Discrete Variable Distribution: Binomial vs Negative Binomial Distribution

I am self learning the discrete random variable distributions, and I think I have some problems understanding the difference between these two distributions.

For what I understand at this stage, I know in binomial distribution, we are curious about the probability of doing n trials with r times of success, and in negative binomial distributions, we are curious about the probability of a "SPECIFIC order of success" in the total of n trials, like the second success in the total of n trials that kind of stuff. (I am not sure if my understanding about this is correct, but this is what comes to my mind after reading Wikipedia).

I have a practice question in my textbook that states the following:

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My friend and I are arguing whether this is a binomial distribution or a negative binomial distribution. In my opinion, I think this is a binomial distribution, since we only care about like the best four out of seven, but not the 4th win in the seven.

For example: if I have a die that, when rolled, shows 6 with probability $$p$$, then I can ask about the probability of observing $$k$$ sixes in $$n$$ rolls. That's a binomial model, and $$k$$ is a realization of the underlying random variable. I could also ask about the probability that it takes $$n$$ rolls to observe exactly $$k$$ sixes. That's a negative binomial model, and here, $$n$$ is a realization of the underlying random variable.
Here's how you'd do it using a binomial model. Your choice is between $$n = 5$$ or $$n = 7$$. Let $$X$$ be the random number of wins that $$A$$ achieves. Then in the first case, we need $$X \ge 3$$ (best of 3), and in the second case, we need $$X \ge 4$$ (best of 4). So we want to compare the probabilities of these two events: $$\Pr[X \ge 3 \mid n = 5], \quad \text{vs.} \quad \Pr[X \ge 4 \mid n = 7].$$ In both cases, $$p = 0.4$$. Using the binomial model, $$\Pr[X \ge 3 \mid n = 5] = \binom{5}{3} (0.4)^3 (0.6)^2 + \binom{5}{4} (0.4)^4 (0.6)^1 + \binom{5}{5} (0.4)^5 (0.6)^0,$$ and $$\Pr[X \ge 4 \mid n = 7] = \binom{7}{4} (0.4)^4 (0.6)^3 + \binom{7}{5} (0.4)^5 (0.6)^2 + \binom{7}{6} (0.4)^6 (0.6)^1 + \binom{7}{7} (0.4)^7 (0.6)^0.$$ After computing these, the probability that is larger tells you which choice of $$n$$ is preferable for team $$A$$.
Now, here's how you'd use the negative binomial model. Rather than having two cases based on the total games played, you now have two cases based on the number of wins $$w$$ achieved by A. Let $$Y$$ be the random number of games played. In the first case, what is the probability that it takes $$Y \le 5$$ games for $$A$$ to win exactly $$w = 3$$ games? In the second, what is the probability that it takes $$Y \le 7$$ games for $$A$$ to win exactly $$w = 4$$ games? In other words, now the comparison is $$\Pr[Y \le 5 \mid w = 3] \quad \text{vs.} \quad \Pr[Y \le 7 \mid w = 4].$$ The first probability is $$\Pr[Y \le 5 \mid w = 3] = \binom{2}{2} (0.4)^3 (0.6)^0 + \binom{3}{2} (0.4)^3 (0.6)^1 + \binom{4}{2} (0.4)^3 (0.6)^2.$$ The second is $$\Pr[Y \le 7 \mid w = 4] = \binom{3}{3} (0.4)^4 (0.6)^0 + \binom{4}{3} (0.4)^4 (0.6)^1 + \binom{5}{3} (0.4)^4 (0.6)^2 + \binom{6}{3} (0.4)^4 (0.6)^3.$$ You should get the same results as with the binomial model; i.e., $$\Pr[X \ge 3 \mid n = 5] = \Pr[Y \le 5 \mid w = 3], \\ \Pr[X \ge 4 \mid n = 7] = \Pr[Y \le 7 \mid w = 4].$$ Therefore, both of you are right and both of you are wrong: which model you use depends on your choice of stopping criterion. In the binomial model, the game play stops after a fixed number of games are played. In the negative binomial model, the game play stops after a fixed number of wins are observed. The latter may be more commonly the case in a playoff situation (since typically, once a team wins a decisive number of games, there is no point to continuing to play), but the binomial model is mathematically equivalent.