In a game show, there is a game in which you have to order the value of three prizes in order of least expensive to most expensive. You have to get all three right in order to win.

The only problem is you have your spouse do all the shopping, so you only know that the first prize is between 500 and 2000, the second prize is between 1000 and 2500, and the third prize is between 1500 and 3000.

Which order should you put them in so that you have the highest probability of winning, and what is the probability that you will win using this arrangement?

You should put the prizes in the order they are presented with a chance of winning of 91/162 (around 56.17%).

To achieve this, I put the prices into ranges of $500, like so:

a	500-1000


b	1000-1500


c	1500-2000


d	2000-2500


e	2500-3000

These are all the possible values in the puzzle.
Prize 1 has possible values of a, b, or c; prize 2 has possible values of b, c, or d; and prize 3 has possible values of c, d, or e.

This gives 3³=27 possible results, as far as into which ranges each price could potentially fall into, and 3!=6 possible orderings of prizes 1, 2, and 3.

For each of the 27 sets of range assignments, there are certain possibilities for the prize rankings; for example, if prize 1 was in range c, prize 2 was in range b, and prize 3 was in range e, there is only one possible ordering, 2 is the least expensive, then 1, and then 3. On the other hand, if prize 1 was in range b, prize 2 was in range d, and prize 3 was in range d, there are two possibilities: the prizes are, in ascending order, either 123 or 132. That particular arrangement is not deterministic, but both outcomes are equally likely.

That being said, we can list all 27 outcomes for the ranges, and then determine the possible outcomes for the prize ranks in order. As it turns out, 14 of these are deterministic, 12 have two possible values, and one (all three prices fall into range c) has 6 equally likely outcomes.

Here is the list:

abc	123


abd	123


abe	123


acc	123(1/2)	132(1/2)


acd	123


ace	123


adc		132


add	123(1/2)	132(1/2)


ade	123


bbc	123(1/2)		213(1/2)


bbd	123(1/2)		213(1/2)


bbe	123(1/2)		213(1/2)


bcc	123(1/2)	132(1/2)


bcd	123


bce	123


bdc		132


bdd	123(1/2)	132(1/2)


bde	123


cbc			213(1/2)	231(1/2)


cbd			213


cbe			213


ccc	123(1/6)	132(1/6)	213(1/6)	231(1/6)	312(1/6)	321(1/6)


ccd	123(1/2)		213(1/2)


cce	123(1/2)		213(1/2)


cdc		132(1/2)			312(1/2)


cdd	123(1/2)	132(1/2)


cde	123

Remember that each outcome has an overall probability of 1/27.

Based on that, we can find out how likely each of the six rankings are:

P(123) = (1/27)(10 + 10(1/2) + 1/6) = 91/162
P(132) = (1/27)(2 + 6(1/2) + 1/6) = 31/162
P(213) = (1/27)(2 + 6(1/2) + 1/6) = 31/162
P(231) = (1/27)(1/2 + 1/6) = 4/162
P(312) = (1/27)(1/2 + 1/6) = 4/162
P(321) = (1/27)(1/6) = 1/162


As could be expected from the problem, the order 1, 2, 3 is by far the most likely result, occuring a little more than half of the time.
Edited on October 15, 2003, 4:29 pm

Posted by DJ on 2003-10-15 16:24:49