This dice game is an excellent way to learn the mathematical approach to such problems, spiced up with the few small calculations required. Calculations, of course, after you understand the approach.

SevenEleven is also offered in casinos. I have only played it there once. So I don’t even know the exact rules. That’s why I can only examine the game that is played on the Kiez or in other gambling circles. There is no organiser, people play it among themselves.

SevenEleven is played with two dice. There is one player, a person who throws, and there are the opponents. The opponents are called “the ponte”. The person who throws is also “the banker”. The game gets its name from the fact that you are the winner if you throw a 7 or an 11 on the first throw. So the player whose turn it is, also usually in turn, gets the dice and rolls. The ponte holds its bets against it. It is partly up to the thrower how much money he “accepts”. So from the ponte the question is asked “Will the 100 still go?”. The thrower confirms or not.

So there are the 7 and the 11, which win immediately. But there are also a few throws that lose immediately. It is better to write these throws down like this: It is the double 1, 1-2 and the double 6. These are the so-called “craps”, “rubbish” throws. Then you have lost. Usually, the game is played in such a way that the thrower continues if he wins, and the next one takes his turn if the thrower loses his game.

What happens now if the thrower throws neither 7 nor 11, but also no craps? There are quite a decent number of throws of this, which we will examine in a moment. But the rule says that the game continues if it is not yet decided in the first throw, i.e. in the initial throw. And it continues like this: The thrower must then roll the dice until he either rolls a 7 or rolls the number he presented again. The 7 is the number on which he loses and the number he presented and then confirmed is the number on which he wins.

A concrete example might make it clearer: The thrower throws. He has 5 opponents, each of whom has bet 50 euros. He rolls a double 4. That is a total of 8. So it is neither craps nor a 7 nor an 11. So the game continues. He continues to roll the dice. On the next roll he throws 2-3. That is neither 7 nor 8. So nothing happens. He rolls again. He rolls 3-6. That is neither 7 nor 8. He has to roll again. Now he throws 4-3. That is 7 in total. The thrower has lost and must pay out all the amounts of the ponte with equal money. That makes 5 * 50 euros = 250 euros loss. If only he had rolled an 8!

How do you work out which side has an advantage (if any)? Intuition is of no help here. You have to calculate very hard. The approach to the solution is, of course, to examine the individual throws. As always, it makes no difference whether you throw the dice at once or one after the other. In practice, of course, they are always thrown together. Because you could also throw them one after the other, you see in principle (you already know anyway) that there are 36 combinations, because each individual throw has 6 outcomes, the multiplication of 6 * 6 gives the 36. In this game here, there would of course be some throws that could be combined. But what is the point? We can quite comfortably look at all combinations at a glance in a matrix 6 * 6. The individual throws are then summarised by finding the same entries for the entries that are similar in the way that they offer the same distribution of chances. So here is the matrix:

1 | 2 | 3 | 4 | 5 | 6 | ||

1 | -1.000 | -1.000 | -0.333 | -0.200 | -0.091 | +1.000 | -1.624 |

2 | -1.000 | -0.333 | -0.200 | -0.091 | +1.000 | -0.091 | -0.715 |

3 | -0.333 | -0.200 | -0.091 | +1.000 | -0.091 | -0.200 | +0.085 |

4 | -0.200 | -0.091 | +1.000 | -0.091 | -0.200 | -0.333 | +0.085 |

5 | -0.091 | +1.000 | -0.091 | -0.200 | -0.333 | +1.000 | +1.285 |

6 | +1.000 | -0.091 | -0.200 | -0.333 | +1.000 | -1.000 | +0.376 |

-1.624 | -0.715 | +0.085 | +0.085 | +1.285 | +0.376 | -0.509 | |

-0.509 |

First of all, here are the implications of what happens if…. This is to be distinguished in that it has not yet been multiplied out by the probability of occurrence here. So if you roll 1-1, then you have -1.00 units, so you have lost one unit. But that doesn’t tell us how often it comes. But if we want to calculate an expected value, and we need this for the question “who has the advantage?”, we still have to do the multiplication out. I will do that in a moment. Here I would first like to explain the entries briefly.

If you roll 1-1, 1-2, 2-1, 6-6, you will find the entry -1.00 in the matrix. You have lost one unit immediately with these rolls. However, if you roll the main diagonal, i.e. 1-6, 2-5, 3-4, 4-3, 5-2 or 6-1, you immediately gain one unit. Likewise with 5-6 and with 6-5. Those are the rules.

