Last night I played a game of 40K. I wanted to try out some different missions from the typical 40K book, so I printed off the missions book from Dice Like Thunder’s webpage. http://www.mybattalion.com/home/downloads/dlt-missions . It’s a well done pdf and it’s free, which is a bonus as well. We rolled advance as the objective and corridor of fire as the deployment, which had us play from the short end of the table and to be within 36” of the edge. We talked about the battle at the end of the match and we both agreed that I had made a mistake on deployment by bringing my Daemon Prince too close to the enemy gunline. In the forefront like that, he was a singular target and was quickly brought down in turn 2 by the necrons. If, I had more units within targeting range, my enemy would have had more options for firing and perhaps left my HQ alone longer so that he could get into combat. With the Daemon Prince gone, the necron force shot up my rhinos and all my troops had to run to the middle forest. I played smart when my Lesser Daemons deep striked in and used them as a wall of sorts to protect my advancing forces. They wouldn’t prevent the shot but they would offer cover saves from AP 3 weapons. I also utilized my only moving rhino as a shield to protect my Berzerkers and my Noise Marines during this run as well. During the game, I managed to fix another rhino and it was only today (the day after the battle) that I realized that I could have used it to tank shock some of his units. It’s a strategy that I don’t normally remember and therefore don’t consider. I think it’s viable in some cases, particularly with troops who have weapons of shorter range. This gets them out of range, potentially and provides protection for the advancing forces. Overall, the match was a lot closer than we both thought, I lost by one kill point because I had singular survivors from three squads in a forest.
One of the great things about playing warhammer fantasy is the ability to cast magic. Being a fickle force however, there is great risk to trying to tamper with the sorcerous ways. You may already know that this risk is represented by rolling a miscast, or two ones while casting a spell. I have noticed that some players will not risk rolling a lot of dice when casting a spell, the rationale is that they have more of a chance of rolling two ones. Is this true? Well, lets look at some probability to find out.
When rolling two six-sided dice, there are 36 possible combinations that can come up. To get this number you take the six combinations that can turn up for the first dice and multiply it by the six combinations that can arise from the second. 6 x 6 = 36
Out of those 36 possibilities, two ones can only happen 1 way! So, the probability of rolling two ones on two six-sided dice is 1 out of 36 or 0.03. See the chart below:
| | 1 | 2 | 3 | 4 | 5 | 6 |
| 1 | 1,1 | | | | | |
| 2 | | | | | | |
| 3 | | | | | | |
| 4 | | | | | | |
| 5 | | | | | | |
| 6 | | | | | | |
Now, sometimes you need more dice to cast so that your spell will be successful. Let's see what the chance of rolling two or more one's is now. In order to miscast with 3 dice you need to roll one of the following combinations
| Die 1 | Die 2 | Die 3 |
| Roll 1 | Roll 1 | Not a 1 |
| Roll 1 | Not a 1 | Roll 1 |
| Not a 1 | Roll 1 | Roll 1 |
| Roll 1 | Roll 1 | Roll 1 |
This is four combinations. The total possible roll combination for three dice is 6 x 6 x 6 = 216. So, the new probability of rolling at least two ones is 4/216 = 0.018. This is SMALLER than the previous result. Let's do it one more time for four dice and see if the pattern continues.
| Die 1 | Die 2 | Die 3 | Die 4 |
| Roll 1 | Roll 1 | | |
| Roll 1 | | Roll 1 | |
| Roll 1 | | | Roll 1 |
| | Roll 1 | Roll 1 | |
| | Roll 1 | | Roll 1 |
| | | Roll 1 | Roll 1 |
| Roll 1 | Roll 1 | Roll 1 | |
| Roll 1 | Roll 1 | | Roll 1 |
| | Roll 1 | Roll 1 | Roll 1 |
| Roll 1 | | Roll 1 | Roll 1 |
| Roll 1 | Roll 1 | Roll 1 | Roll 1 |
There are 11 ways of rolling at least two ones now. There are 6 x 6 x 6 x 6 = 1296 combinations total. That means there is a 11/1296 = 0.0085 which is smaller than the two probabilities above. Already, I have debunked the myth that by rolling more dice you will have a greater chance of miscasting. For those who are mathematically inclined the formula for finding the probability based on a number of dice is:
(2n-nC0-nC1)/6n where n is the number of dice rolled and C is the Combinatorics function.
For the probabilities for rolling 2 to 10 dice, I have included them below in percentage form.
| number of dice | probability |
| 2 | 2.78% |
| 3 | 1.85% |
| 4 | 0.85% |
| 5 | 0.33% |
| 6 | 0.12% |
| 7 | 0.04% |
| 8 | 0.01% |
| 9 | 0.00% |
| 10 | 0.00% |
