Game Overview

This game involves players A and B tossing a fair coin sequentially until a head appears. A tosses once, then B tosses twice. This pattern repeats.

• A tosses (1 coin). If H, A wins.
• If A gets T, B tosses (2 coins). If H appears, B wins.
• If B gets TT, the turn returns to A.
• The cycle (A: 1 toss, B: 2 tosses) repeats until a player wins by getting H.

Probability Calculations

Let $P(H) = 1/2$ be the probability of heads and $P(T) = 1/2$ for a fair coin.

Let $P_A$ be the probability A wins, and $P_B$ be the probability B wins. Since the game eventually ends, $P_A + P_B = 1$.

Determining $P_A$ (Player A Wins)

A wins if they get H on their toss.

Consider the probability of the game repeating. This happens if A gets T (prob $1/2$) AND B gets TT (prob $(1/2)*(1/2) = 1/4$).

Probability of the game repeating after one round (A tosses T, B tosses TT) = $P(T) \times P(TT) = (1/2) \times (1/4) = 1/8$.

Let $P_A$ be the probability A wins from the start.

$P_A = P(\text{A wins on first toss}) + P(\text{Game repeats}) \times P_A$

$P_A = \frac{1}{2} + \left( \frac{1}{8} \right) \times P_A$

Solving for $P_A$: $P_A - \frac{1}{8} P_A = \frac{1}{2}$ $\frac{7}{8} P_A = \frac{1}{2}$ $P_A = \frac{1}{2} \times \frac{8}{7} = \frac{4}{7}$

Determining $P_B$ (Player B Wins)

B wins if A gets T, and then B gets H in their two tosses.

Probability B wins on their turn = P(A gets T) $\times$ P(B gets H in 2 tosses)

P(B gets H in 2 tosses) = P(H on 1st) + P(T on 1st, H on 2nd) = $1/2 + (1/2)*(1/2) = 3/4$.

Probability B wins on their first available turn = $(1/2) \times (3/4) = 3/8$.

Using the same logic as for $P_A$: $P_B = P(\text{B wins on first turn}) + P(\text{Game repeats}) \times P_B$

$P_B = \frac{3}{8} + \frac{1}{8} P_B$

Solving for $P_B$: $P_B - \frac{1}{8} P_B = \frac{3}{8}$ $\frac{7}{8} P_B = \frac{3}{8}$ $P_B = \frac{3}{8} \times \frac{8}{7} = \frac{3}{7}$

Alternatively, since $P_A + P_B = 1$, $P_B = 1 - P_A = 1 - 4/7 = 3/7$.

Conclusion on Probabilities

We found $P_A = 4/7$ and $P_B = 3/7$.

• Comparing probabilities: $4/7 > 3/7$, so $P(A \text{ Wins}) > P(B \text{ Wins})$.
• Checking the sum: $P_A + P_B = 4/7 + 3/7 = 1$, so $P(A \text{ Wins}) = 1 - P(B \text{ Wins})$.

Based on these findings, the correct statements are $P(A \text{ Wins}) > P(B \text{ Wins})$ and $P(A \text{ Wins}) = 1 - P(B \text{ Wins})$.