The other entries need a short explanation: If, for example, you roll a 4 in the total (2-2 or 3-1 or 1-3) then there are 36 subsequent rolls, just as before. The only difference is that many of these 36 rolls are irrelevant. Because for most of them you have to roll the dice again anyway. So it is enough to examine the dice at which the game ends. This is sufficient because the game only ends when one of these combinations follows. These are always the combinations that add up to 7, i.e. all 6 possibilities on the main diagonal, and those that add up to 4. These continue to be the 1-3, 2-2 and 3-1. So there are 9 combinations of rolls where the game ends and the remaining 27 continue to be played. We simply ignore these 27 (correctly; they are elements of the event space but have no influence on the winner of the game).

Of the 9 relevant roll combinations, 6 are losing rolls and 3 are winning rolls for the thrower. So you can calculate that with the usual division favourable/possible. Since we are now looking at the perspective of the thrower, we must of course recognise the “favourable” for him and take it as a basis. Favourable for him are the 3 combinations in which he scores a 4. The other 6 are unfavourable. But we have to divide favourable/possible, i.e. 3/9.

Unfortunately, this number only calculates the probability of a favourable throw. For a calculation of the expected value, in case we rolled a (sum) 4 in the initial roll, we now have to multiply with the probability. This is calculated as follows: 3/9 * (+1) – 6/9 * (-1). In three out of 9 cases, i.e. the favourable ones, the caster has won a unit. In the 6 unfavourable cases, however, the thrower has lost one unit. This results in the entry in the matrix, because 3/9 – 6/9 = -3/9 = -1/3 = -0.33.

The throw (sum) 4 is exactly analogous to the throw (sum) 10. There are also 3 combinations, namely 5-5, 4-6 and 6-4. This is, but not coincidentally, exactly the reverse of the dice! Therefore, the entries for throwing (sum) 10 also resemble those of (sum) 4.

The same analogy results for 5 and 9. Only: The number of favourable combinations increases. You then have 4 favourable combinations. A 5 can be achieved with 1-4, 2-3, 3-2 or 4-1 (look at the reverse side for the 9). So 4 are favourable, 6 unfavourable. For the calculation of the expected value, this gives: 4/10 * (+1) – 6/10 * (-1). The only thing to note is that the number of possible events has also increased. So in that sense, the game (and this is really true) ends faster. There are more combinations that can end it, so it goes faster. 4/10 – 6/10 = -2/10 = -0.2.

I’ll save the analogy for the 6 and 8 to apply here. There are 11 combinations that then end the game. Of these, 5 are favourable and the 6 unfavourable ones always remain. So the calculation is 5/11 * (+1) – 6/11 * (-1) = -1/11 = -0.91.

Now we have calculated all the expected values for the individual throws. But this is not yet the expectation for the whole game. Now we just have to multiply it by the probability of occurrence for the individual throw-in. But that is really easy in this case. Of course, I have deliberately chosen the representation form of a matrix for this example, because of course all entries are equally likely. There are 36 entries and they are all equally probable, i.e. each with 1/36. Now, of course, you can either divide the final result, which is the sum of the matrix in the row and column above, by 36, or you can look at a new matrix, within which you multiply out and then add up.

Here is this matrix first:

1 | 2 | 3 | 4 | 5 | 6 | ||

1 | -0.028 | -0.028 | -0.009 | -0.006 | -0.003 | +0.028 | -0.045 |

2 | -0.028 | -0.009 | -0.006 | -0.003 | +0.028 | -0.003 | -0.020 |

3 | -0.009 | -0.006 | -0.003 | +0.028 | -0.003 | -0.006 | +0.002 |

4 | -0.006 | -0.003 | +0.028 | -0.003 | -0.006 | -0.009 | +0.002 |

5 | -0.003 | +0.028 | -0.003 | -0.006 | -0.009 | +0.028 | +0.036 |

6 | +0.028 | -0.003 | -0.006 | -0.009 | +0.028 | -0.028 | +0.010 |

-0.045 | -0.020 | +0.002 | +0.002 | +0.036 | +0.010 | -0.014 | |

-0.014 |

The representation obtained in this way is somewhat unwieldy only insofar as the entries result in smaller numbers. The result is nevertheless obvious: The person who is throwing has a disadvantage of -0.014 units. Expressed as a percentage, this is -1.41%.

The same result is obtained, of course, if you calculate the above as the matrix sum of columns (the number at the very bottom right is only the checksum of the rows). The sum of all entries is -0.59, which divided by 36, again gives -1.41%.

The caster has a disadvantage. Probably quite a few of the gamblers know that. But then, because they have the dice in their hands or whatever, they trust their influence. The game is (or was) played a lot. It really is relatively fair.