Multiple choice

A and B toss a coin alternately till one of them tosses heads and wins the game, their respective probability of winning are

  1. $\dfrac 23$ and $\dfrac 13$
  2. $\dfrac 12$ and $\dfrac 12$
  3. $\dfrac 13$ and $\dfrac 23$
  4. $\dfrac 15$ and $\dfrac 45$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let p = 1/2 be the probability of heads. A wins if they get heads on the 1st, 3rd, 5th... toss. P(A) = p + (1-p)^2 * p + (1-p)^4 * p = p / (1 - (1-p)^2) = (1/2) / (1 - 1/4) = (1/2) / (3/4) = 2/3. B wins the rest, so P(B) = 1 - 2/3 = 1/3.

AI explanation

Since A tosses first, A wins on the first toss with probability 1/2, on the third toss with probability (1/2)^3, on the fifth toss with probability (1/2)^5, and so on. Using the formula for the sum of an infinite geometric series, A's total probability of winning is (1/2) / (1 - 1/4) = 2/3. Since only one person can win, B's probability of winning is the remaining probability, which is 1 - 2/3 = 1/3.

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