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The game can be played at https://xcvd.github.io/dice-game/

Dice game grid Dice game grid Dice game grid Dice game grid

The player gets 12 throws of 3 dice and chooses a grid to place these throws in (there are 6*6*6=216 possible throws). Each throw is place in the leftmost available position of the grid and there are 4 grids to choose from.

After placing a throw, the player may throw and place again until all grids have been filled.

Once each grid is filled, points are awarded for 3 of the same dice face value on any of the paylines in that grid (shown above, there are 5 paylines). Points earned are equal to the face value of the dice on for the payline. Once all 12 combinations have been placed, the amount of points is totalled.

The total points earned by the player are the sum of the points for each individual 3x3 grid.

Wins on separate paylines are added together.

Is the optimal strategy for this game simple to compute? Can anybody give me some pointers in the right direction?

xcvd
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    There are some differences of opinion about how to interpret your question. I, for one, am not sure about the rules of the game because I don't know what is being "ordered" at the outset. What exactly is the "result of the throw": all nine values, sums of three values, subsets of three values? What are the rules and restrictions for "placing" values on the grid? (For instance, can a grid cell be occupied by more than one value?) What is the "face value of the dice": sums of three dice, sums of all nine, or something else? What exactly is a "win line"? Is it related to the line segments? – whuber Mar 15 '18 at 14:48
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    If you are rolling 3 dice at a time, how does the 'ordering' work w/i that set of 3? Is it just whichever lands leftmost to rightmost from the thrower's point of view, or can you shuffle those as you like, but the ordering just prevents you from shuffling the rows? When you decide to go for another grid, do I understand correctly that you forever lose the option of using the score from the current grid? – gung - Reinstate Monica Mar 15 '18 at 15:03
  • @gung The ordering could be whichever lands left to right, or storing the result of throwing each dice individually. So the dice throw results in an ordered set of three elements. The ordering prevents you from shuffling the rows. In the first 'game', the player has no choices to make. When you decide to go for another grid, you do not lose the option of using the score from the current grid. – xcvd Mar 15 '18 at 18:16
  • The edit helps immensely (it totally changed my interpretation of the game). But in the full game, what constitutes a "throw"? Would it be 3, 4, 9, 36, or some other number of dice? Must all dice in a throw be stored in the same grid? Can two or more dice occupy the same cell of a grid? – whuber Mar 15 '18 at 18:17
  • @whuber The result of the throw is the ordered set of three elements (the values of the dice). If you get three ones on a red line, you score one point (the face value of the dice). – xcvd Mar 15 '18 at 18:17
  • @whuber 3 dice per throw. The dice need not be stored on the same grid, but always occupy one column. Two dice can never occupy the same cell of a grid. I will update the description of the full game to be more clear – xcvd Mar 15 '18 at 18:19
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    At https://stats.stackexchange.com/a/155402/919, I outline a general approach for such games. I believe it may apply here--although the coding would be much more complicated due to the complicated rules of your game. – whuber Mar 15 '18 at 18:19
  • @whuber Each throw will occupy the next available column on the grid, in order. That is the first element of the ordered set will take the top position in the column, the second element the middle position and the third element the bottom position. Once a column is placed it cannot be moved. – xcvd Mar 15 '18 at 18:23
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    It's growing clearer. It might help to abstract the rules a little bit. A "throw" is a random variable with values in the ordered multi-3-subsets of $\{1,2,3,4,5,6\}.$ A *decision* is a selection of any unoccupied column among the 12 columns in the game. The game therefore can be modeled as an influence diagram in which throws and decisions alternate. So, apply the standard backwards-forwards procedure to work out the optimal decision for any state of the game and result of a throw. The principal complication concerns computing the value of any final state, but that's an easy program to write. – whuber Mar 15 '18 at 18:33
  • Just for general information, this is related to the [secretary problem](https://en.wikipedia.org/wiki/Secretary_problem). – gung - Reinstate Monica Mar 15 '18 at 18:40
  • @Gung could you elaborate? This doesn't look like optimal stopping on the face of it. – whuber Mar 15 '18 at 18:53
  • @whuber, as I understand it, you either use the score from your current grid, or you go for a different grid. The strategy is to know when to stop & use the current grid instead of taking another chance. – gung - Reinstate Monica Mar 15 '18 at 19:20
  • @whuber Thanks for the pointing me towards influence diagrams and backwards-forwards procedure, both new to me, I have some reading to do. For the avoidance of any doubt in the rules, I've added a javascript version of the game to the question. – xcvd Mar 15 '18 at 20:25
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    @gung I believe it's more complex than that. During game play, one is not stopping: one is *choosing.* The game ultimately is played to completion and optimal play at any stage depends on the prospects across all subsequent decisions. There is a formal relationship--optimal stopping can be expressed in terms of a decision tree where a binary decision is made at each juncture--but the special characteristic of optimal stopping that isn't present here (besides the extreme simplicity of the decisions) is the requirement that *exactly one* decision to stop may be made. – whuber Mar 15 '18 at 20:32
  • @whuber Thanks once again. I think I understand the procedure having calculated the optimal strategy for a similar game with a much smaller state space. The worst case with my naive approach of using all combination of dice is when the last column is available on each grid, there are 216^8 possible configurations of dice. I'm trying to reduce the model so that game states are not represented by the values of the dice, but rather by the 'active' winning lines that can still result in a winning combination. I'm having difficulty with this currently. – xcvd Mar 16 '18 at 10:48
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    I'm thinking of your game as having four components of three columns each. The optimal move depends only on (a) how many moves are left in the game; (b) which columns are occupied in each component; and (c) what values are in those columns. There are only $\binom{6+3-1}{3-1}=56$ possible column vectors and $2^3=8$ subsets of columns in a component (and some of these combinations lead to identical or equivalent positions under symmetry). That means you need to track at most $(56^3)(8)(12)$ states, or approximately 17 million, which is far less than $216^8,$ approximately 50 million million. – whuber Mar 16 '18 at 13:48
  • I've failed at describing the rules of the game, but hopefully the link helps. There are 216 possible column vectors (throws are not elements of the multi-3-subset of $\{1,2,3,4,5,6\}$, (3,1,2) is a different throw to (3,2,1)). There are only 4 subsets of columns in a component, since the grid must be filled left to right.T he possible configurations are, 1. empty, 2. one column filled, 3. two columns filled and 4. full. For the state where there are 8 turns left, and 1 column placed in each grid, do I not need to consider the 216 choose 4 combinations of grids? – xcvd Mar 17 '18 at 09:17

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