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!

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.

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.

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?

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    $\begingroup$ Does two pair beat one pair? For example, does 22113 beat 55432 because 22113 has two pairs and 55432 has only one? Or does 55432 beat 22113 because a pair of 5s beats a pair of 2s? $\endgroup$
    – MJD
    Commented Dec 18, 2013 at 2:27
  • $\begingroup$ @MJD Good question. $55432$ scores nothing, since it doesn't contain a $1$. $22113$ scores as four 3's, since $1$ is wild. Other than that, 2 pair, full houses, and straights don't count. $\endgroup$ Commented Dec 18, 2013 at 2:29
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    $\begingroup$ This is going to be quite complicated, because the optimal strategy probably depends on what the opponent's optimal strategy is, and there may not be an equilibrium. $\endgroup$
    – MJD
    Commented Dec 18, 2013 at 2:39
  • $\begingroup$ @MJD Agreed. Perhaps you can assume you're playing against a bunch of drunks ; ) $\endgroup$ Commented Dec 18, 2013 at 2:41
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    $\begingroup$ @Euler....IS_ALIVE: Easier just to play sheepshead! :-) $\endgroup$ Commented Dec 18, 2013 at 3:09

3 Answers 3


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.
  • $\begingroup$ 2. Technically speaking, the last person is playing the game, since the lowest score goes first in the next round. Perhaps they could try to get the lowest score if there's some advantage to doing so. Of course that's 'house rules' that just adds some more complications, we can remove that if necessary. 5. I like this suggestion. $\endgroup$ Commented Dec 19, 2013 at 6:01
  • $\begingroup$ I meant, last person in the last round. I edited it to make clear. $\endgroup$ Commented Dec 19, 2013 at 20:43

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.

  • $\begingroup$ The problem is well posed if we assume everyone is using an optimal strategy. $\endgroup$ Commented Dec 18, 2013 at 13:26
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    $\begingroup$ @ Sergio Parreiras : I fully agree with you. I did not mean to say that the problem was misspecified. I just meant to say one might want to give an answer where not everyone plays optimally and that, given the tone of the question and my reading of it, it would satisfy the OP. I tried to reformulate part of my answer to make it clear. Does my answer still suggest that I think the problem is misspecified? $\endgroup$ Commented Dec 18, 2013 at 13:59
  • $\begingroup$ No Martin, it is clear. $\endgroup$ Commented Dec 18, 2013 at 21:09

In practice the rules for this games are what is wild.

Could there be an optimal strategy for a simple two-player game where: Five dice, three throws, ones are wild, but no one is required. You can save any number of dice on any throw, but you are not required to save any and once saved a die cannot be rethrown. Five sixes is the best hand, only the number of like dice, including wild ones, count. No poker hands.

There was once a computer vs player game like this, programmed in basic, on Compuserve. Obviously the computer had to have a strategy. I have never found it in the internet iphone world, just dozens of different sets of rules for like play in bars in Wisconsin.


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