A and B take turns to draw a card from a well-shuffled pack of 52 playing cards, A starting first. After each draw the card is put back and the pack is shuffled again. The first one to draw a spade wins the game. The probability that A wins the game is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(\displaystyle\frac{4}{7}\)
There are 13 spades in 52 cards, so on each draw
\( \displaystyle P(\text{spade}) = \frac{13}{52} = \frac{1}{4},\quad P(\text{not a spade}) = \frac{3}{4} \)
The card is put back each time, so the draws are independent. A wins on the 1st, 3rd, 5th, … draw, and before each of these every earlier draw must fail:
\( \displaystyle P(\text{A wins}) = \frac{1}{4}+\left(\frac{3}{4}\right)^{2}\frac{1}{4}+\left(\frac{3}{4}\right)^{4}\frac{1}{4}+\cdots \)
This is an infinite geometric series with first term \(\frac{1}{4}\) and common ratio \(\frac{9}{16}\):
\( \displaystyle P(\text{A wins}) = \frac{\frac{1}{4}}{1-\frac{9}{16}} = \frac{\frac{1}{4}}{\frac{7}{16}} \)
\( \displaystyle = \frac{4}{7} \)
Why the other options are incorrect
Option (1) \(\displaystyle\frac{1}{4}\) — Incorrect. This is only the chance that A wins on his first draw; A can also win on later turns.
Option (2) \(\displaystyle\frac{3}{7}\) — Incorrect. This is the probability that B wins, \(1-\frac{4}{7}\).
Option (3) \(\displaystyle\frac{1}{2}\) — Incorrect. The chances are not equal, because A draws first and can win before B gets a turn.