# Can You Even the Odds?

## This Week's Fiddler

There are effectively two states at any point: the current player is about to go once, or the current player is about to go twice. Let $P_1$ be the chance the current player wins in the former case, and let $P_2$ be the chance in the latter case.

For $P_1$, the chance is $\frac{1}{6}$ of winning outright. Upon the $\frac{5}{6}$ chance of not winning outright, the current player switches and gets two rolls, so the chance of winning given that it gets to that point is $1-P_2$. In other words:

$P_1=\frac{1}{6}+\frac{5}{6}(1-P_2)$For $P_2$, the chance is $\frac{1}{6}$ of winning outright. Upon the $\frac{5}{6}$ chance of not winning outright, the current player stays the same with one roll left, so the chance of winning is $P_1$. In other words:

$P_2=\frac{1}{6}+\frac{5}{6}P_1$Hence:

$P_1=\frac{1}{6}+\frac{5}{6}(1-P_2) \\ P_1=\frac{1}{6}+\frac{5}{6}(1-(\frac{1}{6}+\frac{5}{6}P_1)) \\ P_1=\frac{31}{36}-\frac{25}{36}P_1 \\ P_1=\frac{31}{36}\cdot\frac{36}{61} \\ P_1=\frac{31}{61}$Which gives us our answer, $\frac{31}{61}$.

An alternative solution is to split the chance of A winning ABBA into the sum of two geometric sequences: every fourth term in the first position, and every fourth term in the fourth position. The first term is $\frac{1}{6}$, and the fourth term is $\frac{1}{6}(\frac{5}{6})^3=\frac{125}{1296}$. The ratio between every fourth term would be $(\frac{5}{6})^4=\frac{625}{1296}$. This means the sum is:

$\frac{\frac{1}{6}+\frac{125}{1296}}{1-\frac{625}{1296}}=\frac{341}{1296}\cdot\frac{1296}{671}=\frac{31}{61}$Which gives us the same answer!

**Answer**: 31/61 (about 50.82%)

## Extra Credit

Interesting problem! The Thue-Morse sequence is so beautiful, and I expect the answer to be so as well. Simulations say it's around 50.16% for A to win; about even with an ever-so-slight edge to A.

**Answer**: ¯\_(ツ)_/¯