Diane has decided to play the following game of chance. She places a $1 bet on each repeated play of the game in which the probability of her winning $1 is 0.5. She has further decided to continue playing the game until she has either accumulated a total of $4 or has lost all her money.
What is the probability that Diane will eventually leave the game a winner if she started with a capital of $3?

A) 0.47
B) 0.77
C) 0.96
D) 0.75

Question

Diane has decided to play the following game of chance. She places a $1 bet on each repeated play of the game in which the probability of her winning $1 is 0.5. She has further decided to continue playing the game until she has either accumulated a total of $4 or has lost all her money. ​
What is the probability that Diane will eventually leave the game a winner if she started with a capital of $3?
​
A) 0.47
B) 0.77
C) 0.96
D) 0.75

Quizplus · Accepted Answer

The probability of Diane winning when she starts with $3 and aims to reach $4 (with a win probability of 0.5 per game) can be calculated using the concept of absorbing Markov chains or simpler, by recognizing this as a problem of biased random walks with absorbing barriers. Since the game is fair (winning and losing probabilities are equal), the probability of reaching her goal before losing all her money is directly proportional to her starting position relative to her goal and the losing point. Starting with $3 and aiming for $4 means she needs to win once before losing three times. The probability of reaching $4 before going bankrupt is equivalent to the probability of winning one game out of one, which is 0.5 raised to the power of the number of games she needs to win (1), but more intuitively, it's easier to understand as her being 1 step away from winning and 3 steps away from losing, making her chances 1 out of 1+3 steps, which simplifies to 0.75 or 75%.

Diane Has Decided to Play the Following Game of Chance