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?
Proper order: A < B < C
Likelihood of winning: 91/162
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Proof:
Let's call the first prize A, the second prize B, and the third prize C.
Assumption: each prize is equally likely to fall anywhere within its range
A:
1 - 0500-1000
2 - 1000-1500
3 - 1500-2000
B:
4 - 1000-1500
5 - 1500-2000
6 - 2000-2500
C:
7 - 1500-2000
8 - 2000-2500
9 - 2500-3000
There are 3x3x3=27 equally likely
regions that these could fall in, and we can describe each region as an ordered triplet. These regions can be divided up into 4 groups of size 10, 10, 6, and 1.
The first group consists of those 10 regions that unambiguously guarantee a win (
+10).
{1, 4, 7} {1, 4, 8} {1, 4, 9} {1, 5, 8} {1, 5, 9}
{1, 6, 9} {2, 5, 8} {2, 5, 9} {2, 6, 9} {3, 6, 9}
The second group consists of those ten regions where A is DEFINITELY less than B and C
-or- where C is DEFINITELY greater than A or B. In these regions the
other
two will be in the
proper order 1/2 of the time for a win (10/2 =
+5):
{1, 5, 7} {1, 6, 8} {2, 4, 7} {2, 4, 8} {2, 4, 9}
{2, 5, 7} {2, 6, 8} {3, 5, 8} {3, 5, 9} {3, 6, 8}
The third group consists of those six regions that unambiguously guarantee a loss (
+0).
{1, 6, 7} {2, 6, 7} {3, 4, 7} {3, 4, 8} {3, 4, 9} {3, 6, 7}
And the final group consists of the one region where A, B, and C all fall within 1500-2000. It is equally likely that we have them any of the six possible orders:
ABC, ACB, BAC, BCA, CAB, CBA, but only
ABC will give a win (
+1/6):
{3, 5, 7}
So, totalling up, we have: 10 + 5 + 1/6 = 91/6
Putting this over the 27 possible regions we have:
(91/6) / 27 =
91/162
Since this is greater than 50%, any other possibility is less than 50%. Therefore this has the greatest likelihood to occur.
Edited on October 15, 2003, 3:42 pm
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