1

Suppose I throw a dice $n$ times, and sum the results in a variable $A$. Then I repeat and call the second sum $B$.

What is the probability that $A = B$? I assume the answer depends on the number of possible values ($6$ in this case) so let's call that number $v$.

I realize that this is equal to $$ \sum_{i=1..v} (P(A=i) \cdot P(B=i)) $$ but I do not know how to compute $P(A=i)$ and $P(B=i)$.

For the continues case, the Irvin-Hall distribution gives us an answer (the distribution of a sum of random variables), but what do we do in the discrete case ?

Vinyl_cape_jawa
  • 4,794
Ofer strichman
  • 11

1 Answers1

1

This is probably difficult to work out exactly, but it can be done approximately (for sufficient large $n$).

Let $k$ be the number of faces on your die. The variance of $A$ and of $B$ is $n \frac{k^2-1}{12}$, since each is a sum of $n$ independent versions of a discrete uniform random variable. Let $D = A - B$, then $$ E(D) = 0, \, var(D) = n \frac{k^2-1}{6} \, . $$ By the Central Limit Theorem, for sufficiently large $n$, $Z = \sqrt{\frac{6}{n(k^2-1)}} D$ is approximately standard normal. Then $$ P(D = 0) = P(|D| \le \frac{1}{2}) = P(|Z| \le \frac{1}{2} \sqrt{\frac{6}{n(k^2-1)}} \approx \frac{1}{\sqrt{2 \pi}} \cdot \sqrt{\frac{6}{n(k^2-1)}} = \boxed{\, \sqrt{\frac{3}{\pi n(k^2 - 1)}} \, } \, $$ This goes to zero quite slowly, which seems to be related to the birthday paradox.

Hans Engler
  • 15,439