Note: I know there are much easier ways to solve the below problem. I'm asking this question not to get the right answer, but to understand why my approach is wrong.
There are other questions similar to this, but none asked exactly this way.
I recently posted a question here that basically sums up to "how do I know whether to use ordered tuples rather than unordered tuples to calculate probabilities?". What I gathered from the answer to that question was that the probabilities of ordered tuples are always equal, so if a question mentions that the probabilities of some events occurring are "equal" or that events occur "randomly", go with an approach that uses ordered tuples.
Here's the question now:
Two teams play in a seven-game series. The first team to win four games is crowned champion. The teams each have a $\frac{1}{2}$ probability of winning each game. What is the probability that the series will be over after the sixth game?
I've successfully modeled many similar problems by using ordered tuples. My approach is below:
The most a team can play is 4 games. Let's imagine a bucket of 8 balls, each ball corresponding to a win for each team. Let's assign arbitrary numbers to each of these eight balls to make them distinct. There are $8!$ ordered tuples representing all permutations of wins in the series. This is our sample space. Now we have to calculate the probability that 6 games are played. Choose one of the balls from the winning team and place it in the sixth position. Choose two winning balls from the losing team. There should now be 5 balls on the left of the sixth ball and 2 on the right, and all of those balls should be ordered so that we get the right probabilities. So we have the following expression for the probability:
$$\frac{{4\choose 1} \times {4 \choose 2} \times 5! \times 2!}{8!} = \frac{2}{7}$$
When I use the same logic for all the options in the sample space (4, 5, 6, 7), I get 4 probabilities that add to 1. The solution is wrong, though. I can't see the holes in my solution. Where is the flaw?
Best Answer
As stated in the chat, here is my write-up for how you could use ordered strings. Each string has an unordered set of $A$s or $B$s that comprise it.
You have a maximum of seven games. You are looking for the first to four wins. You can assume they play to seven games regardless, but we consider it a victory for one team as soon as they reach four games.
Let's call the teams $A$ and $B$. If team $A$ wins, we mark down $A$. If team $B$ wins, we mark down $B$. So, any outcome can be represented by a string of $A$s and $B$s. But, we cannot choose how many each team gets. We can only mark down the results. This yields strings like:
$$AAAAAAA \\ AAAAAAB \\ AAAAABA \\ AAAAABB \\ \vdots \\ BBBBBAA \\ BBBBBAB \\ BBBBBBA \\ BBBBBBB$$
Each letter has two choices, independent of any previous or future letter. These are all ordered strings. There are a total of $2^7 = 128$ ordered strings. Each string represents an equally likely outcome from the best-of-seven match. WLOG, let's calculate the probability that team $A$'s $4$th win occurs on game number $n$. By symmetry, the same will be true for team $B$.
For team $A$ to win on the fourth game, we must have $AAAA\#\#\#$ where $\#$ can be either $A$ or $B$. The number of ways this can occur is $2^3$ because there are $2^3$ ways to choose the final three games. The first four must all be $A$.
Thus, the probability that $A$ will win in exactly $4$ games is:
$$\dfrac{2^3}{2^7} = \dfrac{1}{16}$$
Now, for team $A$ to win on exactly the $5$th game, the must have their $4$th win on the $5$th game, meaning they have three wins among the first four games, and team $B$ must have one win. So, the last two games could be won by anyone. This means, team $A$ clinches victory a total of $\dbinom{4}{3}2^2$ ways. First we choose three of the four games for a team $A$ victory and the remaining game must be a $B$ victory, then the last two can be either $A$ or $B$. Thus, the probability of a $5$th game victory for team $A$ is:
$$\dfrac{\dbinom{4}{3}2^2}{2^7} = \dfrac{1}{8}$$
Next, for $6$th game team $A$ victory, we have $A$'s fourth win on game $6$, so we must have three $A$s and two $B$'s among the first five games. The seventh game can be anyone's victory. Thus, there are $\dbinom{5}{3}2^1$ ways for $A$ to clinch victory on game $6$, and that gives a probability of:
$$\dfrac{\dbinom{5}{3}2^1}{2^7} = \dfrac{5}{32}$$
Finally, for a $7$th game team $A$ victory, we have $A$'s fourth win on game $7$. This means we have three team $A$ and three team $B$ victories among the first six games. There are $\dbinom{6}{3}$ ways this can happen. The probability $A$ wins on game $7$ is:
$$\dfrac{\dbinom{6}{3}}{2^7} = \dfrac{5}{32}$$
Adding up probabilities, we have team $A$ winning:
$$\dfrac{1}{16}+\dfrac{1}{8}+\dfrac{5}{32}+\dfrac{5}{32} = \dfrac{1}{2}$$
Just as expected. And the probability of either team achieving victory on game $6$ is twice the probability of team $A$ winning on game $6$:
$$\dfrac{10}{32} = \dfrac{5}{16}$$