# Thread: Probability with dice and target numbers

1. ## Probability with dice and target numbers

Hi, this is my first time on the forums and I am having a very hard time with some math questions. I have tried searching the web for my particular question but most sites are concerned with the sums of 2 six-sides dice not target numbers. What I mean by target numbers is I am trying to figure out what the probability of rolling "at least" one 1, 2, 3, etc. on multiple dice. The reason I am doing this is because I am trying to build an excel sheet to help me make better decisions with a table top game I play called "Heavy Gear Blitz". I very fun game, everyone should check it out.

One of the facets of the game is when you roll a double/triple/etc. six instead of getting a six, you get six plus how many other sixes you have rolled. For example, on four dice I rolled a 6, 6, 6, 4. I would then have the final score of 8 because of my first six and then +1 for every six after it. This really puts the math a bit beyond me. I have gotten a bit of the target number math done, but I am really not sure if I have it right or not.

This is what I do:

3 six-sides dice being rolled
chances of not rolling at least a 6 on the first die = (5/6)
chances of not rolling at least a 6 on the second die = (5/6)
chances of not rolling at least a 6 on the third die = (5/6)
I then multiple all the results together = .5787 or 57.87%
I want to find out the result of rolling a 6 so I then subtract the number I found from 1 = .4213 or 42.13%

I hope this is all right. I am not really sure how to do the math for what I was talking about early with "at least" rolling a "7". Thank you for your time and any help you can give!

2. ## Re: Probability with dice and target numbers

Your 0.4213 is the probability of getting at least one 6 when rolling 3 dice

3. ## Re: Probability with dice and target numbers

Close, but not quite. Let x be the number of 6's rolled. You roll a "7" if and only if you roll at least two 6's, or $x \ge 2$. This is equal to 1 - P(x = 0) - P(x = 1).

P(x = 0) you did correctly, $\displaystyle P(x = 0) = (\frac{5}{6})^3 \approx .579$

P(x = 1) is given by $\displaystyle P(x = 1) = {{3}\choose{1}} (\frac{1}{6})(\frac{5}{6})^2 \approx .347$

Therefore, $P(x \ge 2) = 1 - .579 - .347 = .074$ (exact value 2/27).

4. ## Re: Probability with dice and target numbers

@biffboy: I know I was showing my process for finding at least 1 six in three dice to make sure I had that right before I went on to the hard stuff (2 sixes in three dice).

@richard1234: ok, cool. Thanks. I am a bit confused though. Why don't you use the process you used in x=0 or x=1 for x>=2 and what if I wanted to find the percentage on 2 sixes in 4, 5 dice and so on? Also what if I wanted to find a "8", "9", and "10" in multiple dice in the way I got a "7"? Sorry for so many question, I just want a better understanding. Thanks again for you time!