problem 30

1.3.4 problem 30

Suppose the word consists of $7$ letters. Once we choose the first letter, the seventh one has to be the same. Once we choose the second letter, the sixth one has to be the same. In general, we are free to choose $4$ letters. Hence, the probability that a $7$ letter word is a palindrome is

\frac{2 6^{4}}{2 6^{7}} = \frac{1}{2 6^{3}}

If the word consists of $8$ letters, then there are $2 6^{8}$ possible words, but for a palindrome, the number of letters we are free to choose is still $4$ . Hence, the probability is

\frac{2 6^{4}}{2 6^{8}} = \frac{1}{2 6^{4}}

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