Basic Probability theory

I need to first admit that I am a novice war gamer and that I have only 10 or so completed games under my belt. But I have noticed one thing - a lot of gamers have only a rudimentary grasp of probability theory which impact’s their gaming and leads to the cursing of the Dice Gods. In order to avoid making the Dice Gods angry (never a good thing) I thought I should point out some basics in probability theory - one should expect nothing less from an idiot like me who runs around with the moniker of “Uber-Geek”

First off, I am assuming that the outcome of all die rolls are random. Now we all know that, technically, that’s not true because the friction co-efficient for each dice face is different due to the different markings (pips). I think we can dispense with that aspect as the error term introduced doesn’t even round to anywhere near 1%. Second, I hope none of you plan on using any of this information to rush off to a casino and break the bank. If you do, it will end badly - for you. If a fevered brain like mine can babble this stuff out, I’m pretty sure a Casino has done the same and then some. They also tend to be somewhat resistant to the plea “can I have a do-over? I didn’t understand the probabilities”.

So, you need to roll a 6 on a D6 - what are your odds? Well, you need one of six outcomes so 1 divided by 6 equals .167 or 16.7%. Hmm what if you had a +1 modifier, well then your odds are 2 out of 6 (since either a roll of 5 or 6 works) so your odds are 16.7% + 16.7% or 33.4%. Each die outcome level or die modifier impacts the probability by 16.7% - hey wake up the good parts are coming.

What if you still need to roll at least one 6, but now had two shots - you can roll 2 dice, but still only need one 6. If one has a 16.7% on one dice, surely one has double that or 33.4% if 2 dice are rolled. Nope. The odds are a bit less than that.

For two rolls, there is a 1/6 probability of rolling a six on the first roll. If this occurs, we've satisfied our condition. There is a 5/6 probability that the first roll is not a 6. In that case, we need to see if the second roll is a 6. The probability of the second roll being a 6 is 1/6, so our overall probability is 1/6 + (5/6)*(1/6) = 11/36. Why did I multiply the second 1/6 by 5/6? Because I only need to consider the 5/6 of the time that the first roll wasn't a 6. As you can see the cumulative probability is only 30.%, which is slightly less than 33.4% (or 2/6 from the previous paragraph).

For three rolls, there is a 1/6 probability of rolling a six on the first roll. There is a 5/6 probability that the first roll is not a 6. In that case, we need to see if the second roll is a 6. The probability of the second roll being a 6 is 1/6, giving us a probability of 11/36. There is a 25/36 probability that neither of the first two rolls was a 6. In that case, we need to see if the third roll is a 6. The probability of the third roll being a 6 is 1/6, giving us a probability of 1/6 + (5/6)*(1/6) + (25/36)*(1/6) = 91/216 (or 42.2%). Again, this is less than 3/6. Observe that as the number of die are increased the gap from the simple additive assumption grows.

The general formula for rolling at least one 6 in n rolls is 1 - (5/6)^n.

Of course, all of this probability mumbo jumbo in no replacement for the sheer exhilaration one feels after seeing your lowly M4A1 Sherman overcome the thick frontal armor of a dastardly Tiger tank and win the game by rolling a pair of 6’s.

Just remember, sometimes your die rolling luck might just be improved with a better understanding probability theory.

Ok class, lets take roll call:

Adams

Baker

Bueler....., Bueler....., Bueler.....