How to get probability of rolling dice above average?

Question

I've seen someone say that he did $64$ rolls of D$20$ dice with lowest roll $6$ and average $15.5$. How do I calculate probability of this to know how lucky he must be?

I can calculate probability of never rolling lower than 6 by : $$ P(\text{No Bad Rolls}) = P(\text{Good Roll})^{\text{#Rolls}} = (\frac{N_{\text{Possible Rolls}}-N_{\text{Bad Rolls}}}{N_{\text{Possible Rolls}}})^{\text{#Rolls}} = (\frac{20-5}{20})^{64} \approx 1 × 10^{-8}\tag1 $$

I can calculate probability of getting total equal to $64 \cdot 15.5 = 922$ according to this answer as $$ P(\text{Total} = 922) = \frac{N(922,64,20)}{20^{64}} $$ $$N(922,64,20)=\sum_{j\ge0}(-1)^j\binom {64}j\binom{922-1-20 j}{64-1} \approx 3.67 × 10^{74}$$ from a computation in Wolfram Alpha. Thus, $$ P(\text{Total} = 922) \approx 2 × 10^{-9} \tag2 $$ So to get my answer I could just multiply (1) and (2) for a total of $2*10^{-17}$ ? But it seems wrong since these events seem statistically dependant, how should I proceed further?

Answer 1

First, note that a more accurate way of calculating the luck of a given total is by calculating the probability of obtaining that total or higher. For instance, if you flip a fair coin $1000$ times, then you expect to obtain heads $500$ times, but the probability of obtaining exactly $500$ heads is still small. On the other hand, the probability of obtaining $500$ heads or higher is slightly more than $0.5$, which measures luck more accurately.

To remove your issue regarding statistical independence, you can simply calculate the conditional probability that the sum is at least $922$ given that all the rolls are at least $6$. Effectively, you are working with a new kind of dice where the rolls are equally likely to be $6, 7, \ldots, 20$. Now, you can multiply the probability of all rolls being at least $6$ with the conditional probability of obtaining a total of at least $922$.

If you want to use your formula, note that you can equivalently work out the conditional probability where the rolls are $1, 2, \ldots, 15$ and the total is at least $922 - 64(5) = 602$.

Answer 2

Average is 15.5

Say the dice rolls as $ n_1 , n_2 , ...., n_{64} $

Sum of $ n_1 , n_2 , .... $ is 992 since the minimum value any $ n _i $ can have is 6, you can find the number of integral solutions, which would be $ ^{992-6*64+64-1} \mathrm{ C }_{64-1} $ = $ ^{671} \mathrm{ C }_{63} $

The final probability will be $ \frac{^{671} \mathrm{ C }_{63}}{20^{64}} $

Finding number of integral solutions

Now that you know how to find the number of non negative integral solutions, number of solutions where each term is greater than a given number can be determined as follows

For simplicity let us assume the linear equation $$ x + y + z = 8 $$

Say we wanna find number of solutions for the above equation where all terms are greater than or equal to 2

We can write x as X+2 and now X can be ranging from [0,~], similarly for the other 2 numbers, the above equation can be rewritten as $$ X + Y + Z = 2 $$ and then we gotta find the number of possible non negative integral solutions for the new equation.