I’m stuck on what actually counts as the “sample space” in simple probability problems. I keep flip-flopping between listing every tiny outcome and grouping them into bigger buckets, and then my probabilities wobble. My brain wants tidy buckets; math seems to want microscopic detail. I’m trying to reconcile these two vibes.
Example 1: rolling two fair dice and looking at the sum. If I write the sample space as all ordered pairs, S = {(1,1), (1,2), …, (6,6)}, then I’m fine: there are 36 equally likely outcomes, and the event “sum = 7” is E = {(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)}. That gives 6/36, which I know is right.
But sometimes I get lazy and I write the sample space as just the possible sums S’ = {2,3,4,5,6,7,8,9,10,11,12}. Then my gremlin brain tries to do P(sum = 7) = 1/11 because 7 is one of 11 outcomes, which I know is wrong. So… is S’ a valid sample space at all? If so, do I have to attach different probabilities to each sum and stop doing naïve counting? Or is there some rule of thumb that says I should prefer a sample space where all outcomes are equally likely whenever I plan to count? That “equally likely” part is exactly where I keep slipping.
Example 2: drawing two marbles from a bag without replacement. Say the bag has 3 red (R) and 2 blue (B). If I go detailed: S = {RR, RB, BR, BB} (I’m treating order as different because of the without-replacement part), and then the event “exactly one red” is {RB, BR}. That feels okay. But if I decide I don’t care about order and switch to the coarser S’ = {2R, 1R1B, 0R}, my instinct was to say P(1R1B) = 1/3 because it’s one of three outcomes, and I can hear the math gods facepalming. I think the trouble is that the three elements in S’ aren’t equally likely. Am I thinking about this the right way? Is it “legal” to use the coarser sample space as long as I remember to weight the outcomes properly? Or should I always build the fine-grained sample space first and then define events by grouping outcomes?
I think my main confusion is: what makes something an “elementary outcome”? Is it “what physically happens” (ordered stuff), or “what the question observes/records” (like just the sum or the counts)? If a problem only cares about the multiset of colors or the sum of dice, should I change the sample space to those objects, or keep the detailed one and define an event that lumps outcomes together? Follow-up question: is there a standard or preferred way to set this up to avoid mistakes, especially when outcomes aren’t equally likely?
If someone could explain how to choose a sample space on purpose (instead of me guessing and then tripping over the counting), and how that choice affects whether I can just count vs. needing weights, I’d be super grateful. And if my partial attempts above are on the right track, could you point out where I’m nearly there vs. where I’m off?
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3 Responses
Love this question-it’s the “high‑resolution photo vs thumbnail” dilemma of probability! A sample space is any exhaustive, mutually exclusive list of outcomes you choose to distinguish, and there isn’t a single “correct” one; what matters is that the probabilities on it are correct. Naïve counting (probability = 1/number of outcomes) only works when those outcomes are equally likely, so if you want to count, pick a fine, symmetric, equally likely model. For two dice, the ordered pairs {(1,1), …, (6,6)} are uniform, so P(sum = 7) = 6/36. The coarser space {2,…,12} is also valid, but you must carry non‑uniform weights: {1,2,3,4,5,6,5,4,3,2,1}/36, so 7 still has 6/36, not 1/11. For marbles, the safest “counting” setup is to label the marbles and use ordered draws without replacement: there are 5×4 = 20 equally likely ordered pairs; that gives RR = 6/20, RB = 6/20, BR = 6/20, BB = 2/20. If you only care about counts, the coarse space {2R, 1R1B, 0R} is fine too, but now the probabilities are 3/10, 3/5, 1/10-definitely not 1/3 each. Conceptually, the detailed model is your high‑res world; “sum of dice” or “number of reds” is a function of that world (a random variable), and when you compress to that thumbnail, you inherit whatever non‑uniform distribution that function induces. Practical rule of thumb: choose a uniform micro‑model if you want to count; choose the coarser “what we actually record” model if you prefer, but then attach the correct weights. Follow‑up: when you tackle these, do you like starting with a uniform micro‑model and pushing probabilities forward, or working directly on the coarse outcomes-say, how would you set it up for drawing 3 from a bag with 4 red and 3 blue to find “exactly two red”?
