Ship, Captain, Crew

Ship, Captain, Crew is a simple dice game that I thought might make a good minigame.

Rules

On a player’s turn, they roll five six-sided dice, up to three times. After each throw, the player can “bank” a 6, a 5, or a 4 and only re-roll the remaining dice, but they can only bank them in that order, or together. So, for example, a 5 cannot be banked without a 6 already banked or banked alongside it. You can bank any dice once you have a 4 banked.

After three throws, if a player has a 6, 5, and 4 (“ship,” “captain,” and “crew,” respectively), they score the total of the remaining two dice (the “cargo”). Without a crewed and captained ship, the cargo is worth nothing.

Score of seven.

Modeling

This is once again well-modeled as an absorbing Markov process. We consider the four states: “unscored,”¹ “ship,” “captain,” and “crew,” and construct our transition matrix from there.

The odds of changing from “unscored” to “ship” are the odds of rolling exactly one “success” in 5 attempts,² or $P = ( 1 - ) ^5$. As we consider the rest of the transition matrix, (the number of dice or attempts, conveniently a function of the current state), each element is roughly given by the binomial distribution, with the lower triangle zeros (as you cannot progress “backwards”). The final element in each row, however, is simply the remainder, as the exact number of “successes” doesn’t matter past that point. Ultimately, it looks like this:

${\displaystyle P\approx \left[\begin{array}{cccc}0.4019& 0.4019& 0.1608& 0.3549\\ 0& 0.4823& 0.3858& 0.1319\\ 0& 0& 0.5787& 0.4213\\ 0& 0& 0& 1\end{array}\right]}$

First, it is a fun note, if not entirely productive, that the initial odds of rolling no successes and rolling exactly one success are equal:

${\displaystyle {\left(\frac{5}{6}\right)}^5=\left(\genfrac{}{}{0px}{}{5}{1}\right)\left(\frac{1}{6}\right){\left(\frac{5}{6}\right)}^4\approx 0.4019}$

But let’s ask a more interesting question: how likely is it to end the game with a crew? Unlike other absorbing Markov processes I’ve looked at, this one does not iterate until saturation. So instead of extracting the transient matrix $Q$ and from there deriving the fundamental matrix, we can directly iterate the whole transition matrix:

${\displaystyle t_3=P^3\times t_0}$

Where $t_0 =$ is our initial state, and $t_3$ is our state after three rolls. Solving, $t_3$, so roughly 35% of turns should end with at least a crew.

What about player agency though? We can stop iteration at $P^2$ instead, and see that about 17% of turns will have a crew after the second roll, allowing cargo-related decisions.

Expanding

While this analysis may seem disheartening, I would counter that the game is still fun and worthwhile, and might still be useful as a minigame inside a larger structure. That said, I’ve put the code in a gist, as always, and it’s fully parameterized etc. You may be able to tweak the game to your liking by changing the number and kind of dice, the number of throws each turn, or the number of steps required before scoring.

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1. Please forgive this misnomer, as, technically, all results without a crew are unscored. It is too deeply embedded in my notes.↩︎

2. Note that I don’t mean exactly one “six.” These are instead, the odds of rolling one six and no fives (instead of one six and no other sixes). But the odds are equivalent.↩︎

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