Nick Berry asks: you roll a pair of six-sided dice and sum that to get a total. Your good friend does the same. What are the chances that you both get the same total?
My instinct is to say that it should be somewhere between 1/6 and 1/11 – you’d get 1/6 if the question were about one die and 1/11 if it were about picking uniformly a number from 2, 3, …, 12.
Let’s generalize. What if you roll two n-sided dice? In that case, the probabilities of getting 2, 3, …, n+1 are , and the probabilities of getting n+2, n+3, …, 2n are – generalizing the well-known triangular distribution. The probability that you and your friend both roll 2, then, is
or, putting everything over a common denominator,
At this point we can already see that there are order-of-n terms in the numerator, of order , so we should get a result of order 1/n. We can peservere and do the algebra and we get
or just a hair over 2/3n.
Berry also asks what happens if you have more dice. For k six-sided dice he gives the results of a Monte Carlo simulation which has roughly . He writes that “the percentage chance of a match score falls off a lot slower than I would have predicted.”
Why does it fall off so slowly? The mean result from a single die is 7/2, with variance 35/12. So the mean result from k dice is 7k/2, with variance 35k/12 – so only values within a few multiples of are possible with any appreciable frequency. There are on the order of a few square roots of k of these, so the answer should be for some . Like Berry, I am too lazy to find the constant .
Edited to add: Matthew Aldridge gives in the comments an argument (for approximating that for six-sided dice, with $k$ large, the probability of both getting the same total is approximately , by approximating both sums as normal and applying the central limit theorem. This is about . If we bear in mind that the variance of a single roll of an n-sided die is , then we get for rolling k n-sided dice.
This approximation is surprisingly good even for k = 1, where it has no business being good. It gives . This reduces to 1/n, the correct answer, if you let and ignore the -1.