My Math Forum Probability with dice and target numbers

 June 13th, 2012, 09:06 PM #1 Newbie   Joined: Jun 2012 Posts: 1 Thanks: 0 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!
 June 14th, 2012, 12:24 AM #2 Senior Member   Joined: Feb 2009 From: Adelaide, Australia Posts: 1,519 Thanks: 3 Re: Probability with dice and target numbers You are correct. The chance of rolling at least one six is 1-(5/6)³. Now, you want to know the chance of rolling n dice and getting exactly k sixes. Suppose you roll 8 dice, and you are interested in getting 3 sixes. To get the sixes on (for example) the 3rd, 6th, and 8th rolls, the probability is $\frac56\cdot\frac56\cdot\frac16\cdot\frac56\cdot\f rac56\cdot\frac16\cdot\frac56\cdot\frac16=\frac{5^ 5}{6^8}$ and this probability is the same for any choice of three dice out of the eight. But there are ${8\choose 3}=\frac{8!}{5!3!}$ ways to choose those three dice. These are independent roll results, not overlapping, which means we can add up the probabilities. So the probability of exactly 3 sixes out of 8 dice, for a score of 6+1+1 = 8, is ${8\choose 3}\frac{5^5}{6^8}$, about 10.42 %. In general, for n dice and exactly k sixes, the probability is ${n\choose k}\frac{5^{n-k}}{6^n}$.

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