Question 1104726
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In all, there are 8! permutations of 8 cards.


Of them, the "fortunate" permutations are those that have first three cards as #1, #2 and #3; the rest 5 of the cards may go in an arbitrary order.


Therefore, the probability under the question is {{{5!/8!}}} = {{{1/(8*7*6)}}} = 0.002976 = 0.2976%.
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