Roll Me A Square

Relevant wiki: Expected Value - Problem Solving

Let $E_n =$ The expected number of remaining rolls given that $n$ was the last number rolled.

Since $1$ and $4$ are both square, clearly $E_1 = E_4 = 0$ .

Also, the only two digit squares (consisting of the numbers $1$ ... $6$ ) are $25$ , $36$ , and $64$ . But, $64$ doesn't "buy you anything new" since $4$ is also a square, so rolling a $6$ is essentially like starting over. So, $E_6 = E_0$ . Translating these into a set of linear equations:

• $E_0 = 1 + (1/3)*E_0 + (1/6)*E_2 + (1/6)*E_3$
• $E_2 = 1 + (1/6)*E_0 + (1/6)*E_2 + (1/6)*E_3$
• $E_3 = 1 + (1/6)*E_0 + (1/6)*E_2 + (1/6)*E_3$

The above equations are derived by considering states $0$ , $2$ , and $3$ labeled by what the last number you rolled is (except for $0$ which is the state you start in). So, $E_n =$ The expected number of rolls from state $n$ . Then you can derive the above equations, for example for $E_2$ , from state $2$ you use up $1$ roll (the first term, $1$ ) and you have $1/6$ probability of getting to $0$ , $2$ or $3$ . This translates to the second equation in the solution. You set up similar equations for $0$ , $4$ and $6$ .

Solving for $E_0$ , you get, $E_0 = 18/7$

$18+7 = \boxed{25}$