From 11-year-old (!) Kayla Schubmehl comes a question inspired by the book “Math Games with Bad Drawings,” by Ben Orlin:
Suppose you (player A) and a friend (player B) are playing a game in which you alternate rolling a die. So the order of play is AB|AB|AB, and so on. (The vertical bars here are just for organizational purposes, and do not signify anything special that happens.) The first player to roll a five wins the game. As it turns out, whoever goes first has a distinct advantage!
Kayla wondered about other ways you and your friend could take turns, ways that might result in a fairer game. For example, consider the “snake” method, in which the order is reversed after each time you both roll: AB|BA|AB|BA, and so on.
Assuming you are the first to roll, what is the probability you will win the game?
The probability you have won in $4$ rolls is $\frac{341}{1296}$. The probability $B$ has won in $4$ rolls is $\frac{330}{1296}$.
Since this pattern repeats every $4$ rolls, the probability you win:
\begin{align*} &= \frac{341}{1296} \cdot \left(1 + \left(\frac{5}{6}\right)^4 + \left(\frac{5}{6}\right)^8 + \cdots\right) \\ &= \frac{341}{1296} \cdot \frac{1}{1 - \left(\frac{5}{6}\right)^4} \\ &= \frac{341}{1296} \cdot \frac{1296}{1296-625} \\ &= \frac{341}{671} = \boxed{\frac{31}{61}} \\ \end{align*}Similarly, the probability $B$ wins:
\begin{align*} &= \frac{330}{1296} \cdot \left(1 + \left(\frac{5}{6}\right)^4 + \left(\frac{5}{6}\right)^8 + \cdots\right) \\ &= \frac{330}{671} = \boxed{\frac{30}{61}} \\ \end{align*}Kayla also offered some Extra Credit:
Another way to take turns is to use the Thue-Morse sequence, where the entire history of the order is flipped to its complement (from A to B and from B to A) after each round. As an illustration, consider the first few rounds:
Round 1: Player A goes first.
Round 2: Only A went in the first round. So now player B goes.
Round 3: Up to this point, the order has been AB. Flipping this to its complement, round 3’s order is BA.
Round 4: Up to this point, the order has been ABBA. Flipping this to its complement, round 4’s order is BAAB.
Round 5: Up to this point, the order has been ABBABAAB. Flipping this to its complement, round 5’s order is BAABABBA.
Writing this out as a single sequence of turns, the order is A|B|BA|BAAB|BAABABBA, and so on.
Assuming you are the first to roll, what is the probability you will win the game?
The pattern that repeats is ABBABAABBAABABBA.
The probability you win in the first $16$ rolls is
\begin{align*} &= \frac{1}{6} \cdot \left(1 + \left(\frac{5}{6}\right)^3 + \left(\frac{5}{6}\right)^5 + \left(\frac{5}{6}\right)^6 + \left(\frac{5}{6}\right)^9 + \left(\frac{5}{6}\right)^{10} + \left(\frac{5}{6}\right)^{12} + \left(\frac{5}{6}\right)^{15}\right) \\ &= \frac{6^{15} + 5^36^{12} + 5^56^{10} + 5^96^6 + 5^{10}6^5 + 5^{12}6^3 + 5^{15}}{6^{16}} \\ &= \frac{1339018029701}{6^{16}}\\ \end{align*}Since this pattern repeats every $16$ rolls, the probability you win:
\begin{align*} &= \frac{1339018029701}{6^{16}} \cdot \left(1 + \left(\frac{5}{6}\right)^{16} + \left(\frac{5}{6}\right)^{32} + \cdots\right) \\ &= \frac{1339018029701}{6^{16}} \cdot \frac{1}{1 - \left(\frac{5}{6}\right)^{16}} \\ &= \frac{1339018029701}{6^{16}} \cdot \frac{6^{16}}{6^{16} - 5^{16}} \\ &= \frac{1339018029701}{(6-5)(6+5)(6^2+5^2)(6^4+5^4)(6^8+5^8)} \\ &= \frac{11 \cdot 17 \cdot 61 \cdot 113 \cdot 1,038,811}{(1)(11)(61)(1921)(2,070,241)} \\ &= \boxed{\frac{1,038,811}{2,070,241} \approx 0.50178} \\ \end{align*}My recursive code gave me $0.50159$ which is slightly off.