Question

We toss ten fair coins independently. what is the probability that the sequence of tosses is not a palindrome?

Answer 1

The key is realizing that the game only reaches the nth toss if tosses 2 to n-1 were all opposite the first toss:
HH is a palindrom

HTH is a palindrome

HTTH is a palindrome

HTTTH is a palindrome

…

Complex palindromes like HTTHTTH never occur. In this case the game would have ended at the 4th toss.

So the probability that the game continues to the nth toss is P(n) = (1/2)^(n-2). As n approaches infinity, P(n) approaches zero.

(Vihren Lindrov said it in a different way, but definitely said it before I did.)

Answer 2

Answer:

Step-by-step explanation:

The key is realizing that the game only reaches the nth toss if tosses 2 to n-1 were all opposite the first toss:

HH is a palindrom

HTH is a palindrome

HTTH is a palindrome

HTTTH is a palindrome

…

Complex palindromes like HTTHTTH never occur. In this case the game would have ended at the 4th toss.

So the probability that the game continues to the nth toss is P(n) = (1/2)^(n-2). As n approaches infinity, P(n) approaches zero.

(Vihren Lindrov said it in a different way, but definitely said it before I did.)