I’m trying to write the sample space for rolling two fair six-sided dice. I see two options:
1) S1 = all ordered pairs (1,1) through (6,6), which are equally likely.
2) S2 = the possible sums {2,3,…,12}.
If I use S2 and treat those 11 outcomes as equally likely, I get P(sum = 8) = 1/11. But when I use S1 and then look at the sum, I get a different probability, so I think I’m mixing things up.
Is it valid to take S2 as the sample space but assign different probabilities to each sum? Or is the “right” sample space here S1, with the sum as a function of it? When a question asks for “the sample space and the probability the sum is 8,” which representation should I present so it’s technically correct?
Any help appreciated!
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3 Responses
Both views are useful, they just answer slightly different questions. The “fine” sample space S1 lists all 36 ordered pairs, each with probability 1/36. The sum is then a function f(i, j) = i + j from S1 to {2,…,12}. So P(sum = 8) = |{(2,6),(3,5),(4,4),(5,3),(6,2)}| / 36 = 5/36. This approach is standard because it starts from equally likely outcomes and derives everything else cleanly.
You can also take S2 = {2,…,12} as your sample space, but then you must assign a non-uniform probability mass function p(2)=1/36, p(3)=2/36, …, p(7)=6/36, …, p(12)=1/36. For questions only about the sum, this is equivalent to working in S1. The trade-off is that S2 forgets information about order, so you cannot answer questions like “P(first die > second die)” from S2 alone. Strictly speaking, the right sample space here is S1; S2 by itself isn’t a valid sample space unless you also state those non-uniform probabilities.
Simple example: Using S1, P(sum = 8) = 5/36 as above. Using S2 with the pmf, you read off p(8) = 5/36 directly. For a clear walkthrough of this idea, see the dice probability discussion here: https://en.wikipedia.org/wiki/Dice#Probability
Both are valid: S1 is the natural “equally-likely universe,” with the sum as a function on it giving P(sum=8)=5/36 (five pairs land on 8), while S2 works only if you keep non-uniform weights 1,2,3,4,5,6,5,4,3,2,1 out of 36. Which picture feels friendlier to you-the 6×6 checkerboard of pairs or the little dice-mountain of sums?
Great question-I remember tripping over the same thing! The “microscope” view is S1: all 36 ordered pairs are equally likely, and the sum is a function of that space; from there P(sum = 8) = 5/36 because the pairs (2,6), (3,5), (4,4), (5,3), (6,2) work. The “binning” view is S2: the sums 2 through 12; that can also be a valid sample space, but you have to attach the right probabilities to each sum based on how many pairs produce it (1,2,3,4,5,6,5,4,3,2,1 out of 36). I’ll admit I still get a bit unsure because I was taught that a sample space should list equally likely outcomes, which kind of pushes me to say S1 is the “proper” one-but I think that’s not strictly necessary, and S2 with non-uniform weights is fine. Some folks also argue that if the dice are indistinguishable then the sums are sort of the “fundamental” outcomes and could be treated as equally likely by symmetry, but that runs into the obvious snag that 7 shows up more than 2 or 12. So if a problem says “give the sample space and the probability the sum is 8,” I’d usually present S1 as the sample space and define the sum as a function of it, then report 5/36. If your instructor is happy with the coarser view, you could present S2 but make sure to include the differing probabilities next to each sum; taking them all as 1/11 would be modeling a different experiment (like an imaginary 11-sided die). Hope that helps-I might be overthinking it a tad, but that’s how I keep the story straight.