# Probability and Combinations

• Nov 8th 2008, 06:27 AM
RobP
Probability and Combinations
Lucky Seven is a game played with seven coasters numbered one to seven. The object is to turn all seven coasters number-side up. The coasters are shuffled and dealt number-side down; the first player turns over any coaster, and leaves it in place. The exposed number on the coaster determines the position (always reading from left to right) of the next coaster to be turned over, and so on. If the player turns over a coaster that has a number denoting one already turned, he or she is out.

What are the odds of turning all coasters over and how has it been calculated?
• Nov 12th 2008, 05:24 PM
awkward
Quote:

Originally Posted by RobP
Lucky Seven is a game played with seven coasters numbered one to seven. The object is to turn all seven coasters number-side up. The coasters are shuffled and dealt number-side down; the first player turns over any coaster, and leaves it in place. The exposed number on the coaster determines the position (always reading from left to right) of the next coaster to be turned over, and so on. If the player turns over a coaster that has a number denoting one already turned, he or she is out.

What are the odds of turning all coasters over and how has it been calculated?

Hi RobP,

The probability of turning over all the coasters is 1/7.

To see why this is so, consider the numbers on the coasters, read from left to right, as a permutation of 1, 2, 3, ..., 7. Every permutation can be written as a product of disjoint cycles, and the player will turn over all the coasters if and only if the permutation consists of a single cycle. Suppose, for example, the player turns over coasters with numbers 3, 1, 4, and then is sent back to coaster 3 and is out; then the permutation includes the cycle (3 1 4) and must have other cycles of length less than or equal to 4. On the other hand, if the permutation consists of a single cycle then all the coasters will be turned over. (If your goal is to turn over all the coasters, by the way, it doesn't matter which one you turn over first; winning or losing is determined entirely by the order of the coasters.)

A standard theorem in combinatorics states that the number of permutations of 1, 2, ..., n consisting of a single cycle is (n-1)!. So there are 6! permutations of the coasters which consist of a single cycle, and since there are 7! equally likely permutations in all, the probability of success is 6! / 7! = 1/7.