A fair coin is tossed 10 times. Find the probability of satisfying the condition that that the coin does not land on tails twice in a row

Answer 1

Let F_n be the number of n-toss sequences containing no pair of tails.

Then the probability you are looking for is

F₁₀/2¹⁰

because 2¹⁰ is the total number of 10-toss sequences.

For any given n there are two kinds of n-toss sequences without a pair of tails.

1) Sequences starting with heads and followed by an (n-1)-toss sequence without a pair of tails.

There are F_n-1 such sequences.

2) Sequences staring with tails followed immediately by heads, and then followed

by a (n-2)-toss sequence without a pair of tails.

There are F_n-2 such sequences.

Thus we have the recurrence relation

F_n = F_n-1 + F_n-2

This is a Fibonacci sequence with

F₁ = 2

F₂ = 3

F₃ = 5

F₄ = 8

F₅ = 13

F₆ = 21

F₇ = 34

F₈ = 55

F₉ = 89

F₁₀ = 144

The answer to your question is 144/1024 ≈ 14%