# What is the chance of rolling 5 numbers higher than 5 (6+) on ten, ten sided dice (10D10)

I've been trying to understand this but my Maths isn't strong enough. Someone else asked a similar question regarding six-sided dice. I understand this is a binomial distribution and can fill in the formula but can't complete it.

I am trying to figure out the probabilities of success for a roleplaying game, namely the new Vampire the Masquerade 5th edition. To succeed you roll ten-sided dice and need a number of successes (six or more on a die). To attack someone you roll your dice and try to get a number of successes equal or greater than half of theirs.

So I have $10$ dice as do they, we are equal. I need to roll $6$ or more on five of my ten dice.

Thank you so much for any help.

• One way to answer the question is to add the probabilities that exactly six, exactly seven, exactly eight, exactly nine, or exactly ten of the dice display a number that is six or greater. – N. F. Taussig Jun 11 '18 at 20:20
• is it exactly five or five at least? – narek Bojikian Jun 11 '18 at 20:24
• When you say "get a number of successes equal or greater than half of theirs", do you mean they also roll 10D10 and count their successes? – awkward Jun 11 '18 at 20:31
• Sorry if I wasn't that clear. To get a you need a six or more on a ten sided dice, higher than five or 6+. – James Firminger Jun 11 '18 at 20:50
• The rules say you can half their total dice pool and just consider that the number of successes you need to roll. So to beat a person with 10 dice I would need to to roll 6 or more on five dice. – James Firminger Jun 11 '18 at 20:51

Since you didn't mention if it's exactly or at least 5 times, I will assume in my answer that it's exactly 5 dice getting a number higher than 5 and at the end I will discuss how would the "at least" change the answer:

So reformulating the question, we have ten dice, we want for five of them to get a number between 6 and ten (which are half of all possibilities for each) and for the other five to get a number between 1 and 5 (which is also half of the possibilities).

So we only have to choose the five elements which will have our values (regardless of the order) which means we will choose 5 elements out of 10 and for each we have the probability 1/2 to get the right value, and for the other five each will have the also 1/2 probability to get the it's right value.

Hence the answer is $${10\choose5} \cdot (1/2)^5 \cdot (1/2)^5 = \frac{10!}{5!\cdot5!}\cdot \left(\frac{1}{2}\right)^{10} = \frac{252}{1024} = \frac{63}{256}$$

So back to the case of "at least", all we need to do is to sum over all the ways of getting 5, 6, 7, 8, 9 and 10 dice with a number greater than 5. It'd make it easier, if we noticed that getting 6, 7, 8, 9 and 10 dice with a number greater than 5 is a exactly the half of all the possibilities which mean it's probability is $\frac{1}{2}$, and hence the total answer would be $$\frac{1}{2} + \frac{63}{256} = \frac{63 + 128}{256} = \frac{191}{256}$$

• Thank you so much. Yes it was at least five, I should have specified. I'm trying to create a chart of the probabilities of succeeding with different numbers of dice and differemt target numbers. Eg needing to get three (6+)s with five dice or two (6+) with four. I don't understand the section with the exclamation marks, could you perhaps explain to me what you are doing there. When I tried to work this through before I posted here, I got to 1/1024. – James Firminger Jun 11 '18 at 21:11
• Okay sorry I though you knew what that is, n! is called n faculty, it's the product of numbers from 1 to n, and it is the number permutations of the numbers from 1 to n, ie. the number of ways to represent the numbers between 1 and n in different orders. for example: 3! = 1*2*3 = 6 and we have 6 permutation of 6 numbers which are 1,2,3 1,3,2 2,1,3 2,3,1 3,1,2 3,2,1 – narek Bojikian Jun 11 '18 at 21:32
• however, choosing k elements out of n and the order matters would be n*(n-1).. *(n-k+1) = n!/(n-k)! since we have n methods to choose the first number, n-1 methods to choose the second, .. n-k+1 to choose the kth – narek Bojikian Jun 11 '18 at 21:34
• the last part would be the ${n}\choose{k}$ which is the number of ways to choose k elements out of n and the order doesn't matter the value of which is n/(k! * (n-k)!) you can see it in the way that, we are choosing k elements out of n which is n!/(n-k)! but in this case we would habe considered different orderings of the same set as different choices and we have k! of each (permuting over k elements) that's why we divide over k! in addition. – narek Bojikian Jun 11 '18 at 21:36
• In your place however, I would consider reading a basics book about combinatorics before starting doing problems (a high-school book should do in this case you only need the definitions and the intuitions behind them at first). – narek Bojikian Jun 11 '18 at 21:37

Since the actual value of the roll doesn't matter as long as we know whether it's value is greater than 5, we can simplify this problem by denoting E as the event that the roll is greater than 5. This has probability 1/2 assuming a fair dice is being used.

Now, letting z be the number of times E is true, we see that the pgf for one roll of the die is F(z) = ${1\over 2} + {z\over 2}$. Therefore, the pgf after 10 rolls of the die is ${F(z)}^{10}$ = $({1\over 2} + {z\over 2})^{10}$ = $\sum_{k=0}^{10}{10 \choose k}{1\over 2}^{10}z^k$.

In your comment, you mentioned that to win, you need to roll a 5 or more on 5 or dice, therefore, this probability is simply the tail of F(z), $\sum_{k=5}^{10}{10 \choose k}{1\over 2}^{10}z^k$ = $\frac{2^{10}+{10\choose 5}}{2^{11}}$ $\approx{.62}$