Simplest way to calculate dice probabilities

[fnorrll]

Hey guys, i want to better understand calculating probabilities.

As i understand it if i have 2 dice and i need to roll a 5 or a 6 on at least one of them, i have a 20/36 chance of doing so as there are 36 total combinations and 20 of them involve a 5 or a 6 (specifically: 1+5, 2+5, 3+5, 4+5, 5+1, 5+2, 5+3, 5+4, 5+5, 5+6, 6+1, 6+2, 6+3, 6+4, 6+5 6+6, 1+6, 2+6, 3+6, 4+6).

What i want to know is the easiest way to calculate this with multiple dice. I worked it out for 2 dice by listing all possible combinations but that gets tedious for 4 dice, 5 dice etc. Im assuming there must be some easy way to work it out.

For example I roll 3 dice, what are the chances of getting at least 1 5 or 6 result?

To be clear im not asking HOW MANY 5 or 6's i can expect (that seems simpler to work out by combining 1/3+1/3+1/3) but my chances of getting at least ONE result that i want.

Hope this makes sense.

[vectron]

There certainly is, though it does require some basic knowledge about probability and combinatorics(?).

The harsh answer is to tell you to get a relevant text book on the subject and sit down with a pencil and some paper, and work it out yourself.

The slightly friendlier answer is to say that I might look at your specific problem when I get home and aren't limited to my phone. If I remember, I'll post some pointers to get you started, but honestly you do have to put in some effort to be able to solve different kinds of probability problems. Possibly there are lectures on it on khan academy?

Anyway, it's a very interesting subject in general (and quite useful for 40K, of course) so don't be afraid to do some self study

[coredump]

Its actually very easy. The key is, it is easy to determine the probabilty of something always happening, or never happening. It is the 'sometimes' that makes life hard.

So in your case, instead of figuring out the chance of getting a 5/6 "1 or 2 or 3 or 4..." times. We will figure out the chance that it 'never' happens, and take the complementary percentage.

So... Each time you roll a die, there is a 66% (.66 out of 1) chance that it will *not* get a 5 or 6.
So to roll two dice, you have a 66% chance of it not happening on the first roll, and then 66% chance of it not happening on the second roll.
Those will *both* happen only (0.66 x 0.66 = .444) 44.4% of the time. Remember, that is the chance of getting *0* 5's or 6's. Thus the chance of getting "1 or more" would be the 'opposite', or 55.5% of the time. (BTW, 20/36 = .5555 or 55.5%)

This same process can be used for any number of dice. So lets say you want to roll 5 dice, and what to know the chance of getting at least 1 5/6.

.666 x.666 x.666 x .666 x .666 = .1317 So the chance of getting *0* 5/6's is 13.17%. Thus the chance of getting at least 1 5/6 would be 86.8%

Extra Info:
The chance of getting *all* 5/6 would be .333 x .333 x .333 x .333 x .333 = .0041 or 0.4%

If you have a calculator that can deal with exponents... these can be entered as .666^5 or .3333^5.

If you wanted to know the chance of it happening at least 2 times, or exactly 3 times... then it starts getting complicated.

[fragile]

Dec 22, 2013 14:44:41 GMT coredump said:

Extra Info:
The chance of getting *all* 5/6 would be .333 x .333 x .333 x .333 x .333 = .0041 or 0.4%

Then you jump up and scream YAHTZEE !.

I had an opponent do this once. My Biovores dropped 5 wounds on a Terminator squad... five 1's.. Priceless.

[fnorrll]

Thanks guys, especially Core (who i notice has been on top of everything these days, handing out wisdom all over) that sets me on the right track.

[commandersasha]

Coredump knows everything. That's why we pay him so much.

[dakkathrope]

For me a practical way to do it very quickly is to just leave it in fractions. For example: I'm rolling 30 dice to hit and then roll the following to wound. With bs of 3 and a S4. Half would hit (15) and against T4 half would wound (7.5). Against a T5 only 5 wounds, etc. I've found that, for me, that mathing everything out gets tedious and I end up taking longer to figure out the odds while playing. My way is quick, but also can be less accurate (since I'm usually rounding), But I've found it to be sufficient for 40k. I also tend to assume a 10% worse hit/wound rate than what math shows so that my junk rolls don't seem so bad.

[Uhtin]

Dec 23, 2013 9:33:26 GMT commandersasha said:

Coredump knows everything. That's why we pay him so much.

Coredump indeed is a living library of wisdom.

[Hive Bahamut]

My dice rolling odds are easy. If it is one important roll 95% chance of failure with a 5% margin of error. Especially when it comes to Armour Pen rolls. All the rest just happens. I angered the dice gods somewhere along the lines.

[revx]

Dec 23, 2013 9:33:26 GMT commandersasha said:

Coredump knows everything. That's why we pay him so much.

you pay him? man im getting a deal!!!