Probability of unfair coin not giving 2 heads

combinatoricsindependencemathematical modelingprobabilityprobability theory

Problem: An unfair coin with probability $p$ to flip Tails is flipped 4 times independently. What is the probability of no Heads twice in a row?

My answer: I calculated the probability that of the event

$$A = \text{Heads twice in a row}$$

by splitting it into the three events

$$A_i = \text{i Heads, Heads twice in a row}$$

, took the complement and got:

$\mathbb{P}(A^c) = 1- 3p^2(1-p)^2 – 4p(1-p)^3 – (1-p)^4$

Is there a quicker way to derive the answer?

Best Answer

Flipping 4x with no heads twice in a row means either you had 2 heads separated by at least one tail, or 1 head, or no heads. (This is because if you had 3 or 4 heads with 4 coins, you must have had at least 2 in a row.)

So break this down:

  • 2 heads (but no consecutive heads). There are 3 configurations (HTHT, HTTH, THTH). Each has probability $p^2(1-p)^2$.
  • 1 head: There are four configurations, each has probability $p^3(1-p)$.
  • No heads: There is one configuration, with probability $p^4$.

Add these up and you get $$ 3p^2(1-p)^2 + 4p^3(1-p) + p^4 $$

Edit- Your approach is also correct, though you could stand to show your work :) You can verify this by expanding these polynomials in $p$ and showing that they are equal.

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