Rolling for even or odd is fair, no matter how many dice

"Odd is favored because 7 is the most likely outcome."

"Even is favored because you can't roll a 1."

"You're trying to trick me aren't you?"

These are all things that have been said to me at the beginning of Magic: the Gathering games when I offer to roll for who goes first with two dice. For the uninitiated, in order to determine who goes first in turn based tabletop games, a pair of dice are usually rolled. The most common way, historically, has been to say whoever rolls the highest number goes first. But due to the bad feelings that players can get when they roll high but lose anyway, or the rolls come up the same and forces a re-roll, there has been a growing faction of people who prefer to have one person roll the dice, and have the other player call even or odd. If the player who called is correct, they get to play first.

Now everyone seems to agree that rolling one die is fair, but the minute a second die is added, people start to get suspicious. I've heard people who think even is more likely, or that odd is more likely, and they all have their reasons. The only thing they have in common, unfortunately, is that their fears are all unfounded. Rolling two dice is perfectly fair, and indeed rolling any number of dice is fair. And I plan to prove it.

The two dice case is especially easy to prove. We can simply count up the combinations. Let's draw a chart with the possible rolls of one die along one axis and the possible rolls along the other axis. In each place of the chart we will write the corresponding outcome of the dice roll.

1 2 3 4 5 6
1 2 3 4 5 6 7
2 3 4 5 6 7 8
3 4 5 6 7 8 9
4 5 6 7 8 9 10
5 6 7 8 9 10 11
6 7 8 9 10 11 12

When it's laid out like this it's fairly obvious that even and odd outcomes have the same probability. You can just count the number of squares with even vs odd outcomes to figure it out (there's 18 of each, I promise).

But there's an even simpler way to prove it for two dice. Think about the situations that lead to the total of two dice being even vs odd. The only way to roll an even total is for both dice to land on an odd number, or both dice to land on an even number. The actual numbers don't matter, just that their parity is the same. However, as I said earlier, everyone agrees that for a single die, landing on even vs odd is 50/50. So to roll two even numbers is two 50/50 events happening at the same time. The chance of that is exactly 0.5×0.5=0.25. For the exact same reason, the chance of rolling double odds is also 0.25. Therefore the odds of rolling an even total has to be 0.25+0.25=0.5. Cool! Much easier than counting all those squares, right?

But now what do we do when there's three dice being rolled? The number of combinations starts to get out of hand pretty fast as the number of dice grow. Don't worry. There's a simpler way to handle all that.

Instead of thinking of them as three dice, let's think of them as one pair of dice and one lone die. Then rolling all three is the same as rolling the two groups separately and adding the two results up. Now, the only way we can get an even result from the three dice is if we get an odd result from the pair at the same time as an odd result from the singleton, or else an even result from both the pair and the singleton.

We know the result of rolling an odd number from the singleton is 0.5. But wait a second we also just showed that rolling a pair of dice for even vs odd was also fair! That means that the result of rolling an odd number from the pair is also 0.5! For the exact same reasons, the odds of rolling an even number from the singleton and the pair are also 0.5, respectively. This means that again the probability of rolling an even result from three dice is 0.5×0.5+0.5×0.5=0.25+0.25=0.5. Do you see the pattern forming?

We can do this again and again as many times as we want, and we will always get the same outcome. What we have accidentally discovered here is something called the principle of mathematical induction. This is an extremely powerful tool we use in math to prove things about an infinite number of situations in one go.

Think of it like dominoes. We set them up in such a way that if one falls down, they all do. This is a lot like what we did when we were discovering the pattern of adding one die at a time to show that the end result was still a fair roll. We take some number, n, of dice, and we add one. We then say "Well, if n dice are fair, then n+1 dice must also be fair!" because of the arguments we outlined above. But then since n+1 dice is fair, it must also be the case that n+2 dice is fair, and n+3 dice, and on and on.

All it takes now is for someone to knock over the first domino. After all, even if they're all lined up like that, if the first one in the row doesn't fall, then we haven't really achieved anything. This is called the base case of the induction. We need to find some n where we all agree that n dice is fair, no question, independent of the other numbers. But we already have such a number! We all agreed from the jump that a single die was perfectly fair, no ifs, ands, or buts. This knocks down the first domino, which means that the second will fall over, so 2 dice must be fair, and the next, so that 3 dice must be fair, and so on ad infinitum. The end result of this, of course, is that you can roll any number of dice you want, and the roll will be perfectly, 100%, fair.

Hopefully someone googling for a way to explain to their friends at the game table that this method is totally fair found this useful, education, and entertaining. And if your friends still don't believe you, just tell them that they can pick the side they believe is favoured!