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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. :) )

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If you want to roll two dice, and you can't distinguish the order of the two dice, you'll do better to add the squares of the two dice results and comparing. Then two players will get the same value $5.1\%$ of the time. – Thomas Andrews Jan 4 '13 at 18:19
If the goal is to limit the number of comparisons, your husband is right. Rolling one die, two people's results will agree one sixth of the time, or $\approx 16.67\%$. Rolling two dice and adding, two people's results will match about $11.27\%$ of the time. – Thomas Andrews Jan 4 '13 at 18:25
With two die, I get the probability of a tie is 146/1296, which is less than the 1/6 with one die. It seems you have calculated this already. What do you mean by a general solution? – Masked Man Jan 4 '13 at 18:25
I am not only interested in the case of two player, but in the case of any number of players. I want a general solution for any number of players. – Heather Kelly Jan 4 '13 at 19:00

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.

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Thank you. That's the general solution I was looking for in the case of a single die. – Heather Kelly Jan 4 '13 at 19:01
Now, can anyone help me with the two-die case? – Heather Kelly Jan 4 '13 at 19:01
Two player, two $d$-sided dice is easy, we are just summing squares. There is no problem grinding through $3$ players, $2$ standard dice. – André Nicolas Jan 4 '13 at 19:12
Is this like when the math book says, "It is obvious that..."? I got two players with two dice, but got hopelessly stuck on 3 players, 2 dice; and you might as well forget about 4 or 5 players. Any hints? – Heather Kelly Jan 4 '13 at 21:56
Maybe if I have some time I will add the $3$ player $2$ dice. would that be useful to you? About "it is obvious that." Irritating habit, admittedly, but most mathematicians use it correctly, to mean routine argument. – André Nicolas Jan 4 '13 at 22:25

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