Ohhh, I love this question because it’s really about modeling choices! My take: both “microscopic” and “bucketed” sample spaces are valid, but you have to pair the space with the right probabilities. For the two dice, S = {(i,j): i,j in {1,…,6}} is the classic because it’s uniform and counting works cleanly. But the coarser space S′ = {2,…,12} is totally legal if you assign the correct (non-uniform) weights: P(2)=1/36, P(3)=2/36, …, P(7)=6/36, …, P(12)=1/36. Conceptually, you’ve defined a random variable f(i,j)=i+j and “pushed” the uniform measure on ordered pairs onto the set of sums. So 7 isn’t 1/11; it’s 6/36. Rule of thumb (not a law): if you plan to “just count,” pick a space with equally likely outcomes; if you pick a coarser space, remember you’ve squeezed unequal numbers of fine outcomes together, so you must weight them. Khan Academy has a nice gentle intro: https://www.khanacademy.org/math/statistics-probability/probability-library/probability-sample-spaces. I’m 95% sure this framing will keep you out of trouble.
For the marbles, same vibe. If you use ordered draws S = {RR, RB, BR, BB}, the probabilities are RR=3/10, RB=3/10, BR=3/10, BB=1/10, so P(exactly one red)=3/10+3/10=3/5. If you use the coarser S′ = {2R, 1R1B, 0R}, it’s still fine-but the weights are 3/10, 6/10, 1/10 (you can get those either by grouping the ordered outcomes or by the hypergeometric formula: C(3,1)C(2,1)/C(5,2)=6/10 for 1R1B). So yeah, the three elements in S′ aren’t equally likely. Personally, I think of “elementary outcomes” as whatever the model decides to distinguish before you apply the “what do we care about?” function; then you can map to sums/counts afterward. I might be over-formalizing, but if you remember “choose a space where probabilities are easy; if you coarsen, carry the weights along,” you’ll be golden.
I really like the way you framed this. You’ve put your finger on a subtle but important idea: “sample space” isn’t uniquely determined by the physical experiment. It’s a modeling choice. The key is to choose a sample space and probabilities that reflect the mechanism, and then be consistent about how you compute.
Here’s the short headline:
– Counting outcomes works only when those outcomes are equally likely.
– You can absolutely use a coarser sample space (like “just the sum” or “just the counts”)-but then you must carry non-uniform probabilities on that space.
– A safe default is: start with a fine, clearly uniform sample space; then define the thing you care about as a function of that space. If you like, you can then “push” the probabilities to your coarser space and work there.
Let me go step by step and connect this to your two examples.
1) Two dice and the sum
Two legitimate choices:
A. Fine-grained, uniform space
– Sample space: Ω = {(i, j) : i, j ∈ {1,…,6}}. Each outcome has probability 1/36.
– Define the random variable S(i, j) = i + j (the sum).
– Then P(S = 7) = #(pairs that sum to 7)/36 = 6/36.
B. Coarse, non-uniform space
– Sample space: Ω′ = {2, 3, …, 12}, i.e., “the sum.”
– But then you must assign probabilities: P(2)=1/36, P(3)=2/36, …, P(7)=6/36, …, P(12)=1/36. These are not equal.
– Now P(S = 7) = 6/36 again, but you’re not allowed to do 1/11.
What went wrong in the 1/11 temptation? You silently replaced the real probabilities with equal weights on Ω′. The “counting/over total” trick only works when every element of your sample space is equally likely. In Ω′ they are not.
2) Two marbles without replacement (3 red, 2 blue)
Again, multiple valid choices. Let’s show two that are both uniform (so counting is safe), and then your coarser color-category space (not uniform).
