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 $P(A \text{ Wins}) = 1 - P(B \text{ Wins})$.
Important Questions from Discrete Probability
- Let $X$ be a Binomial$(n, p)$ random variable, where $n \in \{5,6\}$ and $p\in \{\frac{1}{4}, \frac{3}{4}\}$. If $X = 3$ is observed, then the maximum likelihood estimate of $(n, p)$ is
- Suppose two fair dice are thrown independently at random. Let $X$ and $Y$ be the numbers on the upper face of the first die and that of the second die, respectively. Then which of the following statements are true?
- A box contains 40 numbered red balls and 60 numbered black balls. From the box, balls are drawn one by one at random without replacement till all the balls are drawn. The probability that the last ball drawn is black equals
Consider the problem of testing $H_0 : \theta = 1$ vs $H_1 : \theta = \frac{1}{2}$ where $\theta$ is the mean of a Poisson random variable. Let $X$ and $Y$ be a random sample from Poisson ($\theta$) distribution. Consider the following test procedure:
Reject $H_0$ if either $X = 0$ or $(X = 1 \text{ and } X + Y \leq 2)$; otherwise accept $H_0$.
Which of the following are true?
In a football league, the goals scored by home teams over 380 matches have the following frequency distribution.
Number of goals 0 1 2 3 4 5 Frequency 92 121 91 50 19 7 The average goals scored by home teams is 1.49. We want to test $H_0$: Goal distribution is Poisson. Based on observations the value of the $\chi^2$-statistic for goodness of fit is 1.27. Given $\chi^2_{0.05, 6} = 1.64, \chi^2_{0.05, 5} = 1.15, \chi^2_{0.95, 6} = 12.59$ and $\chi^2_{0.95, 5} = 11.07$, which of the following are true?
