Suppose n disks, black on one side and white on the other, are laid out in a str

Question

Suppose n disks, black on one side and white on the other, are laid out in a straight line with a random arrangement of black sides up. You are playing a game of solitaire, in which a turn consists of removing a black disk and flipping over its immediate neighbors, if any. (Two disks are not considered immediate neighbors if there used to be a disk between them that is now gone.) A game is winnable if it is possible to remove all n disks.

Describe all winnable games, and devise a strategy which will win all winnable games. [Hint: Once you think you can describe all winnable games, prove your guess using induction. Note that not only must you show you can win all games you think are winnable, but you must also prove that all others aren’t winnable. Do this by induction too.]

Explanation / Answer

Let us consider all the white disks as P(A)=1,2,3.........n

Let us consider all the Black Disks as P(B)=1,2,3,..........n

Given,that these are placed in randomly in a straight tline for playing Solitaire

Probability of removing Black Disks i.e. P(B) from a the random balls are P(B)=1-P(A)

(Since,P(A)+P(B)=1,using Inductive method)

or we can use the other process in Induction method

P(A^B)=P(A)*P(B)==>P(B)=P(A^B)/P(A)

Hence,there are are 'n' number of ways to get a black ball from the set of the line

But,the given condition says that no two balls should be placed conseutively

Lets consider that condition and the probability of drawing a Black Ball is

P(B)=(1-P(A))/P(A^B)

Therefore,by satisying all the conditions the probability will be 'P(B)=(1-P(A))/P(A^B)'

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