Coin Tossing Game 1

Emily and David are playing a coin toss game. They repeatedly flip a coin. Emily wins if 3 heads appear, while David wins if 3 tails appear. The heads and tails do not need to be consecutive. What is the expected number of coin tosses before a winner is determined?

Answer

The first insight is that the number of tosses $$X$$ until a winner is determined can only be 3, 4, or 5 (a winner must be decided by the 5th toss).

$\displaystyle P(X=3)$ : This occurs when the first three tosses are either all heads (HHH) or all tails (TTT), so
$P(X=3)=\frac{2}{8}.$
$\displaystyle P(X=4)$ : For the game to end on the 4th toss, the first three tosses must not be all the same (probability $\frac{6}{8}$ ), and the 4th toss must match the outcome that appears twice in those three tosses (probability $\frac{1}{2}$ ). Thus,
$P(X=4)=\frac{6}{8}\times\frac{1}{2}=\frac{6}{16}=\frac{3}{8}.$
$\displaystyle P(X=5)$ : This occurs when the first four tosses yield exactly 2 heads and 2 tails. There are $\binom{4}{2}=6$ ways to arrange 2 heads and 2 tails, so
$P(X=5)=\frac{6}{16}=\frac{3}{8}.$

Hence, the expected number of tosses is:

E = 3\cdot\frac{2}{8} + 4\cdot\frac{3}{8} + 5\cdot\frac{3}{8} = \frac{6+12+15}{8} = \frac{33}{8} = 4.125.

Notes and comments

Comment 1: This is trivial but note that when we count the number of outcomes with 2 heads and 2 tails using $\binom{4}{2}$ , we are considering the order in which heads and tails appear but also removing the importance of order among the heads themselves—hence the probability is $\frac{6}{16}$ .