If you want to learn to play well at Diceland, you must immediately get used to the idea that (even if it is a “battle game”), it isn’t a simulation game at all. This means that the rules that govern the relationship of strength among the dice are often different from what you might expect.

In reality a battle 6 against one (equal in armament and position) would be destined to inevitably finish with the total destruction of one. Instead with Diceland this is one of the most uncertain “battles” that can happen, because the attacker has the certainty to eliminate the defender, but with ample probabilities to have more losses than those inflicted. For this reason, an attack of this type must be avoided and one of the strategies available to the players is to manage to isolate an enemy dice with a very high value forcing him to attack by himself a dice that has a low value.

The probabilities to obtain a perfect attack (in other words a score that equals the attacker’s dice) is 1 on 6, that is 16,6%, the same that we have to obtain a disastrous score (if we leave 1, we eliminate the enemy’s dice but we loose 5 points).   Also, the probabilities to obtain a score that is lower than 5 (that would make you loose more points than those lost by the enemy) is of 66,6% (4 probabilities on 6). For this reason, when it is possible, it is always better to make a combined attack, in other words use 2 dice against one.  But in this case as well, you should not be too optimistic and most of all, you should not overestimate your attacking strength or make calculations of probabilities “by chance”. One of the most common mistakes is to think that an attack with an enormously superior “strength” to that of the defender, is the best method to obtain an easy win. So this is what happens if I decide to attack a dice valued 1 with two dice valued 6 (this is a theoretically extreme situation in favour of the attacker).  Since it is not possible that this attack fails (we cannot obtain a score that is higher to that of the attacker), I have the certainty to eliminate the enemy’s dice, because any score I might obtain is destined at least to be equal to his. The problem in this case is not in the attack’s result, but the cost in terms of lost points.

We know that the most favourable result from an attack is obtained when the score of a roll is identical to that of the attacker. In the example above, a perfect attack needs to obtain at least a 6 with a roll of the two dice. Obviously the probabilities in our favour are superior to those that we had with only one dice, but do not make the error to think that the chances are double (in other words 33,2%). In reality the percentage is slightly lower 30,5%, as we can easily calculate by looking at the chart on our right. Here we can find all the possible combinations.

In total only 11 over 36, give as a result that is at least a 6, therefore the percentage is slightly lower than what we could expect, but most of all we must consider that the probabilities to have losses are just under 70%. If then we consider (left chart) the possibilities to obtain with both dice a score that is lower than a 5 (which would cause a loss of at least 2 points, therefore superior to that given to the defender), we realise that an attack of this kind is potentially very dangerous.

Indeed, we almost have 44% of probabilities to obtain a score that is not numerically in our favour.  It would be more wise to attack with a dice valued 6 and another 4. In this case the probabilities to obtain a perfect attack would increase to over 55%, in other words 20 useful results out of 36 as you can better see on chart 3:

Without considering the probabilities to obtain a loss that is superior to that of the defender (a situation that, in this case, can happen only if the score is 1 or 2) they go down to a “comfortable” 11,1%.

The combined attack of 6-3 has practically the same possibilities of success and the same percentage of risk.   As we can see in chart 5 that follows, the combinations that involve an obligatory loss of at least 2 points are in quantity the same to that of the previous situation. The only advantage is made by the combination 1-1 that in the case of a combined attack 6-4 will cause a loss of at least 3 points, while if we choose a 6-3 attack, the risk of maximum loss is 2 points. This is not a lot, but in a game where the calculation of probabilities is essential, it can be a determining factor.