A. Ordered, identity-level outcomes (uniform)
– Think of the actual physical marbles as distinct objects. There are 5 choices for the first draw and 4 for the second: 20 equally likely ordered outcomes.
– Color counts by order:
– RR: 3×2 = 6
– RB: 3×2 = 6
– BR: 2×3 = 6
– BB: 2×1 = 2
– Probabilities:
– P(RR) = 6/20 = 3/10
– P(RB) = 6/20 = 3/10
– P(BR) = 6/20 = 3/10
– P(BB) = 2/20 = 1/10
– So “exactly one red” = RB ∪ BR has probability (6+6)/20 = 12/20 = 3/5.
B. Unordered, identity-level outcomes (also uniform)
– If you only care which two physical marbles are drawn, not the order, there are C(5,2) = 10 equally likely unordered pairs.
– Break them by color composition:
– 2R: C(3,2) = 3 pairs
– 1R1B: 3×2 = 6 pairs
– 0R: C(2,2) = 1 pair
– Probabilities:
– P(2R) = 3/10
– P(1R1B) = 6/10 = 3/5
– P(0R) = 1/10
C. Coarse color-category space (not uniform)
– Sample space: {2R, 1R1B, 0R}.
– These are not equally likely (as the counts above show: 3 vs 6 vs 1), so 1/3 is wrong. The right probabilities are 3/10, 3/5, 1/10.
– This space is fine to use, but you must weight it correctly.
Notice something nice: there are at least two “fine enough yet uniform” ways to model the marbles-ordered sequences (20 equiprobable) or unordered distinct pairs (10 equiprobable). Both are safe for counting. The color-category space is coarser still, but once you collapse identity information, the categories no longer have equal weight.
What is an “elementary outcome,” really?
– It’s whatever atomic description you choose so that you can attach probabilities. “Atomic” here is about your model, not a law of nature.
– The thing you actually observe (sum of dice, number of red marbles) is a function of the elementary outcome. In probability, that function is called a random variable.
– If you replace the original sample space by the space of values of that random variable (e.g., sums 2–12), that new space is perfectly valid-but the probabilities on it are whatever the function induces, and they need not be equal.
A small analogy
Think of the fine-grained sample space as a high-resolution photo (every pixel). The “sum of the dice” or “number of reds” is like a blurred thumbnail of that photo. Many different pixel-level photos map to the same thumbnail. If you only count thumbnails equally, you forget that some thumbnails come from many more high-res photos than others.
A simple rule-of-thumb checklist
– If you want to count, make sure what you are counting are equally likely micro-outcomes.
– If you prefer a coarser description (like sums or color-counts), first compute or look up its probability distribution. Then you can work entirely on that coarser space, but don’t assume it’s uniform.
– If there’s any chance you’ll need more detailed information later (e.g., “given the sum is 7, what’s the chance the first die is 1?”), keep the fine-grained model in the background. Coarsening early can throw away information you might need.
– In “draw without replacement” settings, a handy uniform model is “unordered distinct items”: choose k of N objects uniformly from C(N,k). Then translate to your color categories by counting how many unordered pairs land in each category. That’s the hypergeometric distribution:
P(exactly k reds) = C(R, k) C(B, n − k) / C(R + B, n).
Where to read more
– Khan Academy’s sample spaces for compound events gives a gentle intro with dice and marbles: https://www.khanacademy.org/math/statistics-probability/probability-library/probability-sample-spaces/v/sample-spaces-for-compound-events
– A concise explanation of random variables as functions of outcomes (why “sum” or “number of reds” is a function of what really happened): https://www.statlect.com/fundamentals-of-probability/random-variables
Bottom line
– Your S′ spaces are completely “legal.” The only trap is silently treating them as uniform when they aren’t.
– If you want to be error-resistant, start with a fine, uniform model; define the statistic you care about; push probabilities forward; and only then, if you like, switch to working on the coarser space with its (possibly non-uniform) weights.