Gambling with 4 dice, what is the probability of getting an even sum?

1. Translate the problem requirements: We need to find the probability that rolling 4 dice results in a sum that is even (divisible by 2). Each die shows 1-6, and we want \(\mathrm{P(sum\,is\,even)}\).
2. Apply even/odd sum rules: Determine when the sum of 4 numbers is even by analyzing the parity (even/odd nature) of the individual dice outcomes.
3. Count favorable outcomes using symmetry: Use the fact that each die has equal numbers of even and odd outcomes to establish the probability pattern.
4. Calculate final probability: Apply the counting principle to determine the ratio of favorable outcomes to total outcomes.

Execution of Strategic Approach

1. Translate the problem requirements

Let's break down what we're looking for in everyday terms. We have 4 dice, each showing a number from 1 to 6. We roll all 4 dice and add up the numbers that appear. The question is: what's the probability that this total sum is an even number?

For example, if we roll 2, 3, 4, 1, our sum would be \(2 + 3 + 4 + 1 = 10\), which is even. If we roll 1, 1, 1, 1, our sum would be 4, which is also even. But if we roll 1, 2, 3, 5, our sum would be 11, which is odd.

Process Skill: TRANSLATE - Converting the gambling scenario into a clear mathematical question about the parity (even/odd nature) of sums.

2. Apply even/odd sum rules

Here's the key insight we need to understand: when is the sum of several numbers even versus odd?

Let's think about this step by step:
- If I add two even numbers, I get an even result (like \(2 + 4 = 6\))
- If I add two odd numbers, I also get an even result (like \(1 + 3 = 4\))
- If I add one even and one odd number, I get an odd result (like \(2 + 3 = 5\))

The pattern is this: A sum is even when we have an even number of odd addends. So for our 4 dice, the sum will be even when we have 0, 2, or 4 odd numbers showing.

On each die, the odd numbers are \(\{1, 3, 5\}\) and the even numbers are \(\{2, 4, 6\}\). So each die has a 50-50 chance of showing odd versus even.

3. Count favorable outcomes using symmetry

Now let's count the favorable cases where we get an even sum. Remember, this happens when we have an even number of odd dice results.

Think of each die as a coin flip: "odd" or "even" with equal probability of \(\frac{1}{2}\) each.

For 4 dice, we want:
- 0 odd dice (all 4 even): This is like getting 4 "heads" in 4 coin flips
- 2 odd dice (exactly 2 even): This is like getting exactly 2 "heads" in 4 coin flips
- 4 odd dice (all 4 odd): This is like getting 0 "heads" in 4 coin flips

Here's the beautiful symmetry: in any sequence of coin flips, exactly half of all possible outcomes have an even number of heads, and half have an odd number of heads. This is a fundamental property of binary sequences.

4. Calculate final probability

Using the symmetry principle we discovered:

Since each die independently has a \(\frac{1}{2}\) probability of being odd and \(\frac{1}{2}\) probability of being even, and we're looking at 4 independent dice, we can think of this as 4 independent coin flips where "heads" = odd and "tails" = even.

In any sequence of 4 coin flips, exactly half of all 16 possible outcomes will have an even number of heads (0, 2, or 4 heads), and half will have an odd number of heads (1 or 3 heads).

Therefore, the probability of getting an even sum = \(\frac{1}{2}\).

We can verify this makes sense: there's a perfect symmetry between getting an even sum and getting an odd sum, so each should have probability \(\frac{1}{2}\).

Final Answer

The probability of getting an even sum when rolling 4 dice is \(\frac{1}{2}\).

This matches answer choice B. \(\frac{1}{2}\).

Common Faltering Points

Errors while devising the approach

1. Misunderstanding the even/odd sum rule: Students often get confused about when a sum is even versus odd. They might incorrectly think that adding more numbers always makes it more likely to get an even sum, or they might not realize that the key insight is counting how many odd numbers appear (not the actual values on the dice).

2. Attempting to enumerate all possible outcomes: Many students will try to list all \(6^4 = 1,296\) possible dice combinations instead of recognizing the elegant even/odd pattern. This approach is extremely time-consuming and error-prone, leading them to either give up or make counting mistakes.

3. Confusing individual die probabilities with sum probabilities: Students might incorrectly think that since each die has equal numbers of even and odd faces (3 each), the probability should be something other than \(\frac{1}{2}\), not recognizing that what matters is the combination pattern across all 4 dice.

Errors while executing the approach

1. Incorrectly applying the even/odd counting rule: Even if students understand that they need an even number of odd dice, they might miscount the scenarios. For example, they might forget that 0 odd dice (all even) gives an even sum, or they might incorrectly include cases with 1 or 3 odd dice as favorable outcomes.

2. Making errors in the symmetry argument: Students might struggle with the coin-flip analogy and incorrectly calculate probabilities for getting exactly 0, 2, or 4 'heads' in 4 flips. They might add these probabilities incorrectly or use wrong binomial coefficients.

Errors while selecting the answer

1. Second-guessing the elegant \(\frac{1}{2}\) result: The answer \(\frac{1}{2}\) might seem 'too simple' to students, causing them to doubt their work and select a more 'complex-looking' fraction like \(\frac{2}{3}\) or \(\frac{3}{4}\), thinking that a probability problem involving 4 dice should have a more complicated answer.