<Question>
Two candidates, Dan and Donald, run for class president. Two other candidates, Freddie and Bernie, run for vice president. Each candidate has a 50$\%$ chance of winning. What is the probability that both Dan and Freddie win? (Express your answer as a fraction.)
</Question>

<think>
<Plan_1>
Okay, let's try to figure out this probability problem. So, we have two candidates, Dan and Donald, running for class president. Then there are two other candidates, Freddie and Bernie, running for vice president. Each candidate has a 50% chance of winning their respective races. The question is asking for the probability that both Dan and Freddie win. Hmm, I need to express this as a fraction.

First, let me make sure I understand the problem correctly. There are two separate elections here: one for president and one for vice president. Each candidate in each election has an equal chance, 50%, of winning. So, for the president race, Dan and Donald are the candidates, and each has a 50% chance. Similarly, for the vice president race, Freddie and Bernie are the candidates, each with a 50% chance. The events seem to be independent because the outcome of one election doesn't affect the other. That's important because if the events are independent, the probability of both happening is the product of their individual probabilities.

So, the president race: Dan has a 50% chance, which is 1/2. Freddie is a vice president candidate, so his race is separate. Bernie is also running for vice president, but we don't care about his outcome here.
We only care about Freddie winning the vice president race. So, Freddie's probability is also 1/2.

Since the two races are independent, the probability that both Dan wins president and Freddie wins vice president should be the product of their individual probabilities.
That would be 1/2 multiplied by 1/2, which is 1/4.

Wait, but let me double-check to make sure I haven't missed anything. Are there any dependencies between the two races? The problem states that each candidate has a 50% chance of winning.
It doesn't mention any dependencies or influences between the races. So, it's safe to assume they are independent events.

Another way to think about it: the outcome of the president election doesn't affect the vice president election. So, all possible outcomes are combinations of who wins president and who wins vice president.
There are 2 candidates for president and 2 for vice president, so in total, there are 2 x 2 = 4 possible outcomes. Let's list them:

1. Dan president, Freddie vice president
2. Dan president, Bernie vice president
3. Donald president, Freddie vice president
4. Donald president, Bernie vice president

Each of these outcomes is equally likely since each candidate has a 50% chance in their respective races. So, each combination has a probability of 1/2 * 1/2 = 1/4.

Now, the event we're interested in is both Dan and Freddie winning. Looking at the list, that's the first outcome. So, only one out of the four possible outcomes.
Hence, the probability is 1/4.

Alternatively, using probability rules: the probability of two independent events both occurring is the product of their probabilities. So, P(Dan wins) * P(Freddie wins) = (1/2) * (1/2) = 1/4.

Is there any other angle I should consider? For example, does the problem state that there are only two candidates in each race? Yes, it does.
Dan and Donald for president, Freddie and Bernie for vice president. So, no other candidates to consider. Therefore, each has exactly two choices, each with 50% chance.

Another check: if all candidates have equal probability, then for president, the probability Dan wins is 1/2, and similarly for Freddie, it's 1/2.
Since the events are independent, multiplying them gives the joint probability. So, yes, 1/2 * 1/2 = 1/4.

I think that's thorough.
The key points are recognizing the independence of the two races and that each candidate has an equal chance.
</Plan_1>

<Execution_1>
No prompt to execute.
</Execution_1>

<Final_answer>
Therefore, the probability that both Dan and Freddie win their respective races is $\boxed{\frac{1}{4}}$.
</Final_answer>
</think>