Mastermind game states

Question

I'm trying to analyse a game of Mastermind and am having trouble quantifying the amount of possible game states. I know that a code has $\text{# of colors}^{\text{# of pegs per guess}}$ combinations (in my case that would be $6^4=1296$). However, an entire board state also consists of 10 guesses. Each guess has the same amount of combinations, thus my intuition would be that the amount of total states in a game of Mastermind would be $\text{# of rows}^{\text{# of combinations per row}}$. This approach yields $11^ {1296}$ board states which is astronomically large and I'm having a hard time believing this is true.

To clarify what I mean by a board state, I mean any legal state the game board can be in using the standard game rules. Having 3 empty rows, then one guess row and another 6 empty rows is not a legal board state.

How do I go about estimating this number?

Answer 1

Your formula going from rows to the full board is incorrect, and should be $\#\text{combinations per row}^{\#\text{rows}}$, giving $1296^{11}$ which is much less. Substituting in the formula for combinations per row, this is just $$(\#\text{colours}^{\#\text{pegs per row}})^{\#\text{rows}}=\#\text{colours}^{\#\text{pegs per row}\times\#\text{rows}}=\#\text{colours}^{\#\text{total pegs}}$$ which you can get directly as the number of ways to choose a colour for each peg.

As Arthur says, this is the number of possibilities for a completely full board. The number of possibilities for a board with only $10$ rows used is likewise $(6^4)^{10}$, then $(6^4)^9$ for $9$ rows, and so on, giving a total of $$\sum_{i=0}^{11}(6^4)^i=\frac{(6^4)^{12}-1}{6^4-1}$$

Answer 2

Other way round : $\text{# of combinations per row}^{\text{# of rows}}$

If you sum over 0 to 10 rows you get a geometric series, giving a total of $$\frac{\text{# of combinations per row}^{1+\text{# of rows}}-1} {\text{# of combinations per row}-1}$$