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

Question

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.

Answer 1

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 $$\frac{1}{2} + \frac{63}{256} = \frac{63 + 128}{256} = \frac{191}{256}$$

Answer 2

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), $\sum_{k=5}^{10}{10 \choose k}{1\over 2}^{10}z^k$ = $\frac{2^{10}+{10\choose 5}}{2^{11}}$ $\approx{.62}$