# A non standard way to a probability problem

We have teams $$A$$ and $$B$$.Every team has equal chance to win. A team wins after they have won $$4$$ matches. Team $$A$$ won the first match, what is the probability, that team $$B$$ won the whole game.

My try:

So team $$A$$ needs $$3$$ more wins to win the game. Team $$B$$ needs $$4$$ more wins to win the game. So the winner will be determined in maximum of $$7$$ games. So now I tried to do it my own way (I dont know if it is usual to do it this way)

Let's say that we are looking at the wins of team $$A$$. (They need $$3$$ more wins).

$$...A_i,...A_j,...A_n$$, where $$3$$ dots symbolize the wins of team $$B$$. So $$(i-1) + (j-i-1) + (n-j-i-1) \leq 3$$ (Because there must not be more than 3 losses, because at 4 team $$B$$ wins the game ) we get: $$n-i \leq 6 \quad \text{and} \quad 0

If we look what integer tuples can satisfy the upper inequalities we get: $$N = \{(1,6),(2,6), (3,6),(4,6),(5,6),\\(1,5),(2,5),(3,5),(4,5),(1,4),(2,4),(3,4),(1,3),(2,3),(1,2)\}$$

Where $$N$$...possible winnings of team $$A$$

So we can count the number of elements ($$15$$) an that would be our number in the numerator, where in the denominator is the number of elements in our sample space.

To calculate the sample space I would do: $$|\omega| = 1 + (\binom{4}{3} - 1)\\ + (\binom{5}{3} - \binom{4}{3} - 1)\\ + (\binom{6}{3} - (1 + (\binom{4}{3} - 1) + (\binom{5}{3} - \binom{4}{3} - 1)))\\ + (\binom{7}{3} -(1 + (\binom{4}{3} - 1)+ (\binom{5}{3} - \binom{4}{3} - 1) + (\binom{6}{3} - (1 + (\binom{4}{3} - 1)+ (\binom{5}{3} - \binom{4}{3} - 1)))) )$$

I guess we could also write this with a recursive formula. However we get $$|\omega|=35$$.

(What I did here is that $$1$$ is for $$3$$ consecutive wins, $$\binom{4}{3} - 1$$ are other possible matches in 4 games and so on.)

So if my procedure is correct the probability is $$P(\text{team B wins the whole game}) = 1 - \frac{|N|}{|\omega|} = 1 - \frac{15}{35} = \frac{20}{35}$$

However is my process atleas partially correct (I do not know the solution).

• That answer does not make sense. Clearly, $A$ has the advantage here, so the probability that $B$ wins must be $<\frac 12$.
– lulu
Commented Nov 8, 2021 at 12:47
• An easy way to solve the problem: Assume all $7$ games are played (even if a winner is decided ahead of game $7$). $B$ wins the series if and only if $B$ wins at least $4$ out of thee remaining $6$ games.
– lulu
Commented Nov 8, 2021 at 12:49

Your approach looks right, but I think you made a couple mistakes when calculating $$N$$. First, you erred with your initial inequality. It should be $$(i-1)+(j-i-1)+(n-j-1)\leq 3$$ $$\implies n\leq 6$$ This result should make sense because if you have $$7$$ games then clearly $$B$$ would have $$4$$ wins since $$A$$'s $$3$$rd win is always the last game.

The second mistake you made is when calculating $$|N|$$. You must find all triples $$(i,j,n)$$ that satisfy the inequality $$n\leq 6$$ (not just tuples $$(i,n)$$). This can be done with some combinatorics (or just brute force). However, one must consider the fact that two runs of games of different length do not have the same probability of occuring. For example, a game of length $$6$$ would have a probability of $$2^{-6}$$ of occuring while a game of length $$5$$ would have a probability of $$2^{-5}$$ of occuring.

So we will not actually solve for $$|N|$$, but rather a weighted sum of number of triples.

We will first fix $$n$$ as some integer that is at least $$3$$ (this is because we need at least $$3$$ games to get $$3$$ wins) and at most $$6$$. Then there are $$n-1$$ slots left before the $$A_n$$th game. There are $$\binom{n-1}{2}$$ ways to choose which of these slots get the other $$2$$ $$A$$ wins. So our probability becomes $$\sum_{n=3}^6 \binom{n-1}{2}2^{-n}$$

To actually solve this sum, we can use a related identity from putnam 2020 A2. Our sum is equivalent to $$=\sum_{n=0}^3 \binom{n+2}{2}2^{-n-3}$$ $$=\sum_{n=0}^3 \binom{2+n}{n}2^{-n-3}$$ $$=2^{-5}\sum_{n=0}^3 \binom{2+n}{n}2^{2-n}$$ $$=2^{-6}\binom{5}{3}+2^{-5}\sum_{n=0}^2 \binom{2+n}{n}2^{2-n}$$ $$=2^{-6}\binom{5}{3}+2^{-5}4^2$$ $$=\frac{10}{64}+\frac{1}{2}$$ $$=\frac{21}{32}$$ Hence, the probability of $$B$$ winning is $$1-\frac{21}{32}=\boxed{\frac{11}{32}}$$

On a side note, while we never used the value of $$|\omega|$$, I would like to point out a more elegant approach to find its value since you did compute it. Consider all permutations of $$3$$ $$A$$'s and $$4$$ $$B$$'s. For each of these cut off from the $$3$$rd $$A$$ or $$4$$th $$B$$ depending on which comes first. These will each give new unique sequences of wins, so the total is just $$\binom{7}{4}=35$$.

To more simply calculate the odds of this problem, we can relate it to flipping a coin. A heads gives team $$A$$ the victory, while a tails gives team $$B$$ the victory. To solve the problem, we are looking for the probability that at least $$4$$ of $$6$$ flips are tails. This is not difficult to calculate.

Total possible outcomes: $$2^6=64$$

6 tails: $$_6C_6=1$$ way; $$\dfrac1{64}$$

5 tails: $$_6C_5=6$$ ways; $$\dfrac{6}{64}$$

4 tails: $$_6C_4=15$$ ways; $$\dfrac{15}{64}$$

Total: $$\dfrac{1+6+15}{64}=\dfrac{22}{64}=\dfrac{11}{32}$$. This gives a $$34.375\%$$ chance that team B manages to win the tournament.