What are the probabilities of a tie when rolling to see who goes first in a board games for various numbers of dice and various numbers of players? When playing many board games, the first step is to have everyone roll a die to see who goes first, with a roll off in the case of a tie.  While doing that over the Christmas break, my husband suggested that we roll two dice instead of one, with the assertion that this would make ties less likely.  My brother disagreed, claiming it wouldn't make any difference.  I'm interested in investigating this question.
I've been able to calculate the probability of ties for the case of rolling one die for any number of players, and the case for rolling two dice with two players.  However, I haven't actually found a general solution in either case (I mostly used a brute force approach in the one die case).  Is anyone here aware of any sources that have investigated this issue?
(For the record, I'm pretty sure both my husband and brother have forgotten the conversation, so you don't need to worry about hurting anyone's feelings.  :) )
 A: We look only at the simplest case, where there is a single die. But to make things more interesting, let the die be $d$-sided, with equal probabilities for the numbers $1,2,\dots,d$. Suppose there are $m$ players. We find the probability that a tie doesn't occur. 
Assume that "highest number goes first." So there is no tie if for some $k\ge 2$, one players rolls a $k$ and all other players roll numbers $\le k-1$.
Let the number of players be $m$. First we find the probability that Alicia rolls a $k$ and everybody else rolls a number $\le k-1$. 
The probability Alicia rolls a $k$ is $\frac{1}{d}$. The probability everyone else rolls a number $\le k-1$ is $\left(\frac{k-1}{d}\right)^{m-1}$. So the probability of no tie, with Alicia winning, is
$$\sum_{k=2}^d \frac{1}{d} \left(\frac{k-1}{d}\right)^{m-1}.$$
Sum over all players. The probability of no tie is
$$\frac{m}{d^m}\sum_{k=2}^d (k-1)^{m-1}.$$
The remaining sum is a well-known one, with a long history. There are simple formulas for it in the cases $m-1=1,2,3$. For the general case, please see Faulhaber's Formula.
Remark: We can, with some pain, compute the probability of no tie with $3$ players and $2$ dice. We compute the probability that there is no tie and Alicia is the winner, and multiply by $3$. 
Alicia can be the clean winner in any one of $10$ ways. She can throw a $3$ and be the clear winner, or throw a $4$ and be the clear winner, or throw a $5$ and be the clear winner, and so on up to throwing a $12$ and being the clear winner. We start computing these various probabilities.
The probability Alicia throws a $3$ is $\frac{2}{36}$. The probability the other two both throw something $\le 1$ is $\left(\frac{1}{36}\right)^2$. Multiply.
The probability Alicia throws a $4$ is $\frac{3}{36}$. The probability the other two both throw something $\le 3$ is  $\left(\frac{3}{36}\right)^2$. Multiply.
The probability Alicia throws a $5$ is $\frac{4}{36}$. The probability the other two both throw something $\le 4$ is  $\left(\frac{6}{36}\right)^2$. Multiply.
And so on. Add up the $10$ terms we get. One could even extend to $d$-sided dice and $3$ players, and get a closed form formula.  
