Probabilities of rolling with multiple dice.

Hi,

I'm trying (and failing terribly as Mathematics isn't a strong subject for me) to work out probabilities of rolling numbers with d6's.

What I want to know is what is the chance of rolling at least X (1- 6) on individual dice with Y amounts of dice in the roll. I'm going to be putting this calculation into a spreadsheet.

Many thanks in advance.

Dave.

Quote: ICMDWILLIAMS

Hi,

I'm trying (and failing terribly as Mathematics isn't a strong subject for me) to work out probabilities of rolling numbers with d6's.

What I want to know is what is the chance of rolling at least X (1- 6) on individual dice with Y amounts of dice in the roll. I'm going to be putting this calculation into a spreadsheet.

Many thanks in advance.

Dave.

Hi Dave. It's Ahigh.

Y=1 is the easy case:

At least 1: 6/6
At least 2: 5/6
At least 3: 4/6
At least 4: 3/6
At least 5: 2/6
At least 6: 1/6

Y=2 now you have 36 cells instead of just 6
At least 1: 6/6 * 6/6 = 36/36
At least 2: 5/6 * 5/6 = 25/36
At least 3: 4/6 * 4/6 = 16/36
At least 4: 3/6 * 3/6 = 9/36
At least 5: 2/6 * 2/6 = 4/36
At least 6: 1/6 * 1/6 = 1/36

For Y=3 and above you just get same answer as Y=1 but to the power of Y.

1: (6/6)^^Y
2: (5/6)^^Y
etc

I think. I am no math guy, so don't be surprised when somebody says that I'm wrong or if I misunderstood the question. But that's what I think you're saying.

I have a buddy named David Williams that I worked with before, and when I answered this question, I assumed you were him. But no connection.

Quote:

Hi Dave. It's Ahigh.

Y=1 is the easy case:

At least 1: 6/6
At least 2: 5/6
At least 3: 4/6
At least 4: 3/6
At least 5: 2/6
At least 6: 1/6

Y=2 now you have 36 cells instead of just 6
At least 1: 6/6 * 6/6 = 36/36
At least 2: 5/6 * 5/6 = 25/36
At least 3: 4/6 * 4/6 = 16/36
At least 4: 3/6 * 3/6 = 9/36
At least 5: 2/6 * 2/6 = 4/36
At least 6: 1/6 * 1/6 = 1/36

For Y=3 and above you just get same answer as Y=1 but to the power of Y.

1: (6/6)^^Y
2: (5/6)^^Y
etc

I think. I am no math guy, so don't be surprised when somebody says that I'm wrong or if I misunderstood the question. But that's what I think you're saying.

I have a buddy named David Williams that I worked with before, and when I answered this question, I assumed you were him. But no connection.

Hi Ahigh, not that isn't me, it is a very common name :-)

I understand your working out for Y=1 that makes sense, however I may have not described the issue correctly. When there are multiple dice and rolling a minimum of a 4 (as an example) is required, not on all the dice, but just one of the dice, how would that be worked out?

Quote: ICMDWILLIAMS

I understand your working out for Y=1 that makes sense, however I may have not described the issue correctly. When there are multiple dice and rolling a minimum of a 4 (as an example) is required, not on all the dice, but just one of the dice, how would that be worked out?

Notice that the probability of something happening "at least once" = 1 minus the probability of it not happening at all.

The probability of rolling N or higher on at least one out of Y dice = 1 minus the probability of rolling less than N on all of the dice, which is 1 - ((N-1) / 6)^Y.

I like dice questions
I think a better example of what you are after would really help out as I have to try and figure out what you are trying to get to.

Quote: ICMDWILLIAMS

When there are multiple dice

say 3 dice as we can easily list all 216 possible roll sequences
and count

Quote: ICMDWILLIAMS

and rolling a minimum of a 4 (as an example) is required,

I read this and see you want a 6 or a 5 or a 4
true or false

Quote: ICMDWILLIAMS

not on all the dice, but just one of the dice, how would that be worked out?

ah, on one die only
we could have 4,x,y or 5,x,y or 6,x,y
so it could be 3/6 for the one die to be a 6 or 5 or 4
and there are 3 possible dice
the other two dice have to be 3,2 or 1 and that is 3/6 and 3/6

multiply all this together
3*(3/6)^3 = 3/8
by counting I get 81/216

but maybe this is not what you really wants
a few examples of what you are after would help out big time here
Sally

Hi Sally thank for taking the time to answer.

What I'm actually doing is putting together a spreadsheet for a game called Flames of War, a WWII war game, where we roll dice to see if we hit the opposition troops.

We have a target to hit them (1 to 6), you only need to hit them once and anything better than you needed to hit also hits. And a rate of fire (this is the amount of dice used to roll, this can also be from 1 to 6), so I am looking for an equation that I can enter into my spreadsheet to do this.
So if can abbreviate TTH (target to hit) and ROF (rate of fire) this will make my life easier.


Thanks Dave.

Quote: ICMDWILLIAMS

We have a target to hit them (1 to 6), you only need to hit them once and anything better than you needed to hit also hits. And a rate of fire (this is the amount of dice used to roll, this can also be from 1 to 6), so I am looking for an equation that I can enter into my spreadsheet to do this.

So if can abbreviate TTH (target to hit) and ROF (rate of fire) this will make my life easier.

In other words, for example, if the TTH (I always call them To Hit Numbers, but that's probably from decades of ASL) is 5, you need a 5 or 6 to hit, right?
In that case, the probability of a miss with one die is (TTH - 1) / 6, so the probability of all ROF dice missing is ((TTH - 1) / 6)^ROF, and the chance that at least one of these hits is 1 minus the chance of all of them missing, or:
P(hit) = 1 - ((TTH - 1) / 6)^ROF.

Quote: ThatDonGuy

In other words, for example, if the TTH (I always call them To Hit Numbers, but that's probably from decades of ASL) is 5, you need a 5 or 6 to hit, right?
In that case, the probability of a miss with one die is (TTH - 1) / 6, so the probability of all ROF dice missing is ((TTH - 1) / 6)^ROF, and the chance that at least one of these hits is 1 minus the chance of all of them missing, or:
P(hit) = 1 - ((TTH - 1) / 6)^ROF.

Absolutely spot on that seems to be producing the right sort of values, many thanks!

I know it's a common name, but you might be interested about David Williams, the guy I worked with previously. He was the game designer for a card game called "Legend of the Five Rings."

http://www.l5r.com

These guys created these sorts of games with dice and stuff. We had a couple of these designer guys when I was at Red 5 Studios. I don't get into that stuff myself, but they always had all kinds of multi-faceted dice and were discussing this sort of Dungeons and Dragons sorts of things. Just a coincidence I know, but still.