Summer Problem Solving Marathon Solution #42

Let x be the probability of getting heads (H) on the biased coin. Then 1-x is the probability of getting tails (T).

We first calculate the probability of getting exactly one head. There are five outcomes with exactly one head (matching the five ways of picking a location for the H in the sequence). Each of those five ways has probability x(1-x)⁴ of occurring. So p(H=1) = 5x(1-x)⁴.

Next we calculate the probability of getting exactly two heads. There are ten outcomes with exactly two heads (matching the ten ways of picking two locations for the two Hs in the sequence). Each of these ten ways has probability x²(1-x)³ of occurring. So p(H=2)=10x²(1-x)³.

Since p(H=1) = p(H=2), we have:

5x(1-x)⁴ = 10x²(1-x)³

Simplifying (given that 0 < x < 1), we have:

1 - x = 2x
x = 1/3

We now calculate p(H=3). There are ten ways of picking locations for the three heads. Each of those ten ways has probability x³(1-x)² of occurring. So our probability is:

10(1/3)³(2/3)² = 40/243