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<blockquote data-quote="clearstream" data-source="post: 8419172" data-attributes="member: 71699">I mean more this situation. Picture a simplified deck of six cards - three 3s and three 4s - from which we'll draw and sum two cards for each score. (I referred to those two summed cards as a 'pair'.)Say the deck look like this after shuffling - 3, 3, 4, 3, 4, 4 - so that our scores would be 6, 7, 8. For our purposes, that is identical to a deck that looks like this after shuffling - 3, 3, 3, 4, 4, 4 - because that will also make our scores 6, 7, 8. It doesn't matter what order the middle pair are drawn in. It also doesn't matter which 3 and which 4 because for our purposes 3s are all the same, and 4s are all the same.So long as we're agnostic as to where each score is allocated - so 8, 6, 7 is as good as 6, 7, 8 - say because we are allocating-as-desired (or are happy with anything we get), then - 4, 3, 4, 4, 3, 3 - is also identical to the above shuffled decks (for our purposes).If precise card order mattered, we'd have to worry about 720 permutations. Given that the ordering within pairs doesn't matter, we can worry about just 30 permutations. And given that the ordering of pairs doesn't matter, we can worry about just 15 combinations. At least, that is my understanding.Does that sound right? Or have I gone wrong somewhere?</blockquote>

[QUOTE="clearstream, post: 8419172, member: 71699"] I mean more this situation. Picture a simplified deck of six cards - three 3s and three 4s - from which we'll draw and sum two cards for each score. (I referred to those two summed cards as a 'pair'.) Say the deck look like this after shuffling - 3, 3, 4, 3, 4, 4 - so that our scores would be 6, 7, 8. For our purposes, that is identical to a deck that looks like this after shuffling - 3, 3, 3, 4, 4, 4 - because that will also make our scores 6, 7, 8. It doesn't matter what order the middle pair are drawn in. It also doesn't matter which 3 and which 4 because for our purposes 3s are all the same, and 4s are all the same. So long as we're agnostic as to where each score is allocated - so 8, 6, 7 is as good as 6, 7, 8 - say because we are allocating-as-desired (or are happy with anything we get), then - 4, 3, 4, 4, 3, 3 - is also identical to the above shuffled decks (for our purposes). If precise card order mattered, we'd have to worry about 720 permutations. Given that the ordering within pairs doesn't matter, we can worry about just 30 permutations. And given that the ordering of pairs doesn't matter, we can worry about just 15 combinations. At least, that is my understanding. Does that sound right? Or have I gone wrong somewhere? [/QUOTE]

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