Optimal Strategy for Bar Dice

Question

This game is played in bars in Wisconsin, USA, but I'm sure variations are played many places around the world. The game has practical value, since once mathematicians figure out the best strategy, we can probably get free drinks for life!

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The setup

You are playing a game against the bartender and your $N$ friends for a prize (usually shots of some sort of alcohol). Whoever loses must buy for the whole group, including the bartender. If the bartender is the loser, the house buys.

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The rules

(Example can also be found at http://www.thedrinkingsurvey.com/bar-dice-drinking-game.php)

Players sit in a circle. Each round, everyone gets a turn. The first person who shakes is ‘setting the score’. All remaining players try to set a new ‘high score’. The person with the high score wins the round and sits out of all remaining rounds.

The first person to shake for the round shakes 5 dice, and has up to three rolls to set a high score. Each player thereafter must try to set a higher score in the same number of rolls or less.

$\cdot$ 1 ’s are wild, and in order for your hand to count at all, it must contain a 1.

$\cdot$ All scoring is pair based: 5 of a kind beats 4 of a kind; 4 of a kind beats 3 of a kind; and 3 of a kind beats 2 of a kind.

$\cdot$ Scoring is also value based; three 5s beats three 4s, and so on.

$\cdot$ Scoring is also turn based; three 5s in two rolls beats three 5s in three rolls.

$\cdot$ After each roll, you are allowed to keep aside dice that you do not wish to re-shake for the remaining rolls.

Once there are two players left, they play a “best of three” match. The first person to lose 2 games is the loser and must pay for everyone.

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Finally, whoever had the worst score in the previous round rolls first in the next round.

What is the best strategy to play this game?

Answer 1

I am afraid MJD is right when saying that you cannot devise an optimal strategy without some -- at least probabilistic -- information on the type of strategies played by the other players (You might also have to simplify the setting a little). Even with such information, if other players play strategies which are too complex, it will be a nightmare to identify your own optimal strategy.

So I guess the only way to get some analytical insight into optimal strategies is to make some assumptions on other's strategy. I like you idea of assuming that you play a bunch of drunks ;) Assume however that you play smart drunk people, and that your friends are not too screwed. You however are perfectly sober, and you want to take advantage of this.

It might be reasonable to assume that smart drunk people realize that they are too drunk for complicated computations, and that they'd better stick to strategies which are as simple as possible. If they are not completely lobotomized, they know that they should always at least try to beat the former player's score. But if they are wasted enough, they might know that they are not able to effectively devise a more elaborate strategy. So you may want to assume that everyone but you (who is sober) play this strategy : stop playing as soon as you beat the former player.

Of course, this assumes away the problem of what the first player would do, as she has nobody to beat. To make the problem tractable, you might also need to assume that your friends are sober enough to compute 5 dices probabilities and decide optimally what is the best selection of dices to re-roll. In reality, we know that this is really not obvious (well we said they are smart drunks, so maybe they can do that in day life, but after a few drinks, who knows...). Other configurations of their abilities at computing simple probabilities might make things much more complicated.

Finally, addressing your problem in this forms overlooks equilibrium issues (or more generally avoids relying on a canonical game theoretic solution concept). I think this is not necessary in your case though. It might be nice to solve the problem when everyone plays optimally, and the problem is well specified that way. But if I read your question right you don't require this and it might be easier not to look for these kind of solutions.

I realize this is not really an answer to your question, just some ideas on how to get started. Even with these assumptions, the solution might still be quite complicated. At least, it is worth trying to solve the easy case before tackling more complicated ones...

Other possible assumption on others' strategies :

• Other's systematically play until their last throw, each time re-rolling optimally in order to get the best "facial" score possible.
• Other's play in order to reach a probability $x$ that the following players do not beat their score, given that the following player try to reach a probability $x$ that the following players do not beat their score, given that ...

These are only other -- maybe intractable -- ideas just to suggest the scope of possible assumptions, and the apparent necessity to make assumptions to get an answer.

Answer 2

Not really an answer, more of a sketch.

1. This game is a finite game of complete information with random moves by nature so in "theory" we can solve it by backwards induction (Zermelo's algorithm). In practice we may lack computational power to find the backward induction solution (notice the game of Chess also can be "solved in theory" as one can prove there exists an optimal strategy but computing it is another story).
2. The last person in the last round that plays is not playing a game in the technical sense (he or she does not need to worry about the strategies that the other players are using). The last person faces a purely optimization problem. Given the highest existing score and the number of rolls, the last person has to decide which dice to set aside at each roll. I'm not implying this is an easy problem but we can solve for the strategy of the last person.
3. The last-but-one person must use the same strategy as of the last person until the point where she attains the highest score. After that, if she attains the highest score before the max number of rolls then she faces an optimal stopping problem. She has to compute the probability that the last player using the optimal strategy will beat the (possibly higher) score she might get with additional rolls and weight it against the probability the last player will beat her current score using her current number of rolls.
4. After solving for the optimal strategies of the last two people playing we keep moving "backwards" and solve for the optimal strategies of the others.
5. It is a cute game. My suggestion is that perhaps we could gain some insights by looking at a smaller/simplified version of the game with only two players, two dices (instead of 5), and each dice with only three outcomes (say 1, 2, and 3) instead of the usual six.