Know the probability of $A$ beating $B$, also $B$ beating $C$, what's the probability of $A$ beating $C$ three games in a row?

Here's a question from my probability textbook:

When a chess game is played, the better player has a ${3\over4}$ probability of winning and the worse player has a ${1\over4}$ probability of winning. (There are no draws.) $A$ plays three games with $B$, and wins two. It's not too hard to see that the odds are $3$ to $1$ in favor of $A$ being the better player. If also $B$ beat $C$ two games out of three, prove that the chance of $A$ winning the first three games he plays with $C$ is ${{103}\over{352}}$.

Here's what I did. We have the following:

• ${3\over4}$ chance $A$ is better than $B$.
• ${1\over4}$ chance $B$ is better than $A$.
• ${3\over4}$ chance $B$ is better than $C$.
• ${1\over4}$ chance $C$ is better than $B$.

Hence:

• $({3\over4})^2 = {9\over{16}}$ chance of getting the ordering $ABC$.
• $({1\over4})^2 = {1\over{16}}$ chance of getting the ordering $CBA$.

The probabilities of the remaining orderings $ACB$, $BAC$, $BCA$, $CAB$ we can't calculate directly since we don't have any data on games between $A$ and $C$. What do I mean by this? Consider the orderings $ACB$ and $CAB$. There's a ${3\over4}$ chance of $A$ being better than $B$, and a ${1\over4}$ chance of $C$ being better than $B$. Here is where I wave my hands and say that the chances of getting $ACB$ and $CAB$ are equal by symmetry (in the absence of any other information), so:

• $({1\over2})({1\over4})({3\over4}) = {3\over32}$ chance of getting the ordering $ACB$.
• $({1\over2})({1\over4})({3\over4}) = {3\over32}$ chance of getting the ordering $CAB$.

Analogously:

• $({1\over2})({1\over4})({3\over4}) = {3\over32}$ chance of getting the ordering $BAC$.
• $({1\over2})({1\over4})({3\over4}) = {3\over32}$ chance of getting the ordering $BCA$.

As a sanity check, the probabilities of the $6$ orderings adds up to $1$:$${9\over{16}} + {1\over{16}} + 4\left({3\over{32}}\right) = 1.$$So the respective probabilities of $A$ being better than $C$, $C$ being better than $A$ are$${9\over{16}} + {3\over{16}} = {3\over4}, \quad {1\over{16}} + {3\over{16}} = {1\over4}.$$So the probability of $A$ winning the first three games he plays with $C$ is$$\left({3\over4}\right)\left({3\over4}\right)^3 + \left({1\over4}\right)\left({1\over4}\right)^3 = {{41}\over{128}}.$$However, this is the answer in the back of my book:

We have the double event that $A$ has beaten $B$ twice out of $3$ times, and $B$ has beaten $3$ twice out of $3$ times. A priori the order of merit of the three players may equally be$$ABC, \quad ACB, \quad BCA, \quad BAC, \quad CAB, \quad CBA,$$the consequent chances of the observed double event are as$$9: 3 : 3: 3: 3: 1.$$ Hence the chance that $A$ is a better player than $C$ is$$(9 + 3 + 3) ÷ 22 = {{15}\over{22}}.$$Thus the chance of $A$ winning the three games is$$\left({{15}\over{22}}\right)\left({3\over4}\right)^3 + {7\over{22}}\left({1\over4}\right)^3 = {{103}\over{352}}.$$

So where did I go wrong? Or am I right and it's the case the answer in the back of my book is wrong? Admittedly my symmetry argument is handwaving (if not outright wrong) since we don't have any more information. I also don't understand why:

A priori the order of merit of the three players may equally be$$ABC, \quad ACB, \quad BCA, \quad BAC, \quad CAB, \quad CBA,$$

Any help would be well-appreciated.

So where did I go wrong?

Your error was at the beginning, where you misinterpreted the better player having a ${3\over4}$ chance of winning a game (i.e., $A$ winning $75\%$ of infinitely many games if $A$ is better than $B$) as meaning that there is necessarily a $75\%$ chance of $A$ being better than $B.$

I also don't understand why:

A priori the order of merit of the three players may equally be$$ABC, \quad ACB, \quad BCA, \quad BAC, \quad CAB, \quad CBA,$$

Without any (prior) information about the game performance of players $A,B,C,$ each of the six $(3!)$ possible rankings are equally likely.

the consequent chances of the observed double event are as $$9: 3 : 3: 3: 3: 1.$$

This line is wrongly/confusingly phrased. The author means that, with that new information about the relative performances of players $A,B,C$ (i.e., “the observed double event”), we can now update the relative odds of those six possible rankings.

Following Ivan's notation in the comments, let $Q$ be the aforementioned given event. Then $$P(Q|ABC)=\left({3\choose2}\left({3\over4}\right)^2{1\over4}\right)^2={729\over4096}\\ P(Q|CBA)=\left({3\choose2}\left({1\over4}\right)^2{3\over4}\right)^2={81\over4096}\\ P(Q|ACB),P(Q|BCA),P(Q|BAC),P(Q|CAB)\\=\left({3\choose2}\left({3\over4}\right)^2{1\over4}\right)\left({3\choose2}\left({1\over4}\right)^2{3\over4}\right)={243\over4096}.$$ Using Bayes' Theorem and the Law of Total Probability and noting that $P(xyz)={1\over6}$ for each of the rankings $xyz,$ $$P(xyz|Q)\\=P(Q|xyz)P(xyz)\Bigg/\bigg(P(Q|ABC)P(ABC)+P(Q|CBA)P(CBA)+P(Q|ACB)P(ACB)+P(Q|BCA)P(BCA)+P(Q|BAC)P(BAC)+P(Q|CAB)P(CAB)\bigg)\\={{P(Q|xyz)}\over{P(Q|ABC)+P(Q|CBA)+P(Q|ACB)+P(Q|BCA)+P(Q|BAC)+P(Q|CAB)}}.$$ Thus, $$P(ABC|Q):P(ACB|Q):P(BCA|Q):P(BAC|Q):P(CAB|Q):P(CBA|Q)\\=729:243:243:243:243:81\\=9:3:3:3:3:1.$$

Hence the chance that $A$ is a better player than $C$ is$$(9 + 3 + 3) ÷ 22 = {{15}\over{22}}.$$Thus the chance of $A$ winning the three games is$$\left({{15}\over{22}}\right)\left({3\over4}\right)^3 + {7\over{22}}\left({1\over4}\right)^3 = {{103}\over{352}}.$$

This final part is straightforward.