Question
A game consists of spinning the spinner twice and adding the results. Player A wins if the sum is 4. Otherwise Player B wins. Decide whether the game is fair
Answers
GPT-4o mini
To determine if the game is fair, we need to calculate the probabilities of Player A winning (that is, obtaining a sum of 4) and Player B winning (which happens if the sum is not 4).
Let's assume a standard spinner with 6 equal sections numbered 1 through 6. When the spinner is spun twice, there are a total of \(6 \times 6 = 36\) possible outcomes, corresponding to each of the 6 results from the first spin combined with each of the 6 results from the second spin.
Next, we need to identify the outcomes where the sum of the results equals 4. The combinations that yield a sum of 4 are:
1. \(1 + 3 = 4\)
2. \(2 + 2 = 4\)
3. \(3 + 1 = 4\)
This can be represented as the following pairs:
- (1, 3)
- (2, 2)
- (3, 1)
Thus, there are 3 outcomes that result in a sum of 4.
Now, we can calculate the probability of Player A winning (sum is 4):
\[
P(\text{A wins}) = \frac{\text{Number of outcomes where sum is 4}}{\text{Total outcomes}} = \frac{3}{36} = \frac{1}{12}
\]
Since Player B wins if the sum is not 4, we can calculate the probability of Player B winning as:
\[
P(\text{B wins}) = 1 - P(\text{A wins}) = 1 - \frac{1}{12} = \frac{11}{12}
\]
Now we can compare the probabilities:
- \(P(\text{A wins}) = \frac{1}{12}\)
- \(P(\text{B wins}) = \frac{11}{12}\)
Since Player B has a significantly higher probability of winning than Player A, the game is not fair. Player A has a much lower chance of winning, making the game biased in favor of Player B.
Let's assume a standard spinner with 6 equal sections numbered 1 through 6. When the spinner is spun twice, there are a total of \(6 \times 6 = 36\) possible outcomes, corresponding to each of the 6 results from the first spin combined with each of the 6 results from the second spin.
Next, we need to identify the outcomes where the sum of the results equals 4. The combinations that yield a sum of 4 are:
1. \(1 + 3 = 4\)
2. \(2 + 2 = 4\)
3. \(3 + 1 = 4\)
This can be represented as the following pairs:
- (1, 3)
- (2, 2)
- (3, 1)
Thus, there are 3 outcomes that result in a sum of 4.
Now, we can calculate the probability of Player A winning (sum is 4):
\[
P(\text{A wins}) = \frac{\text{Number of outcomes where sum is 4}}{\text{Total outcomes}} = \frac{3}{36} = \frac{1}{12}
\]
Since Player B wins if the sum is not 4, we can calculate the probability of Player B winning as:
\[
P(\text{B wins}) = 1 - P(\text{A wins}) = 1 - \frac{1}{12} = \frac{11}{12}
\]
Now we can compare the probabilities:
- \(P(\text{A wins}) = \frac{1}{12}\)
- \(P(\text{B wins}) = \frac{11}{12}\)
Since Player B has a significantly higher probability of winning than Player A, the game is not fair. Player A has a much lower chance of winning, making the game biased in favor of Player B.