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?

Let X, Y, and Z denote respectively the values of the first, second, and third prizes, expressed in $500 units. So we know that X is in the interval (1,4), Y in (2,5), and Z in (3,6).

Let us assume that these are randomly distributed variables in their respective
intervals. Note that the volume of the cube bounded by the three above intervals is 27. So the probability of any event involving X, Y, and Z can be obtained by finding the volume of the subregion of the cube associated with the event and dividing the volume by 27.

We will try to find the probabilities associated with realizing each of the 6 possible orders
among the 3 prizes, and choose the one order that has the most probability.

For instance, let us consider the ordering where we take the first prize, second prize, and third price in that order. The probability for this to be correct is P( X < Y < Z) which can obtained by splitting this event into several components:

a) P[X in (1,2), Y in (2,3), Z in (3,6)] = (1).(1).(3)/27=3/27
b) P[X in (1,3), Y in (3,5), Z in (3,6), Z > Y] = 8/27, obtained by integrating (1/27).(6-y) over the rectangle y=3 to y=5 and x=1 to x=3.
c) P[X in (2,3), Y in (3,5), Y>X, Z in (3,6)] = (3/2).(1/27), obtained by integrating (1/27).3y over the region y=x to y=3 and x=2 to x=3.
d) P[X in (3,4), Y in (3,5), Y>X, Z in (3,6), Z >Y] = (8/3).(1/27)
obtained by integrating (1/27).(6-y) over the region y=x to y=5 and x=3 to x=4.

Adding these we get P[X
Fortunately for us, we do not need to consider any other ordering and can decide that the above ordering (first,second,third) is the best because its probability is 91/162 and therefore the sum of the probabilities of all other orderings combined is 71/162 which is less than 91/162.

Posted by Prab on 2003-11-05 22:07:57