Bonus post – another way to work out the odds of rolling a 1 with two dice
I was contacted last week by a couple of people about my claim that the odds of rolling a 1 with two dice is 11/36, not 1/3. This post contains two different proofs that it’s 11/36.
An elegant proof
Remember that we were trying to work out the odds of me hitting a counter of Zoe’s in a game of backgammon. The counter was one space away from one of mine. So I have to roll a 1 with two dice. More precisely, I have to roll AT LEAST one 1. That is, I have to roll a 1 with the first die, or a 1 with the second die, or a 1 with both dice.
Rolling AT LEAST one 1 is the complement (“opposite” if you like) of me rolling NO 1 at all. So what is the chance of me NOT rolling a 1 with two dice?
Well, let’s look at the possible combinations:
- The first die would need to roll a 2, 3, 4, 5, or 6. (5 ways)
- The second die would need to roll a 2, 3, 4, 5, or 6. (5 ways)
- That’s 5 x 5 = 25 possible ways to NOT roll a 1.
Now, there are 36 possible pairs the two dice can make (6×6). So the probability of NOT rolling a 1 is 25/36.
The opposite (the “complement”) of “NOT rolling a 1” is “rolling AT LEAST one 1”. Therefore, the probability of rolling at least one 1 is 1-25/36 = 36/36 – 25/36 = 11/36.
(I know it feels counter intuitive. It feels like it should be 1/6 + 1/6. But it isn’t because the dice are effectively “ordered pairs”. It’s as if there’s a little bit of each of the 1/6 chances that “overlaps” and we have to be careful not to double count that little overlapping bit.)
A brute force proof
Of course the other way to prove that it’s 11/36 is to take a “brute force approach” and list all the possible pairs you can roll with two dice and count the pairs that have at least one 1 in them. I’ve done this so you don’t have to!
As you can see there are eleven possible combinations of the thirty six that contain at least one 1.
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