[Math] Optimal Strategy for Bar Dice

dicegame theoryrecreational-mathematics

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?

Best Answer

Not really an answer, more of a sketch.

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).
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.
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.
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.
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.

Related Solutions

[Math] Simple dice game: Optimal strategy

There is no simplified description of the Nash equilibrium of this game.

You can compute the best strategy starting from positions where both players are about to win and going backwards from there. Let $p(Y,O,P)$ the probability that you win if you are at the situation $(Y,O,P)$ and if you make the best choices. The difficulty is that to compute the strategy and probability to win at some situation $(Y,O,P)$, you make your choice depending on the probability $p(O,Y,0)$. So you have a (piecewise affine and contracting) decreasing function $F_{(Y,O,P)}$ such that $p(Y,O,P) = F_{(Y,O,P)}(p(O,Y,0))$, and in particular, you need to find the fixpoint of the composition $F_{(Y,O,0)} \circ F_{(O,Y,0)}$ in order to find the real $p(O,Y,0)$, and deduce everything from there.

After computing this for a 100 points game and some inspecting, there is no function $g(Y,O)$ such that the strategy simplifies to "stop if you accumulated $g(Y,O)$ points or more". For example, at $Y=61,O=62$,you should stop when you have exactly $20$ or $21$ points, and continue otherwise.

If you let $g(Y,O)$ be the smallest number of points $P$ such that you should stop at $(Y,O,P)$, then $g$ does not look very nice at all. It is not monotonous and does strange things, except in the region where you should just keep playing until you lose or win in $1$ move.

Probability – Most Unfair Set of Three Nontransitive Dice

It turns out that you can do very slightly better than the dice in your example. The probability has to be a multiple of $1/36$, your example has $20/36$, and the best possible value is $21/36=7/12$.

From my answer to the question that anon linked to, we know that the probability of one $6$-sided die beating another while having a lower median can be at most $3/4-1/12=2/3=24/36$. Wikipedia gives an example of a cycle of four dice in which each die beats the next one with this probability.

You're looking for a cycle of three dice in which each die beats the next one with the same probability. Without loss of generality we can assume that the dice carry $18$ different numbers. Then there are $18!/6!^3=17153136$ different sets of dice, so it's well within the reach of our electronic friends to try them all out. Here's code that does that (rather inefficiently). The result is that the maximal probability for such a cycle is $21/36=7/12$ and there are eight different sets of dice that attain that probability. Here they are, with consecutive stretches of numbers collapsed into one as in your example:

0 3 3 3 6 6
2 2 2 5 5 5
1 4 4 4 4 4

0 0 3 3 3 6
2 2 2 2 2 5
1 1 1 4 4 4

0 3 5 5 5 5
2 4 4 4 4 7
1 1 1 6 6 6

0 3 5 7 7 7
2 2 4 4 6 9
1 1 1 8 8 8

0 3 3 5 5 5
2 2 2 4 4 7
1 1 1 6 6 6

0 3 5 5 7 7
2 2 2 4 6 9
1 1 1 8 8 8

0 3 3 3 3 5
2 2 2 2 4 7
1 1 1 6 6 6

0 0 3 6 6 6
2 2 2 5 5 5
1 1 1 4 7 7

There are two sets with $7$ different numbers, four sets with $8$ and two with $10$, and your near-optimal example has $9$, whereas the mean number of different numbers after collapsing, averaged over all sets with $18$ different numbers, is $18-17\cdot(5/17)=13$, so consecutive stretches of numbers and thus high collapsibility are a striking feature of these sets.

P.S.: I just realized that there's a symmetry in that you can invert the orders of both the dice and the numbers and get another set with the same probability. The first two sets are related by this symmetry, as are the third and seventh, the fourth and sixth and the fifth and eighth. Thus there are only four really different sets of dice, with $7$, $8$, $10$ and $8$ different numbers, respectively.

Best Answer

Related Solutions

[Math] Simple dice game: Optimal strategy

Probability – Most Unfair Set of Three Nontransitive Dice

Related Question