X (i.e. X > Y) with probability 1 2 . \frac12.

    • Then what's happening must be one of the following three: (1) X > Y > Z, (2) X > Z > Y, (3) Z > X > Y.
      That is, given the information X > Y, the probability of Y being the cheapest is 2 3 , \frac23, while that of Z being the cheapest is 1 3 . \frac13.
    • So, in Case 2, they are guaranteed to get the cheapest price either at Y ( \Big( with probability 2 3 ) \frac23\Big) or at Z ( \Big( with probability 1 3 ) , \frac13\Big), the choice between which can only be determined after they check with Z.

Therefore, the following are the probabilities that they will end up shopping at X, Y, Z, respectively: X : 1 2 , Y : 1 2 × 2 3 = 1 3 , Z : 1 2 × 1 3 = 1 6 , \text{X}: \frac12,\quad \text{Y}: \frac12\times \frac23=\frac13,\quad \text{Z}: \frac12\times \frac13=\frac16, which implies that they are most likely to buy bread at X and least likely to buy it at Z.

So, the answer is XYZ. □ _\square

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