videoguy505

Whenever we're bored, we play a dice game we uncreatively call "The
Dice Game". If there's an actual name for this game, I haven't found
it. The rules are:

Everyone 'antes' a set amount (usually $1).

First shooter rolls six dice. The shooter must remove at least 1 die
and set aside. Shooter rerolls remaining dice until all the dice are
set aside. The shooter must set aside a rolled 1 and a 2 to "qualify"
his score. If the shooter fails to roll a 1 and a 2, he is out and
must pay another ante. If he does roll a 1 and 2, the total of the
remaining 4 dice is his score. The best possible roll is 1 and 2 with
6, 6, 6, 6, for 24 points.

For example, shooter rolls six dice to start. Rolls are 1, 3, 4, 4, 5,
6. Shooter elects to set aside 1 and 6. Rolls 4 remaining dice: 2, 2,
4, 5. Shooter elects to set aside 2 to "qualify" his turn. Rolls 3
remaining dice: 3, 4, 4. Must remove one, selects a 4. Rolls again:
2, 6. Sets aside the 6. Rolls the last die: 3. Dice set aside: 1, 2,
6, 6, 4, 3, for a final score of 19.

The next shooter repeats the process. When the last person has
completed their turn, the highest score wins the pot for that round.
Note a person may be unable to beat the high score, but may still have
to roll to qualify his turn; i.e., a person has to beat a score of 23,
but has so far set aside: 1, 5, 5, 6. The person can only score a max
of 22, but must still continue rolling and hope to get a 2 to avoid
having to pay another ante into the pot.

My question is, is there an optimal strategy for this game...
especially regarding when to save a 5? I know that a later turn is
obviously better, since you know the score to beat. Often one may
elect to discard a 5 (or even a 6?) if one hasn't yet qualifed... to
give a higher chance of rolling a needed 1 or 2 on the next roll. For
example, a person rolls 6, 6, 3, 3, 4, 5 on the first roll. Elects to
save both of the sixes. Rolls 6, 6, 4, 5 on the next roll. Pulling
out both sixes means he must next roll two dice, and hope to get a) a 1
and a 2, b) a 1 on one die, and a 2 on the next roll, or c) a 2 on one
die, then a 1 on the next roll. The person may elect to remove only 1
six, so that he may roll three dice with his next turn, which seemingly
makes it easier to qualify his turn. The person in this example has
chosen to risk not making the highest score on this turn (thus
forfeiting his $1) for a chance at not having to forfeit $2... is this
a profitable long term strategy?

I haven't found any discussions of strategy for this game, if anyone
has leads I'd appreciate it.

arthur j odwyer

[...]

Sounds similar to Pickomino in the general strategy, although the goals
in Pickomino are different and you only need to roll one special number to
qualify your turn. In Pickomino, the special number is basically a 6,
instead of a 1 or 2, but after a little thought I don't think this is relevant.

Obviously.

No, that can't be right. Wouldn't you just try to score as high as
possible all the time, no matter when your turn was? I know I would.

You can work out the odds by brute force, if you really want to know
when to save 5s and so on. But I don't know where you'd look to find out
who's done the math for you already. rec.puzzles sounds like a good bet;
maybe sci.math? (They have some posters who like calculation problems.)

-Arthur

eric sosman

That's a slightly different goal. The strategy that
maximizes your expected score may not be the same as the
strategy that maximizes your likelihood of exceeding a
specified score. For example, suppose an earlier player has
already achieved a perfect score of 24, so your only hope
of sharing in the pot is to match that score. A strategy
that aims to maximize your expected score might tell you
to be content if you roll 1-2-5-6-6-6 for a total of 23,
but if you follow that strategy in this situation you are
guaranteed to lose.

It seems to me the goal of the Kth player should be to
maximize the likelihood that his score exceeds the maximum
of the K-1 preceding scores and the expected maximum of
the N-(K-1) scores that will follow. Since the score that
K actually achieves can influence the expected scores of
the later players (by "raising the bar"), the calculation
seems quite complicated. Perhaps it could be attacked by
starting with the Nth player and computing his expected scores
when he aims to exceed each of the possible preceding maxima,
then work back to the (N-1)st player, and so on.

--
Eric Sosman
esosman@acm-dot-org.invalid

mark j tilford

(I think that this method will analyze a slightly simpler problem, where
draws do not split the pot; they entirely go to the first tying player,
though I may have made an error in it...)

Let NDICE and NSIDES be fixed constants
Define f(total, pot, later_players) -> (avg_payoff, p_later_wins, avg_pot)
- total is the highest total of any previous qualifying player.
- pot is the number of chips in the pot
- later_players is the number of players who play after this player

- avg_payoff is the average win by this player in this situation
- p_later_wins is the probability this player or someone later wins
- avg_pot is the average end value of the pot
Define g(total, has1, has2, pot, later_players, dice_left) ->
(avg_payoff, p_later_wins, avg_pot)
- total is (points by highest previous player) - (points stored by this player)
- has1 is 1 if this player has stored a 1, 0 if otherwise
- has2 is 1 if this player has stored a 2, 0 if otherwise
- pot is the value of the pot initially

Return values are as above
Define h(total, has1, has2, pot, later_players, (dice)) ->
(avg_payoff, p_later_wins, avg_pot, no_dice)
- (dice) is a sorted list of dice values
- no_dice is the number of dice taken by the player


f and g are cached; h is not. (If g is written properly, no value of h
will be evaluated by g more than once; it will only be reevaluated if the
player is looking for a strategy.)
g (A, B, C, D, E, F) ->
weighted_avg (sorted lists L of F numbers from 1 to NSIDES:
weight(multinomial(L)) : h(A, B, C, D, E, L))

h(...) uses rules:
Always take a 1 if there is one and you don't already have one.
Similar with 2.
After that, consider each possibility of taking top N dice.
Return values corresponding to highest avg_payoff.

(Rest of recursion omitted).

--
------------------------
Mark Jeffrey Tilford
tilford@ugcs.caltech.edu