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Consider rolling two fair six-sided dice. Let W be the product of the number showing. What is the probability function of W?

I have a hard time figuring out how to solve this question.

Any help will be appreciated!

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  • $\begingroup$ Use the self study tag. $\endgroup$ – Michael R. Chernick May 28 '18 at 3:34
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    $\begingroup$ You can just enumerate the outcomes. $\endgroup$ – Glen_b May 28 '18 at 7:46
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Each distinct pair has probability 1/36 by fairness and independence of trials. For each pair compute the product and sum together all pairs that give the same product. For example the product 1 only occurs for the pair (1,1). So its probability is 1/36. For 36 only the pair (6,6) yields 36. So the probability of 36 is also 1/36. On the other hand the pairs (1,2) and (2,1) are the only pairs that yield the product 2. Now the product 4 occurs for (1,4), (4,1) and (2,2). So the product 4 has probability 3/36 or 1/12. Proceed in the same way to get the probabilities for the products 3, 5, 6, 8, 9, 10, 12, 15. 16, 18, 20, 24, 25, and 30 to get the full distribution. Note that (1,3) and (3,1) yields 3, (1,5) and (5,1) yields 5, (3,2) and (2,3) and (6,1) and(1,6) yields 6, (4,2) and (2,4) yields 8, (3,3) yields 9, (5,2) and (2,5) yields 10, (4,3) and (3,4) and (6,2), (2,6) yields 12, (3,5) and (5,3) yields 15, (4,4) yields 16 (3,6) and (6,3) yields 18, (5,4) and (4,5) yields 20, (6,4) and (4,6) yields 24, (5,5) yields 25 and (5,6) and (6,5) yields 30.

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    $\begingroup$ Not only is this a difficult and inefficient algorithm, it easily leads to incorrect results. In particular, although $5,$ $8,$ $10,$ $15,$ $18,$ and $24$ are possible values of the product, you seem to have overlooked them, nicely demonstrating my point. $\endgroup$ – whuber May 28 '18 at 13:08
  • $\begingroup$ You are right. The fact that such product would be missing would become evident when the count does not sum up to 36. The pairs (5,1) and (1,5) for 5, (4,2), (2, 4) for 8, (5,3) and (3,5) for 15, (6,3) and (3,6) for 18 and (6,4) and (4, 6) for 24. The approach I am suggesting is brute force but what would be a more efficient way to do it. I am editing my answer to fix it. $\endgroup$ – Michael R. Chernick May 30 '18 at 4:28

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