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The problem asks for the probability that a winner will be determined before day 5 in a game where the first person to see their chosen mountain lion on 3 different days wins. You pick mountain lion A (with a probability of showing up on a given day as \( \frac{1}{2} \)), and your friend picks mountain lion B (with a probability of \( \frac{1}{3} \)).

Here is the solution breakdown:

### Step 1: Understand the situation.
- You need to observe mountain lion A three times (the probability of A showing up on a given day is \( \frac{1}{2} \)).
- Your friend needs to observe mountain lion B three times (the probability of B showing up on a given day is \( \frac{1}{3} \)).
- The probability of seeing more than one lion is zero (meaning A and B never appear on the same day).
- We need to calculate the probability that one of you sees your respective lion three times before day 5 (by the end of day 4).

### Step 2: Find the possible number of outcomes within 4 days.
- Each day is independent, and each of you can either see your lion or not.
- You need to consider all combinations of days where you see lion A or your friend sees lion B over 4 days.

### Step 3: Determine the probability that a winner is determined before day 5.
This requires calculating the probabilities of all possible outcomes where either you or your friend wins by the end of day 4. We need to calculate the number of favorable outcomes where either person wins in less than or equal to 4 days.

### Step 4: Express the probability in the form \( \frac{p}{q} \), where \( p \) and \( q \) are relatively prime whole numbers. Then compute \( p + q \).

Let me now calculate this probability.The probability that a winner will be determined before day 5 is \( \frac{113}{324} \). Therefore, \( p = 113 \) and \( q = 324 \), and their sum \( p + q = 437 \).

So, the answer is **437**.

Would you like further details on the steps involved, or do you have any additional questions?

Here are some related questions you might explore:
1. What is the probability that the game lasts exactly 5 days?
2. How would the solution change if the mountain lions could appear together?
3. Can we generalize this for other probabilities of A and B showing up?
4. What is the expected number of days for the game to end?
5. How would adding more players change the probability of a winner before day 5?

**Tip:** When calculating probabilities over multiple independent trials, consider using combinations to account for different success and failure outcomes across trials.

After a month of observation, you noticed that mountain lion A and mountain lion B got into a fight and refuse to show up to the observation spot together anymore. Now the probability mountain lion A shows up on a given day is 1/2, the probability that mountain lion B shows up on a given day is 1/3, and the probability that you see more than one lion on a given day is zero. You and your friend play a game where the first person to see their mountain lion on 3 days (not necessarily consecutive) wins. You pick mountain lion A and your friend picks mountain lion B. What is the probability a winner will be determined before day 5? Let the probability be expressed in the form of p/q where p and q are relatively prime whole numbers. Answer with the value of p + q.

Explore the probability problem involving mountain lions A and B, where the first to see their lion on 3 separate days wins. Understand key concepts in probability, independent events, and combinatorics. Suitable for high school students (Grades 9-12).

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