Probability of Winning or Losing a Carnival Game in Two Attempts

Understanding the probability of winning or losing in a carnival game, especially when you play multiple times, can significantly enhance your chances of success. This article breaks down the problem of playing a carnival game with a 3/7 probability of winning and explores the probabilities of winning both times, losing both times, or getting a mix of wins and losses.

Probability Basics

A carnival game has a winning probability of 3/7, and a losing probability of 4/7. When you play the game twice, the outcome of each game is independent of the other. This means the probability of any event in one game does not affect the probability of the same event in the subsequent games.

Probability of Winning Both Times

When you want to know the probability of winning two times in a row, you can multiply the probability of winning each individual game. This is because the two events are independent.

The probability of winning the game on the first attempt is 3/7. The probability of winning the game on the second attempt is also 3/7. The combined probability of winning both times is the product of these two probabilities:

P(Win both times) P(Win first time) × P(Win second time)

Substituting the values, we get:

P(Win both times) (3/7) × (3/7) 9/49

Probability of Losing Both Times

To find the probability of losing both times, use the same approach as for winning:

The probability of losing on the first attempt is 4/7. The probability of losing on the second attempt is also 4/7. The combined probability of losing both times is:

P(Lose both times) P(Lose first time) × P(Lose second time)

Substituting the values, we get:

P(Lose both times) (4/7) × (4/7) 16/49

Probability of Winning Both or Losing Both

The total probability of winning both times or losing both times is the sum of the individual probabilities. However, it is important to note that these two events are mutually exclusive, meaning they cannot happen at the same time. Therefore, we can simply add the probabilities:

P(Win both or Lose both) P(Win both times) P(Lose both times)

Substituting the values, we get:

P(Win both or Lose both) 9/49 16/49 25/49

Other Possible Outcomes

Let’s consider the probability of getting a mix of wins and losses in the two attempts:

The probability of winning on the first attempt and losing on the second attempt is:

P(Win first, Lose second) P(Win first) × P(Lose second) (3/7) × (4/7) 12/49

The probability of losing on the first attempt and winning on the second attempt is:

P(Lose first, Win second) P(Lose first) × P(Win second) (4/7) × (3/7) 12/49

The combined probability of getting a mix of wins and losses (either win first, lose second or lose first, win second) is:

P(One win and one loss) 12/49 12/49 24/49

Summary of Probabilities

Here’s a summary of the probabilities for all possible outcomes when playing the game twice:

2 wins: P(Win both times) 9/49 1 win and 1 loss: P(One win and one loss) 24/49 2 losses: P(Lose both times) 16/49

As expected, the total of these probabilities adds up to 1:

9/49 24/49 16/49 49/49 1