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This is related to my previous question here.

The numbers $0, 1, 2, 3, \ldots , 8$ are written on individual cards and placed in a bag. Three cards are chosen at random. What is the probability that their sum is $8$?

The more I read this question, the more I am thinking that the question is ambiguous.

My reasoning:

In general for a probability scenario, the ordering does matter because the more ways an event can occur (more orderings), the more likely it is to occur. But in the question posed above, it is not clear if the cards are chosen one-by-one, or if they are chosen all in one go, or if that even matters. If ordering is taken into account, I get an answer of $1/4$ but if ordering is not taken into account, I get an answer of $1/24$.

My question:

Is the question ambiguous? What would the answer be, in this case?

Trogdor
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    The question isn't ambiguous at all. In principle you could just list every subset of order $3$ and count those that sum to $8$. – lulu Sep 18 '18 at 00:38
  • Also: I don't understand where either of your answers comes from. Unless I am wrong (always possible) the only winning, unordered triples are ${1,2,5}$ and ${1,3,4}$. As there are $\binom 83=56$ unordered triples in general, the answer should be $\frac 2{56}=\frac 1{28}$. – lulu Sep 18 '18 at 00:42
  • My mistake @lulu, I had omitted the $0$ from the set. – Trogdor Sep 18 '18 at 00:43
  • Please edit your post accordingly. Obviously that changes the answer, though the methodology would be the same. – lulu Sep 18 '18 at 00:44
  • If you are searching for ambiguity, you could worry as to whether the selection was made with or without replacement. I assumed from the phrasing that the selection was made without replacement....in ordinary speech if one asks you to choose three objects from a list, that means three different objects. But, if you prefer, you could work the problem both ways. – lulu Sep 18 '18 at 00:47
  • You should really add your work to the post, showing how you got to the two different answers – Bram28 Sep 18 '18 at 00:52

1 Answers1

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The problem is unambiguous. It does not matter whether your three numbers are chosen "all in one go" or "one-by-one," at the end of the day you have three random numbers either way. You are hung up on whether or not order matters. The thing is this: you can choose whether or not order matters when you do your computation, and as long as you are consistent, you will get the same answer.

Essentially, if you do your computations while caring about the order the numbers are drawn, then you are keeping track of extra information. There will be more possibilities in the same space, but also proportionally more elements in the event $\{\text{sum of numbers }=8\}$, so the result is the same.

  • If order does not matter, then the sample space consists of all $\binom{9}3=84$ unordered subsets of $\{0,1,2,\dots,8\}$. The event that the sum is $8$ consists of just $5$ sets, namely $$ \{0,1,7\},\{0,2,6\},\dots,\{1,3,4\} $$ Therefore, the probability is $\frac{5}{84}$.

  • If order does matter, then the sample space consists of all $9\cdot8\cdot 7=504$ ordered lists of three distinct elements of $\{0,1,\dots,8\}$. The even that the sum equals $8$ consists of the $5\cdot 3!$ possible permutations of one of the $5$ sets listed before, like $$ (0,1,7),(0,7,1),\dots(7,1,0),(0,2,6),(0,6,2),\dots $$ Therefore, the probability is $\frac{30}{504}=\frac{5}{84}$.

Mike Earnest
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